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a) Ta có: \(\frac{\left[2\left(y-x\right)^3-2\left(y-x\right)^2+\left(x-y\right)\right]}{y-x}\)
\(=\frac{2\left(y-x\right)^3}{y-x}-\frac{2\left(y-x\right)^2}{y-x}+\frac{x-y}{y-x}\)
\(=2\left(y-x\right)^2-2\left(y-x\right)-1\)
\(=2y^2-4yx+2x^2-2y+2x-1\)
Câu c phải là \(\left(\frac{x}{2}-y\right)^3\) chứ không phải \(\left(\frac{4}{2}-2\right)^3\)
A=x2+y2=x2+2xy+y2-2xy
=(x+y)2-2xy
=32-2.(-2)
=9+4
=13
B= x^3 + y^3
=x3+3x2y+3xy2+y3-3x2y-3xy2
=(x+y)3-3xy.(x+y)
=33-3.(-2).3
=27+18
=45
C= x^4 +y^4
=x4+2x2y2+y4-2x2y2
=(x2+y2)2-2.(xy)2
=132-2.(-2)2
=169-8
=161
D= x^6+ y^6
=x6+2x3y3+y6-2x3y3
=(x3+y3)2-2.(xy)3
=452-2.(-2)3
=2041
\(B=\left(x+y\right)^3+3\left(x-y\right)\left(x+y\right)^2+3\left(x-y\right)^2\left(x+y\right)+\left(x-y\right)^3\)
\(=\left(x+y\right)^3+3\cdot\left(x+y\right)^2\cdot\left(x-y\right)+3\cdot\left(x+y\right)\cdot\left(x-y\right)^2+\left(x-y\right)^3\)
\(=\left[\left(x+y\right)+\left(x-y\right)\right]^3\)
\(=\left(x+y+x-y\right)^3\)
\(=\left(2x\right)^3\)
\(=8x^3\)
\(---\)
\(C=8\left(x+2y\right)^3-6\left(x+2y\right)^2x+12\left(x+2y\right)x^2-8x^3\) (sửa đề)
\(=\left[2\left(x+2y\right)\right]^3-3\cdot\left(x+2y\right)^2\cdot2x+3\cdot\left(x+2y\right)\cdot\left(2x\right)^2-\left(2x\right)^3\)
\(=\left[2\left(x+2y\right)-2x\right]^3\)
\(=\left(2x+4y-2x\right)^3\)
\(=\left(4y\right)^3\)
\(=64y^3\)
\(---\)
\(D=\left(x-y\right)^3-3\cdot\dfrac{\left(x-y\right)^2}{2}\cdot y+3\cdot\dfrac{\left(x-y\right)}{4}\cdot y^2-\dfrac{y^3}{8}\)
\(=\left(x-y\right)^3-3\cdot\left(x-y\right)^2\cdot\dfrac{y}{2}+3\cdot\left(x-y\right)\cdot\left(\dfrac{y}{2}\right)^2-\left(\dfrac{y}{2}\right)^3\)
\(=\left[\left(x-y\right)-\dfrac{y}{2}\right]^3\)
\(=\left(x-y-\dfrac{y}{2}\right)^3\)
\(=\left(x-\dfrac{3}{2}y\right)^3\)
#\(Toru\)
3 x 2 + 6 x y 2 - 3 y 2 + 6 x 2 y = 3 x 2 - 3 y 2 + 6 x y 2 + 6 x 2 y = 3 x 2 - y 2 + 6 x y x + y = 3 x - y x + y + 6 x y x + y = 3 x - y + 6 x y x + y = 3 x - y + 2 x y x + y
Vậy chỗ trống là x - y + 2 x y
Đáp án cần chọn là: B