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a)
=(x-2)3
b)\(\left(2-x\right)^3\)
c)\(\left(x+\dfrac{1}{3}\right)^3\)
d)\(\left(\dfrac{x}{2}+y\right)^3\)
e)
\(=\left(x-1\right)^2\left(x-1-15\right)+25\left[3\left(x-1\right)-5\right]\)
\(=\left(x-1\right)^2\left(x-16\right)+25\left(3x-3-5\right)\)
\(=\left(x-1\right)^2\left(x-16\right)+25\left(3x-8\right)\)
a)\(\left(x-y\right)\left(x^3+x^2y+xy^2+y^3\right)=x^4+x^3y+x^2y^2+xy^3-x^3y-x^2y^2-xy^3-y^4=x^4-y^4\)
b) \(x\left(3x-18\right)-3\left(x-4\right)\left(x-2\right)+8=3x^2-18x-3x^2+18x-24+8=-16\)
Ta có: \(\left(x^3-x^2y+xy^2-y^3\right)\left(x+y\right)\)
\(=\left[x^2\left(x-y\right)+y^2\left(x-y\right)\right]\left(x+y\right)\)
\(=\left(x^2-y^2\right)\left(x^2+y^2\right)\)
\(=x^4-y^4=2^4-\left(\dfrac{1}{2}\right)^4=16-\dfrac{1}{16}=\dfrac{255}{16}\)
\(a,=\left(3x+2y\right)^3\\ b,=\left(4-x\right)^3\\ c,=\left(\dfrac{1}{2}x-3y\right)^3\)
`a)(x-1)(x^2+x+1)`
`=x^3+x^2+x-x^2-x-1`
`=x^3-1`
`b)(x^3+x^2y+xy^2+y^3)(x-y)`
`=x^4+x^3y+x^2y^2+xy^3-x^3y-x^2y^2-xy^3-y^4`
`=x^4-y^4`
a) VT`=(x-1)(x^2+x+1)`
`=x^3 +x^2 +x -x^2-x-1 `
`=x^3-1=` VP.
b) VT `=(x^3+x^2y+xy^2+y^3)(x-y)`
`=x^4+x^3y+x^2y^2+xy^3-x^3y-x^2y^2-xy^3-y^4`
`=x^4-y^4=` VP.
D = ( x 3 + y 3 ) – x y ( x + y ) = ( x + y ) ( x 2 – x y + y 2 ) – x y ( x + y ) = ( x + y ) ( x 2 – x y + y 2 – x y ) = ( x + y ) [ x ( x – y ) – y ( x – y ) ] = ( x + y ) ( x – y ) 2
Vì x = y ó x – y = 0 nên D = ( x + y ) ( x – y ) 2 = 0
Đáp án cần chọn là: D
Ta có
B = x 3 + x 2 y – x y 2 – y 3 = x 2 ( x + y ) – y 2 ( x + y ) = ( x 2 – y 2 ) ( x + y ) = ( x – y ) ( x + y ) ( x + y ) = ( x – y ) ( x + y ) 2
Thay x = 3,25 ; y = 6,57 ta được
B = ( 3 , 25 – 6 , 75 ) ( 3 , 25 + 6 , 75 ) 2 = - 3 , 5 . 10 2 = - 350
Đáp án cần chọn là: B
\(C=xyz+\left(xy+yz+xz\right)+x+y+z-1\)
Ta có ĐT tương đương
\(C=xyz+\left(xy+yz+xz\right)+x+y+z-1=\left(x-1\right)\left(y-1\right)\left(z-1\right)\)
Thay \(x=9\) ; \(y=10\) ; \(z=11\) vào BT có :
\(\left(9-1\right)\left(10-1\right)\left(11-1\right)=720\)
Vậy .........
C = xyz - xy - yz - xz + x + y +z- 1
= xy(z-1) - y(z-1) - x(z-1) + 1(z-1)
(xy-y-x+1)(z-1)