Question

Bài 4. Tính giá trị của biểu thức
a) $A=(2 x-y)\left(4 x^2+2 x y+y^2\right)$, biết $2 x-y=3$.
b) $B=\left(x-\frac{2 y}{3}\right)\left(x^3+\frac{2 x y}{3}+\frac{4 y^2}{9}\right)$, biết $3 x-2 y=-1$.

Answer 1

a: 2x-y=3

=>y=2x-3

\(A=\left(2x-y\right)\left(4x^2+2xy+y^2\right)\)

\(=\left(2x-y\right)\left[\left(2x\right)^2+2x\cdot y+y^2\right]\)

\(=\left(2x\right)^3-y^3=\left(2x-y\right)^3+3\cdot2x\cdot y\cdot\left(2x-y\right)\)

\(=3^3+6xy\cdot3=18xy+27\)

b: 

3x-2y=-1

=>\(3\left(x-\dfrac{2}{3}y\right)=-1\)

=>\(x-\dfrac{2}{3}y=-\dfrac{1}{3}\)

Sửa đề: \(B=\left(x-\dfrac{2y}{3}\right)\left(x^2+\dfrac{2xy}{3}+\dfrac{4y^2}{9}\right)\)

\(=\left(x-\dfrac{2y}{3}\right)\left[x^2+x\cdot\dfrac{2y}{3}+\left(\dfrac{2y}{3}\right)^2\right]\)

 

\(=x^3-\left(\dfrac{2y}{3}\right)^3\)

\(=x^3-\dfrac{8y^3}{27}=\dfrac{27x^3-8y^3}{27}\)

\(=\dfrac{1}{27}\left[27x^3-8y^3\right]=\dfrac{1}{27}\left[\left(3x-2y\right)^3+3\cdot3x\cdot2y\left(3x-2y\right)\right]\)

\(=\dfrac{1}{27}\cdot\left[\left(-1\right)^3+18x\cdot\left(-1\right)\right]=\dfrac{1}{27}\left(-18x-1\right)=\dfrac{-18x-1}{27}\)