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

\(=x^4+\frac{2}{5}x^2y-\frac{2}{5}x^2y-\left(\frac{2}{5}y\right)^2\)

\(=x^4-\frac{4}{25}y^2\)

a/ \(\left(x^2+\frac{2}{5}y\right)\left(x^2-\frac{2}{5}y\right)=\left(x^2\right)^2-\left(\frac{2}{5}y\right)^2=x^4-\frac{4}{25}y^2\)

b/ (x - 3y)(x² + 3xy + 9y²) = x³ - 27y³

c/ (5 + 3x)³ = 125 + 225x + 135x² + 27x³

Theo đề ta có: \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2\)

\(\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{1}{ab}+\frac{1}{ac}+\frac{1}{bc}\right)=4\)

\(\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=4-2\left(\frac{1}{ab}+\frac{1}{ac}+\frac{1}{bc}\right)\)

\(\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=4-2.\frac{a+b+c}{abc}=4-2.\frac{abc}{abc}=4-2=2\left(đpcm\right)\)

                                                              Vậy \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=2\)

Ta có :  \(M=\left(x^2+3x+2\right)\left(x^2+7x+12\right)+1=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)+1\)

\(=\left[\left(x+1\right)\left(x+4\right)\right].\left[\left(x+2\right)\left(x+3\right)\right]+1=\left(x^2+5x+4\right)\left(x^2+5x+6\right)+1\)

Đặt \(t=x^2+5x+5\) \(\Rightarrow M=\left(t-1\right)\left(t+1\right)+1=t^2-1+1=t^2\)

Vậy \(M=\left(x^2+5x+5\right)^2\)