Rút gọn rồi tính
A= \(-x\left(x-y\right)^2+\left(x-y\right)^3+y^2\left(x-2x\right)\) tại |2x-1| = 1, y=2
B= \(-\left(2x-y\right)^3-x\left(2x-y\right)^2-y^3\) tại \(\left(x-2\right)^2+y^2=0\)
C= \(\left(x+y\right)\left(x^2-xy+y^2\right)+3\left(2x-y\right)\left(4x^2+2xy+y^2\right)\) tại x + y = 2, y = -3
D= \(\left(x+3y\right)\left(x^2-3xy+9y^2\right)+\left(3x-y\right)\left(9x^2+3xy+y^2\right)\) tại 3x - y = 5, x = 2
E= \(\left(x+1\right)^3+6\left(x+1\right)^2+12x+20\) tại x = 5
F = \(\left(-x-2\right)^3+\left(2x-4\right)\left(x^2+2x+4\right)-x^2\left(x-6\right)\) tại x = -2
a: \(A=-x\left(x-y\right)^2+\left(x-y\right)^3+y^3\left(x-2x\right)\)
\(=\left(x-y\right)^2\left(-x+x-y\right)+y^3\cdot\left(-x\right)\)
\(=\left(x-y\right)^2\cdot\left(-y\right)+y^3\cdot\left(-x\right)\)
\(=-y\left(x^2-2xy+y^2\right)-xy^3\)
\(=-x^2y+2xy^2-y^3-xy^3\)
|2x-1|=1
=>\(\left[{}\begin{matrix}2x-1=1\\2x-1=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=0\end{matrix}\right.\)
Khi x=2 và y=2 thì \(A=-2^2\cdot2+2\cdot2\cdot2^2-2^3-2\cdot2^3\)
=-8+16-8-16
=-16
Khi x=0 và y=2 thì \(A=-0^2\cdot2+2\cdot0\cdot2^2-2^3-0\cdot2^3\)
=-8
b:
\(\left(x-2\right)^2+y^2=0\)
=>\(\left\{{}\begin{matrix}x-2=0\\y=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x=2\\y=0\end{matrix}\right.\)
\(B=-\left(2x-y\right)^3-x\left(2x-y\right)^2-y^3\)
Khi x=2;y=0 thì \(B=-\left(2\cdot2-0\right)^3-2\cdot\left(2\cdot2-0\right)^2-0^3\)
=-64-2*4^2
=-64-32
=-96
c: \(C=\left(x+y\right)\left(x^2-xy+y^2\right)+3\left(2x-y\right)\left(4x^2+2xy+y^2\right)\)
\(=x^3+y^3+3\left(8x^3-y^3\right)\)
\(=x^3+y^3+24x^3-3y^3=25x^3-2y^3\)
x+y=2
=>x-3=2
=>x=2+3=5
Khi x=5; y=-3 thì \(C=25\cdot5^3-2\cdot\left(-3\right)^3=3206\)
d: \(D=\left(x+3y\right)\left(x^2-3xy+9y^2\right)+\left(3x-y\right)\left(9x^2+3xy+y^2\right)\)
\(=x^3+27y^3+27x^3-y^3\)
\(=28x^3-26y^3\)
3x-y=5 nên 6-y=5
=>y=1
Khi x=2;y=1 thì \(D=28\cdot2^3-26\cdot1^3=28\cdot8-26=198\)
e: \(E=\left(x+1\right)^3+6\left(x+1\right)^2+12x+20\)
\(=\left(x+1\right)^3+6\left(x+1\right)^2+12\left(x+1\right)+8\)
\(=\left(x+1+2\right)^3=\left(x+3\right)^3\)
Khi x=5 thì \(E=\left(5+3\right)^3=8^3=512\)
f: ta có: \(F=\left(-x-2\right)^3+\left(2x-4\right)\left(x^2+2x+4\right)-x^2\left(x-6\right)\)
\(=-x^3-6x^2-12x-8+2\left(x-2\right)\left(x^2+2x+4\right)-x^3+6x^2\)
\(=-2x^3-12x-8+2\left(x^3-8\right)\)
\(=-2x^3-12x-8+2x^3-16\)
=-12x-24
Khi x=-2 thì \(F=\left(-12\right)\cdot\left(-2\right)-24=24-24=0\)