Question

Tính

\(1.\left(x+2y\right)^2\)

\(2.\left(2x+3y\right)^2\)

\(3.\left(x+\frac{1}{3}\right)^4\)

\(4.\left(2x+y^2\right)^3\)

\(5.\left(\frac{x}{2}-2y\right)\)

\(6.\left(\sqrt{2x-y}\right)^4\)

\(7.\left(x+1\right)\left(x^2-x+1\right)\)

\(8.\left(x-3\right)\left(x^2+3x+9\right)\)

Answer 1

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

2) \(\left(2x+3y\right)^2=4x^2+12xy+9y^2\)

3) \(\left(x+\frac{1}{3}\right)^4=\left[\left(x+\frac{1}{3}\right)^2\right]^2=\left(x^2+\frac{2}{3}x+\frac{1}{9}\right)^2=x^4+\frac{4}{9}x^2+\frac{1}{81}+\frac{4}{3}x^3+\frac{4}{27}x+\frac{2}{9}x^2=x^4+\frac{2}{3}x^2+\frac{1}{81}+\frac{4}{3}x^3+\frac{4}{27}x\)

4) \(\left(2x+y^2\right)^3=8x^3+12x^2y^2+6xy^4+y^6\)

5) Sửa đề: \(\left(\frac{x}{2}-2y\right)^3=\frac{x^3}{8}-\frac{3x^2}{2}+6xy^2-8y^3\)

6) \(\left(\sqrt{2x-y}\right)^4=\left(2x-y\right)^2=4x^2-4xy+y^2\)

7) \(\left(x+1\right)\left(x^2-x+1\right)=x^3+1\)

8) \(\left(x-3\right)\left(x^2+3x+9\right)=x^3-27\)