Tính gt biểu thức C = (x/2 -y)^3-6(y- x/2)^2+12(y- x/2) -8
mn giup em với ạ, mong ko trình bày tắt
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a) Ta có: \(4\left(x-2\right)^2+xy-2y\)
\(=4\left(x-2\right)^2+y\left(x-2\right)\)
\(=\left(x-2\right)\left(4x-8+y\right)\)
b) Ta có: \(x\left(x-y\right)^3-y\left(y-x\right)^2-y^2\left(x-y\right)\)
\(=x\left(x-y\right)^3-y\left(x-y\right)^2-y^2\left(x-y\right)\)
\(=\left(x-y\right)\left[x\left(x-y\right)^2-y\left(x-y\right)-y^2\right]\)
Ta có: \(\left(x-3\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+9\left(x+1\right)^2=15\)
\(\Leftrightarrow x^3-9x^2+27x-27-x^3+27+9\left(x^2+2x+1\right)=15\)
\(\Leftrightarrow-9x^2+27x+9x^2+18x+9=15\)
\(\Leftrightarrow45x=6\)
hay \(x=\dfrac{2}{15}\)
Lời giải:
a.
$27A=x^3-9x^2+162x-27=(x-3)^3+135x$
$=(303-3)^3+135.303=27040905$
$A=1001515$
b.
$B=2[(x+y)^3-3xy(x+y)]-3[(x+y)^2-2xy]$
$=2(1-3xy)-3(1-2xy)=2-6xy-3+6xy=-1$
c.
$C=x^3+y^3+3xy(x+y)=(x+y)^3=1^3=1$
\(1,P=\left(x+y+x-y\right)\left(x+y-x+y\right)+2\left(x^2-y^2\right)-4y^2\\ P=4xy+2x^2-6y^2\)
Bài 1:
\(P=2\left(x+y\right)\left(x-y\right)-\left(x-y\right)^2+\left(x+y\right)^2-4y^2\)
\(=2\left(x^2-y^2\right)-\left(x^2-2xy+y^2\right)+\left(x^2+2xy+y^2\right)-4y^2\)
\(=2x^2-2y^2-x^2+2xy-y^2+x^2+2xy+y^2-4y^2\)
\(=2x^2+4xy-7y^2\)
Ta có: \(C=\left(\dfrac{x}{2}-y\right)^3-6\left(y-\dfrac{x}{2}\right)^2+12\left(y-\dfrac{x}{2}\right)-8\)
\(=\left(\dfrac{x}{2}-y\right)^3-3\cdot\left(\dfrac{x}{2}-y\right)^2\cdot2-3\cdot\left(\dfrac{x}{2}-y\right)\cdot2^2-2^3\)
\(=\left(\dfrac{x}{2}-y\right)^3-8-6\left(\dfrac{x}{2}-y\right)\left(\dfrac{x}{2}-y-2\right)\)
\(=\left(\dfrac{x}{2}-y-2\right)\left[\left(\dfrac{x}{2}-y\right)^2+2\left(\dfrac{x}{2}-y\right)+2^2\right]-6\left(\dfrac{x}{2}-y\right)\left(\dfrac{x}{2}-y-2\right)\)
\(=\left(\dfrac{x}{2}-y-2\right)\left[\left(\dfrac{x}{2}-y\right)^2-4\left(\dfrac{x}{2}-y\right)+4\right]\)
\(=\left(\dfrac{x}{2}-y-2\right)^3\)