Tính:
a) \((4{x^3}):( - 2{x^2})\);
b) \(( - 7{x^2}):(6x)\);
c) \(( - 14{x^4}):( - 8{x^3})\).
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a) \( - 2{x^2} + 6{x^2} = ( - 2 + 6).{x^2} = 4{x^2}\);
b) \(4{x^3} - 8{x^3} = (4 - 8).{x^3} = - 4{x^3}\);
c) \(3{x^4}( - 6{x^2}) = 3.( - 6).{x^4}.{x^2} = - 18{x^{4 + 2}} = - 18{x^6}\);
d) \(( - 24{x^6}):( - 4{x^3}) = ( - 24: - 4).({x^6}:{x^3}) = 6{x^{6 - 3}} = 6{x^3}\).
`x + 2 3/4 = 5 2/3`
`=> x + 11/4 = 17/3`
`=> x= 17/3 -11/4`
`=>x=35/12`
__
`x - 1 4/5 = 3 2/7`
`=> x- 9/5 = 23/7`
`=> x= 23/7 +9/5`
`=>x=178/35`
__
`x xx 3 1/2 =4 3/4`
`=> x xx 7/2 =19/4`
`=> x= 19/4 : 7/2`
`=> x= 19/4 xx 2/7`
`=> x= 19/14`
__
`x : 2 2/3 = 4 1/3`
`=> x : 8/3 = 13/3`
`=> x= 13/3 xx 8/3`
`=>x=104/9`
a: =>x+11/4=17/3
=>x=17/3-11/4=68/12-33/12=35/12
b: =>x-9/5=23/7
=>x=23/7+9/5=178/35
c: =>x*7/2=4,75
=>x=19/4:7/2=19/14
d: =>x:8/3=13/3
=>x=13/3*8/3=104/9
a) \(\dfrac{1}{2}x(6x - 4) = \dfrac{1}{2}x.6x + \dfrac{1}{2}x.( - 4) = 3{x^2} - 2x\).
b) \(\begin{array}{l} - {x^2}(\dfrac{1}{3}{x^2} - x - \dfrac{1}{4}) = - {x^2}.\dfrac{1}{3}{x^2} + - {x^2}. - x + - {x^2}. - \dfrac{1}{4}\\ = - \dfrac{1}{3}{x^4} + {x^3} + \dfrac{1}{4}{x^2}\end{array}\)
\(\dfrac{3}{7}\times\dfrac{7}{9}\times\dfrac{1}{2}\)
\(=\dfrac{3\times7\times1}{7\times9\times2}\)
\(=\dfrac{21}{126}\)
\(=\dfrac{1}{6}\)
\(\dfrac{5}{8}\times4\times\dfrac{1}{2}\\ =\dfrac{5}{8}\times\dfrac{4}{1}\times\dfrac{1}{2}\\ =\dfrac{5\times4\times1}{8\times1\times2}\\ =\dfrac{20}{16}\\ =\dfrac{5}{4}\)
\(4\times\dfrac{1}{24}\times3\\ =\dfrac{4}{1}\times\dfrac{1}{24}\times\dfrac{3}{1}\\ =\dfrac{4\times1\times3}{1\times24\times1}\\ =\dfrac{12}{24}\\ =\dfrac{1}{2}\)
Bài 5:
a. 1 - 2y + y2
= (1 - y)2
b. (x + 1)2 - 25
= (x + 1)2 - 52
= (x + 1 - 5)(x + 1 + 5)
= (x - 4)(x + 6)
c. 1 - 4x2
= 12 - (2x)2
= (1 - 2x)(1 + 2x)
d. 8 - 27x3
= 23 - (3x)3
= (2 - 3x)(4 + 6x + 9x2)
e. (đề hơi khó hiểu ''x3'' !?)
g. x3 + 8y3
= (x + 2y)(x2 - 2xy + y2)
a: \(=\dfrac{2x^3-3x^2+4x^2-6x-2x+3}{2x-3}=x^2+2x-1\)
b: \(=\dfrac{x-5}{\left(x-2\right)^2}\cdot\dfrac{2\left(x-2\right)}{\left(x-5\right)\left(x+5\right)}=\dfrac{2}{\left(x-2\right)\left(x+5\right)}\)
\(a,\left(x-2\right)\left(x+3\right)-x\left(x-5\right)=x^2-2x+3x-6-x^2+5x=6x-6\)
\(b,\dfrac{1}{x-2}+\dfrac{-2}{x+2}+\dfrac{2x-8}{x^2-4}=\dfrac{x+2}{\left(x-2\right)\left(x+2\right)}-\dfrac{2\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}+\dfrac{2x-8}{\left(x+2\right)\left(x-2\right)}=\dfrac{x+2}{\left(x-2\right)\left(x+2\right)}-\dfrac{2x-4}{\left(x+2\right)\left(x-2\right)}+\dfrac{2x-8}{\left(x+2\right)\left(x-2\right)}=\dfrac{x+2-2x+4+2x-8}{\left(x+2\right)\left(x-2\right)}=\dfrac{x-2}{\left(x+2\right)\left(x-2\right)}=\dfrac{1}{x+2}\)
a) \((4{x^3}):( - 2{x^2})\\= [4: (- 2)].({x^3}:{x^2})\\ = - 2.{x^{3 - 2}}\\ = - 2x\);
b) \(( - 7{x^2}):(6x) \\= ( - 7:6).({x^2}:x) \\= - \dfrac{7}{6}.{x^{2 - 1}}\\ = - \dfrac{7}{6}.x\);
c) \(( - 14{x^4}):( - 8{x^3}) \\= ( - 14: - 8).({x^4}:{x^3})\\= \dfrac{7}{4}.{x^{4 - 3}} \\= \dfrac{7}{4}.x\).