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\(\left(\frac{1}{x+1}-\frac{3}{\left(x+1\right)\left(x^2-x+1\right)}+\frac{3}{x^2-x+1}\right).\frac{3\left(x^2-x+1\right)}{\left(x+1\right)\left(x+2\right)}-\frac{2\left(x-1\right)}{x+1}\)
\(\left(\frac{x^2-x+1}{x^3+1}-\frac{3}{x^3+1}+\frac{3\left(x+1\right)}{x^3+1}\right).\frac{3\left(x^2-x+1\right)}{\left(x+1\right)\left(x+2\right)}-\frac{2\left(x-1\right)}{x+1}\)
\(\left(\frac{x^2-x+1-3+3x+3}{x^3+1}\right).\frac{3\left(x^2-x+1\right)}{\left(x+1\right)\left(x+2\right)}-\frac{2\left(x-1\right)}{x+1}\)
tới đây bạn biến đổi tiếp, gõ = cái này lâu quá, gõ mathtype nhanh hơn
\(\left(\frac{x+1}{2\left(x-1\right)}+\frac{3}{x^2-1}-\frac{x+3}{2\left(x+1\right)}\right)\frac{4x^2-4}{5}\)
\(=\left(\frac{x+1}{2\left(x-1\right)}+\frac{3}{\left(x-1\right)\left(x+1\right)}-\frac{x+3}{2\left(x+1\right)}\right)\frac{4x^2-4}{5}\)
\(=\left[\frac{\left(x+1\right)^2}{2\left(x-1\right)\left(x+1\right)}+\frac{6}{2\left(x-1\right)\left(x+1\right)}-\frac{\left(x+3\right)\left(x-1\right)}{2\left(x-1\right)\left(x+1\right)}\right]\frac{4x^2-4}{5}\)
\(=\left(\frac{x^2+2x+1+6-x^2+x-3x+3}{2\left(x-1\right)\left(x+1\right)}\right)\frac{4\left(x^2-1\right)}{5}\)
\(=\frac{10}{2\left(x-1\right)
\left(x+1\right)}.\frac{4\left(x-1\right)\left(x+1\right)}{5}\)
\(=4\)
Vậy giá trị của biểu thức là 4
\(M=\left(\frac{\left(x^2-1\right)\left(x^2+1\right)-\left(x^4-x^2+1\right)}{\left(x^2+1\right)\left(x^4-x^2+1\right)}\right)\left(x^4-\frac{\left(x^2-1\right)\left(x^2+1\right)}{x^2+1}\right)\)
\(=\left(\frac{x^4-1-x^4+x^2-1}{\left(x^2+1\right)\left(x^4-x^2+1\right)}\right)\left(x^4-x^2+1\right)\)
\(=\frac{x^2-2}{x^2+1}\)
b/ \(M=\frac{x^2+1-3}{x^2+1}=1-\frac{3}{x^2+1}\)
Do \(x^2+1\ge1\Rightarrow\frac{3}{x^2+1}\le3\Rightarrow1-\frac{3}{x^2+1}\ge1-3=-2\)
\(\Rightarrow M_{min}=-2\) khi \(x=0\)