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\(Q=\left(\frac{x^2+1}{x+1}\right)\left(\frac{4}{x-1}-\frac{2}{x}\right)=\left(\frac{x^2+1}{x+1}\right)\left(\frac{4x-2\left(x-1\right)}{x\left(x-1\right)}\right)\)
\(=\left(\frac{x^2+1}{x+1}\right)\left(\frac{4x-2x+2}{x\left(x-1\right)}\right)=\left(\frac{x^2+1}{x+1}\right)\left(\frac{2\left(x+1\right)}{x\left(x-1\right)}\right)\)
\(=\frac{2\left(x^2+1\right)\left(x+1\right)}{x\left(x+1\right)\left(x-1\right)}=\frac{2\left(x^2+1\right)}{x\left(x-1\right)}=\frac{2\left(x^2+1\right)}{x^2-x}\)
Chết rồi mk nhầm đề bài
\(Q=\left(\frac{x^2+1}{x+1}-1\right)\left(\frac{4}{x-1}-\frac{2}{x}\right)\)
\(\frac{1}{x\left(x-1\right)}+\frac{1}{\left(x-1\right)\left(x-2\right)}+\frac{1}{\left(x-2\right)\left(x-3\right)}+...+\frac{1}{\left(x-4\right)\left(x-5\right)}\)
\(=\frac{1}{x}-\frac{1}{x-1}+\frac{1}{x-1}-\frac{1}{x-2}+\frac{1}{x-2}-\frac{1}{x-3}+...+\frac{1}{x-4}-\frac{1}{x-5}\)
\(=\frac{1}{x}-\frac{1}{x-5}=\frac{x-5}{x\left(x-5\right)}-\frac{x}{x\left(x-5\right)}=\frac{-5}{x\left(x-5\right)}\)
\(\frac{1}{x\left(x-1\right)}+\frac{1}{\left(x-1\right)\left(x-2\right)}+\frac{1}{\left(x-2\right)\left(x-3\right)}+...+\frac{1}{\left(x-4\right)\left(x-5\right)}\)
\(=\frac{1}{x}-\frac{1}{x-1}+\frac{1}{x-1}-\frac{1}{x-2}+...+\frac{1}{x-4}-\frac{1}{x-5}\)
\(=\frac{1}{x}-\frac{1}{x-5}\)
\(=\frac{x-5}{x\left(x-5\right)}-\frac{x}{x\left(x-5\right)}\)
\(=\frac{x-5-x}{x\left(x-5\right)}\)
\(=-\frac{5}{x\left(x-5\right)}\)
\(=\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+5}\)
1/ (x+1)(x+2) +1/ (x+2)(x+3) +1/ (x+3)(x+4) +1/ (x+4)(x+5)
=1/x+1 -1/x+2 +1/x+2 -1/x+3 +1/x+3 -1/x+4 +1/x+4 -1/x+5
=1/x+1 -1/x+5
=4/(x+1)(x+5)
a) \(\left(x-1\right)^2-\left(x-2\right)\left(x+2\right)=x^2-2x+1-x^2+4=5-2x\)
mình nghĩ là câu b bạn ghi đề sai vì như thế không có hằng đẳng thức nhé
b)\(\left(x^2+\frac{1}{3}x+\frac{1}{9}\right)\left(x-\frac{1}{3}\right)-\left(x-\frac{1}{3}\right)^3=x^3-\frac{1}{27}-x^3+\frac{1}{27}+x^2-\frac{1}{3}x=x^2-\frac{1}{3}x\)
b,\(\left(x^2+\frac{1}{x}+\frac{1}{9}\right)\left(x-\frac{1}{3}\right)-\left(x-\frac{1}{3}\right)^3\)
\(=\)\(\left(x-\frac{1}{3}\right)\left[\left(x^2+\frac{1}{x}+\frac{1}{9}\right)-\left(x-\frac{1}{3}\right)^2\right]\)
\(=\)\(\left(x-\frac{1}{3}\right)\left(x^2+\frac{1}{x}+\frac{1}{9}-x^2+\frac{2}{3}x-\frac{1}{9}\right)\)
\(=\left(x-\frac{1}{3}\right)\left(\frac{1}{x}+\frac{2}{3}x\right)\) \(=1+\frac{2}{3}x^2-\frac{1}{3x}-\frac{2}{9}x\)
\(P=\left(\frac{x-4}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}\right):\left(\frac{x^2+x+1}{x^2+x+1}-\frac{x-8}{x^2+x+1}\right)\)
\(=\left(\frac{x-4+x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}\right):\left(\frac{x^2+x+1-x+8}{x^2+x+1}\right)\)
\(=\left(\frac{x^2+2x-3}{\left(x-1\right)\left(x^2+x+1\right)}\right):\left(\frac{x^2+9}{x^2+x+1}\right)\)
\(=\left(\frac{\left(x+3\right)\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\right).\left(\frac{x^2+x+1}{x^2+9}\right)\)
\(=\frac{\left(x+3\right)\left(x-1\right)}{\left(x-1\right)\left(x^2+9\right)}\)
\(=\frac{x+3}{x^2+9}\)với \(x\ne1\)
Ta có: P = \(\left(\frac{x-4}{x^3-1}+\frac{1}{x-1}\right):\left(1-\frac{x-8}{x^2+x+1}\right)\)
P = \(\left(\frac{x-4}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}\right):\left(\frac{x^2+x+1-x+8}{x^2+x+1}\right)\)
P = \(\left(\frac{x-4+x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}\right):\left(\frac{x^2+9}{x^2+x+1}\right)\)
P = \(\frac{x^2+2x-3}{\left(x-1\right)\left(x^2+x+1\right)}\cdot\frac{x^2+x+1}{x^2+9}\)
P = \(\frac{x^2+3x-x-3}{\left(x-1\right)\left(x^2+x+1\right)}\cdot\frac{x^2+x+1}{x^2+9}\)
P = \(\frac{\left(x+3\right)\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\cdot\frac{x^2+x+1}{x^2+9}\)
P = \(\frac{x+3}{x^2+9}\)
\(=3x^3-\frac{3}{2}x^2-x^3-\frac{1}{2}x+\frac{1}{2}x+2\)
\(=2x^3-\frac{3}{2}x^2+2\)
=( \(\frac{x^2+1-x-1}{x+1}\))\(\left(\frac{2x+2}{x^2-x}\right)\)
= \(\frac{2\left(x^2-x\right)\left(x+1\right)}{\left(x+1\right)\left(x^2-x\right)}\)
=2
vậy ....
hok tốt
.....
=2 bạn nhé