Thực hiện phép tính:a) \(\dfrac{{5{\rm{x}} - 4}}{9} \dfrac{{4{\rm{x}} 4}}{9}\) b) \(\dfrac{{{x^2}y - 6}}{{2{{\rm{x}}^2}y}} \dfrac{{6 - x{y^2}}}{{2{{\rm{x}}^2}y}}\)c) \(\dfrac{{x 1}}{{...

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19 tháng 7 2023

Thực hiện phép tính:a) \(\dfrac{{5{\rm{x}} - 4}}{9} + \dfrac{{4{\rm{x}} + 4}}{9}\) b) \(\dfrac{{{x^2}y - 6}}{{2{{\rm{x}}^2}y}} + \dfrac{{6 - x{y^2}}}{{2{{\rm{x}}^2}y}}\)c) \(\dfrac{{x + 1}}{{{x^2} - 5{\rm{x}}}} + \dfrac{{x - 18}}{{{x^2} - 5{\rm{x}}}} + \dfrac{{x + 2}}{{{x^2} - 5{\rm{x}}}}\) d) \(\dfrac{{7y}}{3} - \dfrac{{7y - 5}}{3}\)e) \(\dfrac{{4{\rm{x}} - 1}}{{3{\rm{x}}{y^2}}} - \dfrac{{7{\rm{x}} -...

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#Toán lớp 8

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10 tháng 1

a)

\(\dfrac{{5{\rm{x}} - 4}}{9} + \dfrac{{4{\rm{x}} + 4}}{9} \\= \dfrac{{5{\rm{x}} - 4 + 4{\rm{x}} + 4}}{9} \\= \dfrac{{9{\rm{x}}}}{9} \\= x\)

b)

\(\dfrac{{{x^2}y - 6}}{{2{{\rm{x}}^2}y}} + \dfrac{{6 - x{y^2}}}{{2{{\rm{x}}^2}y}} \\= \dfrac{{{x^2}y - 6 + 6 - x{y^2}}}{{2{{\rm{x}}^2}y}} \\= \dfrac{{{x^2}y - x{y^2}}}{{2{{\rm{x}}^2}y}} \\= \dfrac{{xy\left( {x - y} \right)}}{{2{{\rm{x}}^2}y}} \\= \dfrac{{x - y}}{{2{\rm{x}}}}\)

c)

\(\dfrac{{x + 1}}{{{x^2} - 5{\rm{x}}}} + \dfrac{{x - 18}}{{{x^2} - 5{\rm{x}}}} + \dfrac{{x + 2}}{{{x^2} - 5{\rm{x}}}} \\= \dfrac{{x + 1 + x - 18 + x + 2}}{{{x^2} - 5{\rm{x}}}} \\= \dfrac{{3{\rm{x}} - 15}}{{x\left( {x - 5} \right)}} \\= \dfrac{{3\left( {x - 5} \right)}}{{x\left( {x - 5} \right)}} \\= \dfrac{3}{x}\)

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d)

\(\dfrac{{7y}}{3} - \dfrac{{7y - 5}}{3} \\= \dfrac{{7y - 7y + 5}}{3} \\= \dfrac{5}{3}\)

e)

\(\dfrac{{4{\rm{x}} - 1}}{{3{\rm{x}}{y^2}}} - \dfrac{{7{\rm{x}} - 1}}{{3{\rm{x}}{y^2}}} \\= \dfrac{{4{\rm{x}} - 1 - 7{\rm{x}} + 1}}{{3{\rm{x}}{y^2}}} \\= \dfrac{{-3{\rm{x}}}}{{3{\rm{x}}{y^2}}} \\= \dfrac{-1}{{{y^2}}}\)

g)

\(\dfrac{{3y - 2{\rm{x}}}}{{x - 2y}} - \dfrac{{x - y}}{{2y - x}} \\= \dfrac{{3y - 2{\rm{x}}}}{{x - 2y}} + \left( { - \dfrac{{x - y}}{{2y - x}}} \right) \\= \dfrac{{3y - 2{\rm{x}}}}{{x - 2y}} + \dfrac{{x - y}}{{x - 2y}} \\= \dfrac{{3y - 2{\rm{x}} + x - y}}{{x - 2y}} \\= \dfrac{{2y - x}}{{ - \left( {2y - x} \right)}} \\= - 1\)

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Thực hiện phép tính:

\(a)\dfrac{{3{\rm{x}} + 6}}{{4{\rm{x}} - 8}}.\dfrac{{2{\rm{x}} - 4}}{{x + 2}}\)

\(b)\dfrac{{{x^2} - 36}}{{2{\rm{x}} + 10}}.\dfrac{{x + 5}}{{6 - x}}\)

\(c)\dfrac{{1 - {y^3}}}{{y + 1}}.\dfrac{{5y + 5}}{{{y^2} + y + 1}}\)

\(d)\dfrac{{x + 2y}}{{4{{\rm{x}}^2} - 4{\rm{x}}y + {y^2}}}.\left( {2{\rm{x}} - y} \right)\)

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10 tháng 1

\(a)\dfrac{{3{\rm{x}} + 6}}{{4{\rm{x}} - 8}}.\dfrac{{2{\rm{x}} - 4}}{{x + 2}} = \dfrac{{3\left( {x + 2} \right).2\left( {x - 2} \right)}}{{4.\left( {x - 2} \right).\left( {x + 2} \right)}} = \dfrac{3}{2}\)

\(b)\dfrac{{{x^2} - 36}}{{2{\rm{x}} + 10}}.\dfrac{{x + 5}}{{6 - x}} = \dfrac{{\left( {x - 6} \right)\left( {x + 6} \right)\left( {x + 5} \right)}}{{2\left( {x + 5} \right).\left( { - 1} \right)\left( {x - 6} \right)}} = \dfrac{{x + 6}}{{ - 2}} = \dfrac{{-x- 6}}{{ 2}}\)

\(c)\dfrac{{1 - {y^3}}}{{y + 1}}.\dfrac{{5y + 5}}{{{y^2} + y + 1}} = \dfrac{{\left( {1 - y} \right)\left( {1 + y + {y^2}} \right).5\left( {y + 1} \right)}}{{\left( {y + 1} \right).\left( {{y^2} + y + 1} \right)}} = 5\left( {1 - y} \right)\)

\(d)\dfrac{{x + 2y}}{{4{{\rm{x}}^2} - 4{\rm{x}}y + {y^2}}}.\left( {2{\rm{x}} - y} \right) = \dfrac{{\left( {x + 2y} \right).\left( {2{\rm{x}} - y} \right)}}{{{{\left( {2{\rm{x}} - y} \right)}^2}}} = \dfrac{{x + 2y}}{{2{\rm{x}} - y}}\)

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19 tháng 7 2023

Thực hiện phép tính:\(a)\dfrac{{4{\rm{x}} + 2}}{{4{\rm{x - 4}}}} + \dfrac{{3 - 6{\rm{x}}}}{{6{\rm{x}} - 6}}\) \(b)\dfrac{y}{{2{{\rm{x}}^2} - xy}} + \dfrac{{4{\rm{x}}}}{{{y^2} - 2{\rm{x}}y}}\)\(c)\dfrac{x}{{x - y}} + \dfrac{y}{{x + y}} + \dfrac{{2{y^2}}}{{{x^2} - {y^2}}}\) \(d)\dfrac{{{x^2} + 2}}{{{x^3} - 1}} + \dfrac{x}{{{x^2} + x + 1}} + \dfrac{1}{{1 -...

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#Toán lớp 8

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10 tháng 1

\(\begin{array}{l}a)\dfrac{{4{\rm{x}} + 2}}{{4{\rm{x - 4}}}} + \dfrac{{3 - 6{\rm{x}}}}{{6{\rm{x}} - 6}} = \dfrac{{2\left( {2x + 1} \right)}}{{4\left( {x - 1} \right)}} + \dfrac{{3\left( {1 - 2x} \right)}}{{6\left( {x - 1} \right)}}\\ = \dfrac{{2x + 1}}{{2\left( {x - 1} \right)}} + \dfrac{{1 - 2x}}{{2\left( {x - 1} \right)}} = \dfrac{{2x + 1 + 1 - 2x}}{{2\left( {x - 1} \right)}} = \dfrac{2}{{2\left( {x - 1} \right)}} = \dfrac{1}{{x - 1}}\end{array}\)

\(\begin{array}{l}b)\dfrac{y}{{2{{\rm{x}}^2} - xy}} + \dfrac{{4{\rm{x}}}}{{{y^2} - 2{\rm{x}}y}} = \dfrac{y}{{x\left( {2{\rm{x}} - y} \right)}} + \dfrac{{4{\rm{x}}}}{{y\left( {y - 2{\rm{x}}} \right)}}\\ = \dfrac{y}{{x\left( {2{\rm{x}} - y} \right)}} - \dfrac{{4{\rm{x}}}}{{y\left( {2{\rm{x}} - y} \right)}} = \dfrac{{{y^2}}}{{xy\left( {2{\rm{x}} - y} \right)}} - \dfrac{{4{{\rm{x}}^2}}}{{xy\left( {2{\rm{x}} - y} \right)}}\\ = \dfrac{{{y^2} - 4{{\rm{x}}^2}}}{{xy\left( {2{\rm{x}} - y} \right)}} = \dfrac{{\left( {y - 2{\rm{x}}} \right)\left( {y + 2{\rm{x}}} \right)}}{{ - xy\left( {y - 2{\rm{x}}} \right)}} = \dfrac{{ - \left( {y + 2{\rm{x}}} \right)}}{{xy}}\end{array}\)

\(\begin{array}{l}c)\dfrac{x}{{x - y}} + \dfrac{y}{{x + y}} + \dfrac{{2{y^2}}}{{{x^2} - {y^2}}}\\ = \dfrac{x}{{x - y}} + \dfrac{y}{{x + y}} + \dfrac{{2{y^2}}}{{\left( {x - y} \right)\left( {x + y} \right)}}\\ = \dfrac{{x\left( {x + y} \right)}}{{\left( {x - y} \right)\left( {x + y} \right)}} + \dfrac{{y\left( {x - y} \right)}}{{\left( {x - y} \right)\left( {x + y} \right)}} + \dfrac{{2{y^2}}}{{\left( {x - y} \right)\left( {x + y} \right)}}\\ = \dfrac{{{x^2} + xy + {\rm{yx}} - {y^2} + 2{y^2}}}{{\left( {x - y} \right)\left( {x + y} \right)}} = \dfrac{{{x^2} + 2{\rm{x}}y + {y^2}}}{{\left( {x - y} \right)\left( {x + y} \right)}} = \dfrac{{{{\left( {x + y} \right)}^2}}}{{\left( {x - y} \right)\left( {x + y} \right)}} = \dfrac{{x + y}}{{x - y}}\end{array}\)

\(\begin{array}{l}d)\dfrac{{x{}^2 + 2}}{{{x^3} - 1}} + \dfrac{x}{{{x^2} + x + 1}} + \dfrac{1}{{1 - x}}\\ = \dfrac{{x{}^2 + 2}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}} + \dfrac{x}{{{x^2} + x + 1}} - \dfrac{1}{{x - 1}}\\ = \dfrac{{x{}^2 + 2}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}} + \dfrac{{x\left( {x - 1} \right)}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}} - \dfrac{{{x^2} + x + 1}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}}\\ = \dfrac{{{x^2} + 2 + {x^2} - x - {x^2} - x - 1}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}} = \dfrac{{{x^2} - 2{\rm{x}} + 1}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}} = \dfrac{{{{\left( {x - 1} \right)}^2}}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}} = \dfrac{{x - 1}}{{{x^2} + x + 1}}\end{array}\)

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10 tháng 1

Thực hiện phép tính:

\(a)\dfrac{{20{\rm{x}}}}{{3{y^2}}}:\left( { - \dfrac{{15{{\rm{x}}^2}}}{{6y}}} \right)\)

\(b)\dfrac{{9{{\rm{x}}^2} - {y^2}}}{{x + y}}:\dfrac{{3{\rm{x}} + y}}{{2{\rm{x}} + 2y}}\)

\(c)\dfrac{{{x^3} + {y^3}}}{{y - x}}:\dfrac{{{x^2} - xy + {y^2}}}{{{x^2} - 2{\rm{x}}y + {y^2}}}\)

\(d)\dfrac{{9 - {x^2}}}{x}:\left( {x - 3} \right)\)

#Toán lớp 8

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\(a)\dfrac{{20{\rm{x}}}}{{3{y^2}}}:\left( { - \dfrac{{15{{\rm{x}}^2}}}{{6y}}} \right) = \dfrac{{20{\rm{x}}}}{{3{y^2}}}.\left( { - \dfrac{{6y}}{{15{{\rm{x}}^2}}}} \right) = \dfrac{{20{\rm{x}}.\left( { - 6y} \right)}}{{3{y^2}.15{{\rm{x}}^2}}} = \dfrac{{ - 8}}{{3{\rm{x}}y}}\)

\(b)\dfrac{{9{{\rm{x}}^2} - {y^2}}}{{x + y}}:\dfrac{{3{\rm{x}} + y}}{{2{\rm{x}} + 2y}} = \dfrac{{\left( {3{\rm{x}} - y} \right)\left( {3{\rm{x}} + y} \right)}}{{x + y}}.\dfrac{{2{\rm{x}} + 2y}}{{3{\rm{x}} + y}} = \dfrac{{\left( {3{\rm{x}} - y} \right)\left( {3{\rm{x}} + y} \right).2.\left( {x + y} \right)}}{{(x + y).\left( {3{\rm{x}} + y} \right)}} = 2\left( {3{\rm{x}} - y} \right)\)

\(\begin{array}{l}c)\dfrac{{{x^3} + {y^3}}}{{y - x}}:\dfrac{{{x^2} - xy + {y^2}}}{{{x^2} - 2{\rm{x}}y + {y^2}}} = \dfrac{{\left( {x + y} \right)\left( {{x^2} - xy + {y^2}} \right)}}{{y - x}}.\dfrac{{{x^2} - 2{\rm{x}}y + {y^2}}}{{{x^2} - xy + {y^2}}}\\ = \dfrac{{\left( {x + y} \right)\left( {{x^2} - xy + {y^2}} \right).{{\left( {x - y} \right)}^2}}}{{ - (x - y)\left( {{x^2} - xy + {y^2}} \right)}} = \left( {x + y} \right)\left( {y - x} \right) = {{y^2} - {x^2}} \end{array}\)

\(d)\dfrac{{9 - {x^2}}}{x}:\left( {x - 3} \right) = \dfrac{{\left( {3 - x} \right)\left( {3 + x} \right)}}{x}.\dfrac{1}{{x - 3}} = \dfrac{{ - \left( {x - 3} \right)\left( {3 + x} \right)}}{{x.\left( {x - 3} \right)}} = \dfrac{{ - \left( {3 + x} \right)}}{x}.\)

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19 tháng 7 2023

Tính một cách hợp lí:

\(a)\dfrac{{y + 6}}{{{x^2} - 4{\rm{x}} + 4}}.\dfrac{{{x^2} - 4}}{{x + 1}}.\dfrac{{x - 2}}{{y + 6}}\)

\(b)\left( {\dfrac{1}{{x - 4}} + \dfrac{{2{\rm{x}} + 1}}{{{x^2} - 8{\rm{x}} + 16}}} \right).\dfrac{{x - 4}}{{2{\rm{x}} + 1}}\)

#Toán lớp 8

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10 tháng 1

\(\begin{array}{l}a)\dfrac{{y + 6}}{{{x^2} - 4{\rm{x}} + 4}}.\dfrac{{{x^2} - 4}}{{x + 1}}.\dfrac{{x - 2}}{{y + 6}}\\ = \dfrac{{y + 6}}{{{x^2} - 4{\rm{x}} + 4}}.\dfrac{{x - 2}}{{y + 6}}.\dfrac{{{x^2} - 4}}{{x + 1}}\\ = \dfrac{{\left( {y + 6} \right).\left( {x - 2} \right).\left( {{x^2} - 4} \right)}}{{\left( {{x^2} - 4{\rm{x}} + 4} \right).\left( {y + 6} \right).\left( {x + 1} \right)}}\\ = \dfrac{{\left( {y + 6} \right).\left( {x - 2} \right).\left( {x - 2} \right)\left( {x + 2} \right)}}{{{{\left( {x - 2} \right)}^2}.\left( {y + 6} \right).\left( {x + 1} \right)}} = \dfrac{{x + 2}}{{x + 1}}\end{array}\)

\(\begin{array}{l}b)\left(\frac{2x+1}{{x - 3}} + \frac{2x+1}{x+3}\right ) .\dfrac{{x^2 - 9}}{{2{\rm{x}} + 1}} \\ = (2x+1) \left ( \frac {1}{x-3} + \frac {1}{x+3} \right ) . \frac {(x-3)(x+3)}{2x + 1} \\ = (2x+1) \frac {x+3 + x - 3}{(x-3)(x+3)} . \frac {(x-3)(x+3)}{2x + 1} \\ = \frac {2x(2x+1)}{(x-3)(x+3)} . \frac {(x-3)(x+3)}{2x +1} \\= 2x \end{array}\)

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19 tháng 7 2023

Thực hiện phép tính:

\(a)\dfrac{{4{\rm{x}} + 3y}}{{{x^2} - {y^2}}} - \dfrac{{3{\rm{x}} + 4y}}{{{x^2} - {y^2}}}\)

\(b)\dfrac{{2{\rm{x}}y - 3{y^2}}}{{{x^2} - 3{\rm{x}}y}} - \dfrac{x}{{3{\rm{x}} - 9y}}\)

#Toán lớp 8

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10 tháng 1

\(a)\dfrac{{4{\rm{x}} + 3y}}{{{x^2} - {y^2}}} - \dfrac{{3{\rm{x}} + 4y}}{{{x^2} - {y^2}}} = \dfrac{{\left( {{\rm{4x}} + 3y} \right) - \left( {3{\rm{x}} + 4y} \right)}}{{{x^2} - {y^2}}} = \dfrac{{4{\rm{x}} + 3y - 3{\rm{x}} - 4y}}{{{x^2} - {y^2}}} = \dfrac{{x - y}}{{{x^2} - {y^2}}} = \dfrac{{x - y}}{{\left( {x - y} \right)\left( {x + y} \right)}} = \dfrac{1}{{x + y}}\)

\(\begin{array}{l}b)\dfrac{{2{\rm{x}}y - 3{y^2}}}{{{x^2} - 3{\rm{x}}y}} - \dfrac{x}{{3{\rm{x}} - 9y}}\\ = \dfrac{{2{\rm{x}}y - 3{y^2}}}{{x\left( {x - 3y} \right)}} - \dfrac{{{x^2}}}{{3\left( {x - 3y} \right)}}\\ = \dfrac{{3\left( {2{\rm{x}}y - 3{y^2}} \right)}}{{3{\rm{x}}\left( {x - 3y} \right)}} - \dfrac{{{x^2}}}{{3{\rm{x}}\left( {x - 3y} \right)}}\\ = \dfrac{{6{\rm{x}}y - 9{y^2} - {x^2}}}{{3{\rm{x}}\left( {x - 3y} \right)}} = \dfrac{{ - \left( {{x^2} - 6{\rm{x}}y + 9{y^2}} \right)}}{{3{\rm{x}}\left( {x - 3y} \right)}} = \dfrac{{ - {{\left( {x - 3y} \right)}^2}}}{{3{\rm{x}}\left( {x - 3y} \right)}} = \dfrac{{ - \left( {x - 3y} \right)}}{{3{\rm{x}}}}\end{array}\)

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Buddy

19 tháng 7 2023

Tính một cách hợp lí:

\(\dfrac{{x - 5y}}{{2{\rm{x}} - 3y}} - \dfrac{{24{\rm{x}}y}}{{4{{\rm{x}}^2} - 9{y^2}}} - \dfrac{{x + 8y}}{{3y - 2{\rm{x}}}}\)

#Toán lớp 8

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10 tháng 1

Ta có:

\(\begin{array}{l}\dfrac{{x - 5y}}{{2{\rm{x}} - 3y}} - \dfrac{{24{\rm{x}}y}}{{4{{\rm{x}}^2} - 9{y^2}}} - \dfrac{{x + 8y}}{{3y - 2{\rm{x}}}}\\ = \dfrac{{x - 5y}}{{2{\rm{x}} - 3y}} - \dfrac{{24{\rm{x}}y}}{{{{\left( {2{\rm{x}}} \right)}^2} - {{\left( {3y} \right)}^2}}} + \left( { - \dfrac{{x + 8y}}{{3y - 2{\rm{x}}}}} \right)\\ = \dfrac{{x - 5y}}{{2{\rm{x}} - 3y}} - \dfrac{{24{\rm{x}}y}}{{\left( {2{\rm{x}} - 3y} \right)\left( {2{\rm{x}} + 3y} \right)}} + \dfrac{{x + 8y}}{{2{\rm{x}} - 3y}}\\ = \dfrac{{x - 5y}}{{2{\rm{x}} - 3y}} + \dfrac{{x + 8y}}{{2{\rm{x}} - 3y}} - \dfrac{{24{\rm{x}}y}}{{\left( {2{\rm{x}} - 3y} \right)\left( {2{\rm{x}} + 3y} \right)}}\\ = \dfrac{{2{\rm{x}} + 3y}}{{2{\rm{x}} - 3y}} - \dfrac{{24{\rm{x}}y}}{{\left( {2{\rm{x}} - 3y} \right)\left( {2{\rm{x}} + 3y} \right)}}\\ = \dfrac{{{{\left( {2{\rm{x}} + 3y} \right)}^2}}}{{\left( {2{\rm{x}} - 3y} \right)\left( {2{\rm{x}} + 3y} \right)}} - \dfrac{{24{\rm{x}}y}}{{\left( {2{\rm{x}} - 3y} \right)\left( {2{\rm{x}} + 3y} \right)}}\\ = \dfrac{{4{{\rm{x}}^2} + 12{\rm{x}}y + 9{y^2} - 24{\rm{x}}y}}{{\left( {2{\rm{x}} - 3y} \right)\left( {2{\rm{x}} + 3y} \right)}}\\ = \dfrac{{4{{\rm{x}}^2} - 12{\rm{x}}y + 9{y^2}}}{{\left( {2{\rm{x}} - 3y} \right)\left( {2{\rm{x}} + 3y} \right)}} = \dfrac{{{{\left( {2{\rm{x}} - 3y} \right)}^2}}}{{\left( {2{\rm{x}} - 3y} \right)\left( {2{\rm{x}} + 3y} \right)}} = \dfrac{{2{\rm{x}} - 3y}}{{2{\rm{x}} + 3y}}\end{array}\)

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19 tháng 7 2023

Thực hiện phép tính:\(a)\dfrac{x}{{xy + {y^2}}} - \dfrac{y}{{{x^2} + xy}}\) \(b)\dfrac{{{x^2} + 4}}{{{x^2} - 4}} - \dfrac{x}{{x + 2}} - \dfrac{x}{{2 - x}}\)\(c)\dfrac{{{a^2} + ab}}{{b - a}}:\dfrac{{a + b}}{{2{{\rm{a}}^2} - 2{b^2}}}\) \(d)\left( {\dfrac{{2{\rm{x}} + 1}}{{2{\rm{x}} - 1}} - \dfrac{{2{\rm{x}} - 1}}{{2{\rm{x}} + 1}}} \right):\dfrac{{4{\rm{x}}}}{{10{\rm{x}} -...

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#Toán lớp 8

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10 tháng 1

\(\begin{array}{l}a)\dfrac{x}{{xy + {y^2}}} - \dfrac{y}{{{x^2} + xy}}\\ = \dfrac{x}{{y\left( {x + y} \right)}} - \dfrac{y}{{x\left( {x + y} \right)}}\\ = \dfrac{{{x^2} - {y^2}}}{{xy\left( {x + y} \right)}} = \dfrac{{\left( {x - y} \right)\left( {x + y} \right)}}{{xy\left( {x + y} \right)}} = \dfrac{{x - y}}{{xy}}\end{array}\)

\(\begin{array}{l}b)\dfrac{{{x^2} + 4}}{{{x^2} - 4}} - \dfrac{x}{{x + 2}} - \dfrac{x}{{2 - x}}\\ = \dfrac{{{x^2} + 4}}{{\left( {x - 2} \right)\left( {x + 2} \right)}} - \dfrac{x}{{x + 2}} + \dfrac{x}{{x - 2}}\\ = \dfrac{{{x^2} + 4 - x\left( {x - 2} \right) + x\left( {x + 2} \right)}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}\\ = \dfrac{{{x^2} + 4 - {x^2} + 2{\rm{x}} + {x^2} + 2{\rm{x}}}}{{\left( {x - 2} \right)\left( {x + 2} \right)}} = \dfrac{{{x^2} + 4{\rm{x}} + 4}}{{\left( {x - 2} \right)\left( {x + 2} \right)}} = \dfrac{{{{\left( {x + 2} \right)}^2}}}{{\left( {x - 2} \right)\left( {x + 2} \right)}} = \dfrac{{x + 2}}{{x - 2}}\end{array}\)

\(\begin{array}{l}c)\dfrac{{{a^2} + ab}}{{b - a}}:\dfrac{{a + b}}{{2{{\rm{a}}^2} - 2{b^2}}}\\ = \dfrac{{a\left( {a + b} \right)}}{{b - a}}.\dfrac{{2{{\rm{a}}^2} - 2{b^2}}}{{a + b}}\\ = \dfrac{{a\left( {a + b} \right).2.\left( {{a^2} - {b^2}} \right)}}{{ - \left( {a - b} \right).\left( {a + b} \right)}}\\ = \dfrac{{a\left( {a + b} \right).2.\left( {a - b} \right).\left( {a + b} \right)}}{{ - \left( {a - b} \right)\left( {a + b} \right)}} = - 2{\rm{a}}\left( {a + b} \right)\end{array}\)

\(\begin{array}{l}d)\left( {\dfrac{{2{\rm{x}} + 1}}{{2{\rm{x}} - 1}} - \dfrac{{2{\rm{x}} - 1}}{{2{\rm{x}} + 1}}} \right):\dfrac{{4{\rm{x}}}}{{10{\rm{x}} - 5}}\\ = \dfrac{{{{\left( {2{\rm{x}} + 1} \right)}^2} - {{\left( {2{\rm{x}} - 1} \right)}^2}}}{{\left( {2{\rm{x}} + 1} \right)\left( {2{\rm{x}} - 1} \right)}}.\dfrac{{10x - 5}}{{4{\rm{x}}}}\\ = \dfrac{{\left( {2{\rm{x}} + 1 + 2{\rm{x}} - 1} \right)\left( {2{\rm{x}} + 1 - 2{\rm{x}} + 1} \right)}}{{\left( {2{\rm{x}} + 1} \right)\left( {2{\rm{x}} - 1} \right)}}.\dfrac{{5.\left( {2{\rm{x}} - 1} \right)}}{{4{\rm{x}}}}\\ = \dfrac{{4{\rm{x}}.2.5\left( {2{\rm{x}} - 1} \right)}}{{\left( {2{\rm{x}} + 1} \right)\left( {2{\rm{x}} - 1} \right).4{\rm{x}}}} = \dfrac{{10}}{{2{\rm{x}} + 1}}\end{array}\)

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Buddy

19 tháng 7 2023

a) Trong các biểu thức sau, biểu thức nào là đơn thức:

\(\dfrac{1}{5}x{y^2}{z^3};3 - 2{{\rm{x}}^3}{y^2}z; - \dfrac{3}{2}{x^4}{\rm{yx}}{{\rm{z}}^2};\dfrac{1}{2}{x^2}\left( {{y^3} - {z^3}} \right)\)

b) Trong những biểu thức sau, biểu thức nào là đa thức:

\(2 - x + y; - 5{{\rm{x}}^2}y{z^3} + \dfrac{1}{3}x{y^2}z + x + 1;\dfrac{{x - y}}{{x{y^2}}};\dfrac{1}{x} + 2y - 3{\rm{z}}\)

#Toán lớp 8

1

HQ

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Giáo viên

10 tháng 1

a) Các biểu thức: \(\dfrac{1}{5}x{y^2}{z^3}; - \dfrac{3}{2}{x^4}{\rm{yx}}{{\rm{z}}^2}\) là đơn thức

b) Các biểu thức: \(2 - x + y; - 5{{\rm{x}}^2}y{z^3} + \dfrac{1}{3}x{y^2}z + x + 1\) là đa thức

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B

Buddy

19 tháng 7 2023

Thực hiện phép tính:

\(a)\dfrac{{{x^3} + 1}}{{{x^2} - 2{\rm{x}} + 1}}.\dfrac{{x - 1}}{{{x^2} - x + 1}}\)

\(b)\left( {{x^2} - 4{\rm{x}} + 4} \right).\dfrac{2}{{3{{\rm{x}}^2} - 6{\rm{x}}}}\)

#Toán lớp 8

1

HQ

Hà Quang Minh

Giáo viên

10 tháng 1

\(\begin{array}{l}a)\dfrac{{{x^3} + 1}}{{{x^2} - 2{\rm{x}} + 1}}.\dfrac{{x - 1}}{{{x^2} - x + 1}} = \\ = \dfrac{{\left( {{x^3} + 1} \right)\left( {x - 1} \right)}}{{\left( {{x^2} - 2{\rm{x}} + 1} \right).\left( {{x^2} - x + 1} \right)}}\\ = \dfrac{{\left( {x + 1} \right)\left( {{x^2} - x + 1} \right)\left( {x - 1} \right)}}{{{{\left( {x - 1} \right)}^2}.\left( {{x^2} - x + 1} \right)}} = \dfrac{{x + 1}}{{x - 1}}\end{array}\)

\(\begin{array}{l}b)\left( {{x^2} - 4{\rm{x}} + 4} \right).\dfrac{2}{{3{{\rm{x}}^2} - 6{\rm{x}}}}\\ = \dfrac{{\left( {{x^2} - 4{\rm{x}} + 4} \right).2}}{{3{{\rm{x}}^2} - 6{\rm{x}}}} = \dfrac{{{{\left( {x - 2} \right)}^2}.2}}{{3{\rm{x}}\left( {x - 2} \right)}} = \dfrac{{2\left( {x - 2} \right)}}{{3{\rm{x}}}}\end{array}\)

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