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\(A=\dfrac{x^2-y^2+2y^2}{y\left(x-y\right)}\cdot\dfrac{-\left(x-y\right)}{x^2+y^2}+\dfrac{2x^2+2-2x^2+x}{2\left(2x-1\right)}\cdot\dfrac{-\left(2x-1\right)}{x+2}\)
\(=\dfrac{-1}{y}+\dfrac{-1}{2}=\dfrac{-2-y}{2y}\)
a: ĐKXĐ: \(x\notin\left\{0;1\right\}\)
b: ĐKXĐ: \(x\notin\left\{2;-2\right\}\)
\(a,ĐK:x\ne0;x-1\ne0\Leftrightarrow x\ne0;x\ne1\\ b,ĐK:x^2-4=\left(x-2\right)\left(x+2\right)\ne0\Leftrightarrow x\ne2;x\ne-2\)
a) Ta có:
\(f\left( {\dfrac{1}{5}} \right) = \dfrac{5}{{4.\dfrac{1}{5}}} = \dfrac{5}{{\dfrac{4}{5}}} = 5:\dfrac{4}{5} = 5.\dfrac{5}{4} = \dfrac{{25}}{4};\)
\(f\left( { - 5} \right) = \dfrac{5}{{4.\left( { - 5} \right)}} = \dfrac{5}{{ - 20}} = \dfrac{{ - 1}}{4};\)
\(f\left( {\dfrac{4}{5}} \right) = \dfrac{5}{{4.\dfrac{4}{5}}} = \dfrac{5}{{\dfrac{{16}}{5}}} = 5:\dfrac{{16}}{5} = 5.\dfrac{5}{{16}} = \dfrac{{25}}{{16}}\)
b) Ta có:
\(f\left( { - 3} \right) = \dfrac{5}{{4.\left( { - 3} \right)}} = \dfrac{5}{{ - 12}} = \dfrac{{ - 5}}{{12}};\)
\(f\left( { - 2} \right) = \dfrac{5}{{4.\left( { - 2} \right)}} = \dfrac{5}{{ - 8}} = \dfrac{{ - 5}}{8};\)
\(f\left( { - 1} \right) = \dfrac{5}{{4.\left( { - 1} \right)}} = \dfrac{5}{{ - 4}} = \dfrac{{ - 5}}{4};\)
\(f\left( { - \dfrac{1}{2}} \right) = \dfrac{5}{{4.\left( { - \dfrac{1}{2}} \right)}} = \dfrac{5}{{\dfrac{{ - 4}}{2}}} = \dfrac{5}{{ - 2}} = \dfrac{{ - 5}}{2}\);
\(f\left( {\dfrac{1}{4}} \right) = \dfrac{5}{{4.\dfrac{1}{4}}} = \dfrac{5}{{\dfrac{4}{4}}} = \dfrac{5}{1} = 5\);
\(f\left( 1 \right) = \dfrac{5}{{4.1}} = \dfrac{5}{4}\);
\(f\left( 2 \right) = \dfrac{5}{{4.2}} = \dfrac{5}{8}\)
Ta có bảng sau:
\(x\) | –3 | –2 | –1 | \( - \dfrac{1}{2}\) | \(\dfrac{1}{4}\) | 1 | 2 |
\(y = f\left( x \right) = \dfrac{5}{{4x}}\) | \(\dfrac{{ - 5}}{{12}}\) | \(\dfrac{{ - 5}}{8}\) | \(\dfrac{{ - 5}}{4}\) | \(\dfrac{{ - 5}}{2}\) | 5 | \(\dfrac{5}{4}\) | \(\dfrac{5}{8}\) |
a) \(f\left( 1 \right) = 3.1 = 3;f\left( { - 2} \right) = 3.\left( { - 2} \right) = - 6;f\left( {\dfrac{1}{3}} \right) = 3.\dfrac{1}{3} = 1\).
b) Ta có: \(f\left( { - 3} \right) = 3.\left( { - 3} \right) = - 9;f\left( { - 1} \right) = 3.\left( { - 1} \right) = - 3\)
\(f\left( 0 \right) = 3.0 = 0;f\left( 2 \right) = 3.2 = 6;f\left( 3 \right) = 3.3 = 9\);
Ta lập được bảng sau
\(x\) | –3 | –2 | –1 | 0 | 1 | 2 | 3 |
\(y\) | –9 | -6 | –3 | 0 | 3 | 6 | 9 |
a)
\(\begin{array}{l}A = 0,2\left( {5{\rm{x}} - 1} \right) - \dfrac{1}{2}\left( {\dfrac{2}{3}x + 4} \right) + \dfrac{2}{3}\left( {3 - x} \right)\\A = x - 0,2 - \dfrac{1}{3}x - 2 + 2 - \dfrac{2}{3}x\\ = \left( {x - \dfrac{1}{3}x - \dfrac{2}{3}x} \right) + \left( {\dfrac{{ - 1}}{2} - 2 + 2} \right)\\ = - \dfrac{1}{2}\end{array}\)
Vậy \(A = - \dfrac{1}{2}\) không phụ thuộc vào biến x
b)
\(\begin{array}{l}B = \left( {x - 2y} \right)\left( {{x^2} + 2{\rm{x}}y + 4{y^2}} \right) - \left( {{x^3} - 8{y^3} + 10} \right)\\B = \left[ {x - {{\left( {2y} \right)}^3}} \right] - {x^3} + 8{y^3} - 10\\B = {x^3} - 8{y^3} - {x^3} + 8{y^3} - 10 = - 10\end{array}\)
Vậy B = -10 không phụ thuộc vào biến x, y.
c)
\(\begin{array}{l}C = 4{\left( {x + 1} \right)^2} + {\left( {2{\rm{x}} - 1} \right)^2} - 8\left( {x - 1} \right)\left( {x + 1} \right) - 4{\rm{x}}\\{\rm{C = 4}}\left( {{x^2} + 2{\rm{x}} + 1} \right) + \left( {4{{\rm{x}}^2} - 4{\rm{x}} + 1} \right) - 8\left( {{x^2} - 1} \right) - 4{\rm{x}}\\C = 4{{\rm{x}}^2} + 8{\rm{x}} + 4 + 4{{\rm{x}}^2} - 4{\rm{x}} + 1 - 8{{\rm{x}}^2} + 8 - 4{\rm{x}}\\C = \left( {4{{\rm{x}}^2} + 4{{\rm{x}}^2} - 8{{\rm{x}}^2}} \right) + \left( {8{\rm{x}} - 4{\rm{x}} - 4{\rm{x}}} \right) + \left( {4 + 1 + 8} \right)\\C = 13\end{array}\)
Vậy C = 13 không phụ thuộc vào biến x
Bài 1:
\(a,ĐK:x\ne\pm5\\ b,P=\dfrac{x-5+2x+10-2x-10}{\left(x-5\right)\left(x+5\right)}=\dfrac{x-5}{\left(x-5\right)\left(x+5\right)}=\dfrac{1}{x+5}\\ c,P=-3\Leftrightarrow x+5=-\dfrac{1}{3}\Leftrightarrow x=-\dfrac{16}{3}\\ d,P\in Z\Leftrightarrow x+5\inƯ\left(1\right)=\left\{-1;1\right\}\\ \Leftrightarrow x\in\left\{-6;-4\right\}\)
Bài 2:
\(a,\Leftrightarrow\dfrac{3\left(x^2+2x+4\right)}{\left(x-2\right)\left(x^2+2x+4\right)}=\dfrac{3}{x-2}=0\Leftrightarrow x\in\varnothing\\ b,\Leftrightarrow\dfrac{x\left(2-x\right)}{\left(x-2\right)\left(x+2\right)}=0\Leftrightarrow\dfrac{-x}{x+2}=0\Leftrightarrow x=0\)
A=(xy2+xy−x−yx2+xy)(xy2+xy−x−yx2+xy) : (y2x3−xy2+1x+y):xy
A=( \(\dfrac{x}{y\left(x+y\right)}\) - \(\dfrac{x-y}{x\left(x+y\right)}\)) : (\(\dfrac{y^2}{x\left(x-y\right)\left(x+y\right)}\)+\(\dfrac{1}{x+y}\)) : \(\dfrac{x}{y}\)
A=\(\dfrac{x^2-y\left(x-y\right)}{xy\left(x+y\right)}\) : \(\dfrac{y^2+x\left(x-y\right)}{x\left(x-y\right)\left(x+y\right)}\) : \(\dfrac{x}{y}\)
A = \(\dfrac{x^2-xy+y^2}{xy\left(x+y\right)}\) : \(\dfrac{y^2-xy+x^2}{x\left(x-y\right)\left(x+y\right)}\):\(\dfrac{x}{y}\)
A = \(\dfrac{x^2-xy+y^2}{xy\left(x+y\right)}\). \(\dfrac{x\left(x-y\right)\left(x+y\right)}{x^2-xy+y^2}\):\(\dfrac{x}{y}\)
A = \(\dfrac{x-y}{y}\) : \(\dfrac{x}{y}\)
A = \(\dfrac{x-y}{x}\)
A= 1 - \(\dfrac{y}{x}\)>1
=> y/x <0
=> xy<0 , x+y khác 0
a: Để \(f\left(x\right)=\dfrac{2}{\left|x\right|-2}\) có nghĩa thì \(\left|x\right|-2\ne0\)
=>\(\left|x\right|\ne2\)
=>\(x\in R\backslash\left\{2;-2\right\}\)
b: Để \(f\left(x\right)=\dfrac{1}{2-x}+\dfrac{1}{x+3}\) có nghĩa thì \(\left\{{}\begin{matrix}2-x\ne0\\x+3\ne0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x\ne2\\x\ne-3\end{matrix}\right.\)