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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
\(\frac{1}{x\left(x-1\right)}+\frac{1}{\left(x-1\right)\left(x-2\right)}+\frac{1}{\left(x-2\right)\left(x-3\right)}+...+\frac{1}{\left(x-4\right)\left(x-5\right)}\)
\(=\frac{1}{x}-\frac{1}{x-1}+\frac{1}{x-1}-\frac{1}{x-2}+...+\frac{1}{x-4}-\frac{1}{x-5}\)
\(=\frac{1}{x}-\frac{1}{x-5}\)
\(=\frac{x-5}{x\left(x-5\right)}-\frac{x}{x\left(x-5\right)}\)
\(=\frac{x-5-x}{x\left(x-5\right)}\)
\(=-\frac{5}{x\left(x-5\right)}\)
11 tháng 3 2020
I am➻Minh Ừ nhỉ,mình sai bảo đề sai,vc,éo bt đầu óc thế nào -_-
\(ĐKXĐ:x\ne\pm1;x\ne\pm\sqrt{2};x\ne0\)
\(P=\left(\frac{x^3-1}{x-1}+x\right)\left(\frac{x^3+1}{x+1}-x\right):\frac{x\left(1-x^2\right)^2}{x^2-2}\)
\(=\left[\frac{\left(x-1\right)\left(x^2+x+1\right)}{x-1}+x\right]\left[\frac{\left(x+1\right)\left(x^2-x+1\right)}{x+1}-x\right]:\frac{x\left(1-x^2\right)^2}{x^2-2}\)
\(=\left(x^2+x+1+x\right)\left(x^2-x+1-x\right)\cdot\frac{x^2-2}{x\left(1-x^2\right)^2}\)
\(=\left(x+1\right)^2\left(x-1\right)^2\cdot\frac{x^2-2}{x\left(x-1\right)^2\left(x+1\right)^2}\)
\(=\frac{x^2-2}{x}\)
11 tháng 3 2020
\(ĐKXĐ:x\ne\pm1;x\ne\pm\sqrt{2};x\ne0\)
\(P=\left(\frac{x^3-1}{x-1}+x\right)\left(\frac{x^3+1}{x+1}-x\right):\frac{x\left(1-x^2\right)^2}{x^2-2}\)
\(=\left[\frac{\left(x-1\right)\left(x^2+x+1\right)}{x-1}+x\right]\left[\frac{\left(x+1\right)\left(x^2-x+1\right)}{x+1}-x\right]:\frac{x\left(1-x^2\right)^2}{x^2-2}\)
\(=\left(\frac{x^2+x+1}{x-1}+x\right)\left(\frac{x^2-x+1}{x+1}-x\right):\frac{x\left(1-x^2\right)^2}{x^2-2}\)
\(=\frac{x^2+x+1+x^2-x}{x-1}\cdot\frac{x^2-x+1-x^2-x}{x+1}:\frac{x\left(1-x^2\right)^2}{x^2-2}\)
\(=\frac{\left(2x^2+1\right)\left(1-2x\right)\left(x^2-2\right)}{\left(x-1\right)\left(x+1\right)x\left(1-x^2\right)^2}\)
lại gặp một con đề sai ?????? Toàn gặp sai đề ??
31 tháng 3 2020
\(\frac{1}{\left(x+1\right)\left(x+2\right)}-\frac{2}{\left(x+2\right)^2}+\frac{1}{\left(x+2\right)\left(x+3\right)}\)
\(=\frac{\left(x+2\right)\left(x+3\right)-2\left(x+1\right)\left(x+3\right)+\left(x+1\right)\left(x+2\right)}{\left(x+1\right)\left(x+2\right)^2\left(x+3\right)}\)
\(=\frac{\left(x+3\right)\left(x+2-2x-2\right)+x^2+2x+x+2}{\left(x+1\right)\left(x+2\right)^2\left(x+3\right)}\)
\(=\frac{\left(x+3\right)\left(-x\right)+x^2+3x+2}{\left(x+1\right)\left(x+2\right)^2\left(x+3\right)}\)
\(=\frac{-x^2-3x+x^2+3x+2}{\left(x+1\right)\left(x+2\right)^2\left(x+3\right)}=\frac{2}{\left(x+1\right)\left(x+2\right)^2\left(x+3\right)}\)
1 tháng 4 2020
ĐKXD: x\(\ne\)-1,-2,-3
Ta có
\(\frac{1}{\left(x+1\right)\left(x+2\right)}\)-\(\frac{2}{\left(x+2\right)^2}\)+\(\frac{1}{\left(x+2\right)\left(x+3\right)}\)
=\(\frac{\left(x+2\right)\left(x+3\right)-2\left(x+1\right)\left(x+3\right)+\left(x+1\right)\left(x+2\right)}{\left(x+1\right)\left(x+2\right)^2\left(x+3\right)}\)
=\(\frac{\left(x+2\right)\left(x+3+x+1\right)-2\left(x^2+4x+3\right)}{\left(x+1\right)\left(x+2\right)^2\left(x+3\right)}\)
=\(\frac{\left(x+2\right)\left(2x+4\right)-2x^2-8x-6}{\left(x+1\right)\left(x+2\right)^2\left(x+3\right)}\)
=\(\frac{2x^2+8x+8-2x^2-8x-6}{\left(x+1\right)\left(x+2\right)^2\left(x+3\right)}\)
=\(\frac{2}{\left(x+1\right)\left(x+2\right)^2\left(x+3\right)}\)
Chúc bạn học tốt
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)
2 tháng 10 2020
MTC: (x+y)(x+1)(1-y)
\(=\frac{x^2\left(1+x\right)-y^2\left(1-y\right)-x^2y^2\left(x+y\right)}{\left(x+y\right)\left(1+x\right)\left(1-y\right)}=\frac{\left(x+y\right)\left(1+x\right)\left(1-y\right)\left(x-y+xy\right)}{\left(x+y\right)\left(1+x\right)\left(1-y\right)}\)
\(=x-y+xy\)
Với \(x\ne-1;x\ne-y;y\ne1\)thì giá trị biểu thức được xác định
26 tháng 3 2018
\(A=\left(\frac{x+1}{x}\right)^2:\left[\frac{x^2+1}{x^2}+\frac{2}{x+1}\cdot\frac{x+1}{x}\right]\)
\(A=\left(\frac{x+1}{x}\right)^2:\left[\frac{x^2+1}{x^2}+\frac{2}{x}\right]\)
\(A=\left(\frac{x+1}{x}\right)^2:\left(\frac{x^2+1+2x}{x^2}\right)\)
\(A=\left(\frac{x+1}{x}\right)^2:\left(\frac{x+1}{x}\right)^2=1\)