Rút gọn \(P=\left(\frac{x^3-1}{x-1}+x\right)\left(\frac{x^3+1}{x+1}-x\right):\frac{x\left(1-x^2\right)^2}{x^2-2}\)
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27 tháng 11 2015
\(=\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+5}\)
27 tháng 11 2015
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)
16 tháng 12 2020
\(\frac{1}{x\left(x-1\right)}+\frac{1}{\left(x-1\right)\left(x-2\right)}+\frac{1}{\left(x-2\right)\left(x-3\right)}+...+\frac{1}{\left(x-4\right)\left(x-5\right)}\)
\(=\frac{1}{x}-\frac{1}{x-1}+\frac{1}{x-1}-\frac{1}{x-2}+\frac{1}{x-2}-\frac{1}{x-3}+...+\frac{1}{x-4}-\frac{1}{x-5}\)
\(=\frac{1}{x}-\frac{1}{x-5}=\frac{x-5}{x\left(x-5\right)}-\frac{x}{x\left(x-5\right)}=\frac{-5}{x\left(x-5\right)}\)
NC
16 tháng 12 2020
\(\frac{1}{x\left(x-1\right)}+\frac{1}{\left(x-1\right)\left(x-2\right)}+\frac{1}{\left(x-2\right)\left(x-3\right)}+...+\frac{1}{\left(x-4\right)\left(x-5\right)}\)
\(=\frac{1}{x}-\frac{1}{x-1}+\frac{1}{x-1}-\frac{1}{x-2}+...+\frac{1}{x-4}-\frac{1}{x-5}\)
\(=\frac{1}{x}-\frac{1}{x-5}\)
\(=\frac{x-5}{x\left(x-5\right)}-\frac{x}{x\left(x-5\right)}\)
\(=\frac{x-5-x}{x\left(x-5\right)}\)
\(=-\frac{5}{x\left(x-5\right)}\)
BM
27 tháng 10 2018
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)\(\left(x^2+\frac{1}{3}x+\frac{1}{9}\right)\left(x-\frac{1}{3}\right)-\left(x-\frac{1}{3}\right)^3=x^3-\frac{1}{27}-x^3+\frac{1}{27}+x^2-\frac{1}{3}x=x^2-\frac{1}{3}x\)
MC
27 tháng 10 2018
b,\(\left(x^2+\frac{1}{x}+\frac{1}{9}\right)\left(x-\frac{1}{3}\right)-\left(x-\frac{1}{3}\right)^3\)
\(=\)\(\left(x-\frac{1}{3}\right)\left[\left(x^2+\frac{1}{x}+\frac{1}{9}\right)-\left(x-\frac{1}{3}\right)^2\right]\)
\(=\)\(\left(x-\frac{1}{3}\right)\left(x^2+\frac{1}{x}+\frac{1}{9}-x^2+\frac{2}{3}x-\frac{1}{9}\right)\)
\(=\left(x-\frac{1}{3}\right)\left(\frac{1}{x}+\frac{2}{3}x\right)\) \(=1+\frac{2}{3}x^2-\frac{1}{3x}-\frac{2}{9}x\)
27 tháng 2 2020
\(\left(\frac{1}{x+1}-\frac{3}{x^3+1}+\frac{3}{x^2-x+1}\right):\frac{3x^2-3x+3}{\left(x+1\right)\left(x+2\right)}-\frac{2x-2}{x^2+2x}\left(x\ne-1;x\ne0;x\ne-2\right)\)
\(=\left(\frac{1}{x+1}-\frac{3}{\left(x+1\right)\left(x^2-x+1\right)}+\frac{3}{x^2-x+1}\right):\frac{3x^3-3x+3}{\left(x+1\right)\left(x+2\right)}-\frac{2\left(x-1\right)}{x\left(x+2\right)}\)
\(=\left(\frac{x^2-x+1}{\left(x+1\right)\left(x^2-x+1\right)}-\frac{3}{\left(x+1\right)\left(x^2-x+1\right)}+\frac{3x+3}{\left(x+1\right)\left(x^2-x+1\right)}\right)\)\(:\frac{3x^2-3x+3}{\left(x+1\right)\left(x+2\right)}-\frac{2\left(x-1\right)}{x\left(x+2\right)}\)
\(=\frac{x^2-x+1-3+3x+3}{\left(x+1\right)\left(x^2-x+1\right)}:\frac{3x^2-3x+3}{\left(x+1\right)\left(x+2\right)}-\frac{2\left(x-1\right)}{x\left(x+2\right)}\)
\(=\frac{x^2+2x+1}{\left(x+1\right)\left(x^2-x+1\right)}:\frac{3\left(x^2-x+1\right)}{\left(x+1\right)\left(x+1\right)}-\frac{2\left(x-1\right)}{x\left(x+2\right)}\)
\(=\frac{\left(x+1\right)\left(x+2\right)}{\left(x+1\right)\left(x^2-x+1\right)}\cdot\frac{\left(x+1\right)\left(x+2\right)}{3\left(x^2-x+1\right)}-\frac{2\left(x-1\right)}{x\left(x+2\right)}\)
\(=\frac{\left(x+2\right)^2\left(x+1\right)}{3\left(x^2-x+1\right)^2}-\frac{2\left(x-1\right)}{x\left(x+2\right)}\)
I am➻Minh Ừ nhỉ,mình sai bảo đề sai,vc,éo bt đầu óc thế nào -_-
\(ĐKXĐ:x\ne\pm1;x\ne\pm\sqrt{2};x\ne0\)
\(P=\left(\frac{x^3-1}{x-1}+x\right)\left(\frac{x^3+1}{x+1}-x\right):\frac{x\left(1-x^2\right)^2}{x^2-2}\)
\(=\left[\frac{\left(x-1\right)\left(x^2+x+1\right)}{x-1}+x\right]\left[\frac{\left(x+1\right)\left(x^2-x+1\right)}{x+1}-x\right]:\frac{x\left(1-x^2\right)^2}{x^2-2}\)
\(=\left(x^2+x+1+x\right)\left(x^2-x+1-x\right)\cdot\frac{x^2-2}{x\left(1-x^2\right)^2}\)
\(=\left(x+1\right)^2\left(x-1\right)^2\cdot\frac{x^2-2}{x\left(x-1\right)^2\left(x+1\right)^2}\)
\(=\frac{x^2-2}{x}\)
\(ĐKXĐ:x\ne\pm1;x\ne\pm\sqrt{2};x\ne0\)
\(P=\left(\frac{x^3-1}{x-1}+x\right)\left(\frac{x^3+1}{x+1}-x\right):\frac{x\left(1-x^2\right)^2}{x^2-2}\)
\(=\left[\frac{\left(x-1\right)\left(x^2+x+1\right)}{x-1}+x\right]\left[\frac{\left(x+1\right)\left(x^2-x+1\right)}{x+1}-x\right]:\frac{x\left(1-x^2\right)^2}{x^2-2}\)
\(=\left(\frac{x^2+x+1}{x-1}+x\right)\left(\frac{x^2-x+1}{x+1}-x\right):\frac{x\left(1-x^2\right)^2}{x^2-2}\)
\(=\frac{x^2+x+1+x^2-x}{x-1}\cdot\frac{x^2-x+1-x^2-x}{x+1}:\frac{x\left(1-x^2\right)^2}{x^2-2}\)
\(=\frac{\left(2x^2+1\right)\left(1-2x\right)\left(x^2-2\right)}{\left(x-1\right)\left(x+1\right)x\left(1-x^2\right)^2}\)
lại gặp một con đề sai ?????? Toàn gặp sai đề ??