\(A=\left(\frac{\left(1-x\right)\left(1+x+x^2\right)}{1-x}-x\right):\frac{\left(1-x^2\right)}{\left(1-x^2\right)\left(1-x\right)}\)

\(\Leftrightarrow A=\left(1+x+x^2-x\right):\frac{1}{1-x}=\left(1+x^2\right)\left(1-x\right)\)

\(8^5+2^{11}=\left(2^3\right)^5+2^{11}=2^{15}+2^{11}=2^{11}\left(2^4+1\right)=2^{11}.17⋮17\)

bài làm 

x^2 + 5 x^4 + 2x^3 + x - 3 x^2 + 2x - 5 x^4 + 5x^2 2x^3 - 5x^2 2x^3 + 10x -5x^2 - 9x -5x^2 - 25 -9x + 22

a, \(\frac{x^2}{x+1}+\frac{2x}{x^2-1}+\frac{1}{x+1}+1\)

\(=\frac{x^2\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}-\frac{2x}{\left(x-1\right)\left(x+1\right)}+\frac{x-1}{\left(x+1\right)\left(x-1\right)}+\frac{\left(x+1\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}\)

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

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

\(C=\left(\frac{1+a^3}{1+a}-a\right)\left(\frac{2a^2+4}{a^3-8}-\frac{a}{a^2+2a+4}\right)\)

\(=\left(\frac{\left(a+1\right)\left(a^2-a+1\right)}{1-a}-\frac{\left(1-a\right)a}{1-a}\right)\left(\frac{2a^4}{\left(a-2\right)\left(a^2+2a+4\right)}-\frac{a}{a^2+2a+4}\right)\)

\(=\left(\frac{a^3+1-a+a^2}{1-a}\right)\left(\frac{2a^4}{\left(a-2\right)\left(a^2+2a+4\right)}-\frac{a\left(a-2\right)}{\left(a-2\right)\left(a^2+2a+4\right)}\right)\)

\(=\left(\frac{a^3+1-a+a^2}{1-a}\right)\left(\frac{2a^4-a^2+2a}{\left(a-2\right)\left(a^2-2a+4\right)}\right)\)

\(=\left(\frac{a^3+1-a+a^2}{-\left(a-1\right)}\right)\left(\frac{2a\left(a^3-1\right)}{\left(a-2\right)\left(a^2-2a+4\right)}\right)\)

tình nốt nhé, thấy sai sai ở đâu á, kiểm tra lại zùm mk