Với \(x\ne\pm3\)ta có : \(A=\left(\frac{21}{x^2-9}-\frac{x-4}{3-x}-\frac{x-1}{3+x}\right):\left(1-\frac{1}{x+3}\right)\)

\(=\left(\frac{21}{\left(x-3\right)\left(x+3\right)}+\frac{\left(x+3\right)\left(x-4\right)}{\left(x-3\right)\left(x+3\right)}-\frac{\left(x-3\right)\left(x-1\right)}{\left(x-3\right)\left(x+3\right)}\right):\frac{x+2}{x+3}\)

\(=\frac{x^2-x-12-\left(x^2-4x+3\right)+21}{\left(x-3\right)\left(x+3\right)}.\frac{x+3}{x+2}=\frac{3x+6}{\left(x-3\right)\left(x+3\right)}.\frac{x+3}{x+2}\)

\(=\frac{3\left(x+2\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)\left(x+2\right)}=\frac{3}{x-3}\)

\(A=\left(\frac{21}{x^2-9}-\frac{x-4}{3-x}-\frac{x-1}{3+x}\right)\div\left(1-\frac{1}{x+3}\right)\)

\(=\left(\frac{21}{\left(x-3\right)\left(x+3\right)}+\frac{x-4}{x-3}-\frac{x-1}{x+3}\right)\div\left(1-\frac{1}{x+3}\right)\)

\(=\left(\frac{21}{\left(x-3\right)\left(x+3\right)}+\frac{\left(x-4\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\frac{\left(x-1\right)\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\right)\div\left(\frac{x+3}{x+3}-\frac{1}{x+3}\right)\)

\(=\left(\frac{21+x^2-x-12-x^2+4x-3}{\left(x-3\right)\left(x+3\right)}\right)\div\left(\frac{x+3-1}{x+3}\right)\)

\(=\frac{3x+6}{\left(x-3\right)\left(x+3\right)}\div\frac{x+2}{x+3}\)

\(=\frac{3\left(x+2\right)}{\left(x-3\right)\left(x+3\right)}\div\frac{x+2}{x+3}\)

\(=\frac{3\left(x+2\right)}{\left(x-3\right)\left(x+3\right)}\times\frac{x+3}{x+2}\)

\(=\frac{3\left(x+2\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)\left(x+2\right)}=\frac{3}{x-3}\)

Có:\(x^4+64y^4\)

\(=\left(x^4+16x^2y^2+64y^4\right)-16x^2y^2\)

\(=\left(x^2+8y^2\right)^2-\left(4xy\right)^2\)

\(=\left(x^2+4xy+8y^2\right)\left(x^2-4xy+8y^2\right)\)

Linz

= 64y⁴ + 32xy³ + 8y²x²  - 32xy³  -16x²y² -  4x³y + 8x²y² +4x³y +x⁴

= 8y² ( 8y² + 4xy + x² ) - 4xy ( 8y² + 4xy + x² ) + x²  ( 8y² + 4xy + x² )

= ( 8y² - 4xy + x² ) ( 8y² + 4xy + x² )

a) Ta có:

\(A=1+2+2^2+2^3+...+2^{2015}\)

\(A=\left(1+2+2^2\right)+...+\left(2^{2014}+2^{2015}+2^{2016}\right)-2^{2016}\)

\(A=7+...+7\cdot2^{2014}-2^{2016}\)

\(A=7\cdot\left(1+...+2^{2014}\right)-2^{2016}\)

Lại có: \(2^4\equiv2\left(mod7\right)\Leftrightarrow\left(2^4\right)^{504}=2^{2016}\equiv2\left(mod7\right)\)

\(\Rightarrow A\equiv-2\left(mod7\right)\)

Vậy A chia 7 dư -2 hoặc 5

b) \(PT\Leftrightarrow x\left(x+2\right)\left(x+1\right)=0\)

\(\Rightarrow x\in\left\{0;-2-;-1\right\}\)

=> Tổng các nghiệm là: -3