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2 tháng 10 2020
\(\left(\frac{x^2-1}{x^4-x^2+1}-\frac{1}{x^2+1}\right)\times\left(x^4+\frac{1-x^4}{1+x^2}\right)\)
ĐK : ...
\(=\left(\frac{\left(x^2-1\right)\left(x^2+1\right)}{\left(x^4-x^2+1\right)\left(x^2+1\right)}-\frac{x^4-x^2+1}{\left(x^4-x^2+1\right)\left(x^2+1\right)}\right)\times\left(x^4+\frac{\left(1-x^2\right)\left(1+x^2\right)}{\left(1+x^2\right)}\right)\)
\(=\left(\frac{x^4-1}{\left(x^4-x^2+1\right)\left(x^2+1\right)}-\frac{x^4-x^2+1}{\left(x^4-x^2+1\right)\left(x^2+1\right)}\right)\times\left(x^4+1-x^2\right)\)
\(=\left(\frac{x^4-1-x^4+x^2-1}{\left(x^4-x^2+1\right)\left(x^2+1\right)}\right)\times\left(x^4-x^2+1\right)\)
\(=\frac{x^2-2\left(x^4-x^2+1\right)}{\left(x^4-x^2+1\right)\left(x^2+1\right)}\)
\(=\frac{x^2-2}{x^2+1}\)
2 tháng 10 2020
Mình sửa dòng 5 một chút nhé
\(=\frac{\left(x^2-2\right)\left(x^4-x^2+1\right)}{\left(x^4-x^2+1\right)\left(x^2+1\right)}\)( như kia dễ bị nhầm )
26 tháng 3 2020
\(\frac{1}{1-x}+\frac{1}{1+x}+\frac{2}{1+x^2}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)
\(=\frac{1+x+1-x}{\left(1+x\right)\left(1-x\right)}+\frac{2}{1+x^2}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)
\(=\frac{2}{1-x^2}+\frac{2}{1+x^2}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)
\(=\frac{2+2x^2+2-2x^2}{\left(1-x^2\right)\left(1+x^2\right)}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)
\(=\frac{4}{1-x^4}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)
\(=\frac{4+4x^4+4-4x^4}{\left(1-x^4\right)\left(1+x^4\right)}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)
\(=\frac{8}{1-x^8}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)
\(=\frac{8+8x^8+8-8x^8}{\left(1-x^8\right)\left(1+x^8\right)}+\frac{16}{1+x^{16}}\)
\(=\frac{16}{1-x^{16}}+\frac{16}{1+x^{16}}\)
\(=\frac{16+16x^{16}+16-16x^{16}}{\left(1-x^{16}\right)\left(1+x^{16}\right)}\)
\(=\frac{32}{1-x^{32}}\)
BT
9 tháng 11 2017
\(A=\left(\frac{x^2-1}{x^4-x^2+1}-\frac{1}{x^2+1}\right).\left(x^4+\frac{1-x^4}{1+x^2}\right)\)
\(=\left(\frac{\left(x^2-1\right)\left(x^2+1\right)-\left(x^4-x^2+1\right)}{\left(x^4-x^2+1\right)\left(x^2+1\right)}\right).\left(x^4+\frac{\left(1+x^2\right)\left(1-x^2\right)}{1+x^2}\right)\)
\(=\frac{x^4-1-x^4+x^2-1}{\left(x^2+1\right)\left(x^4-x^2+1\right)}\left(x^4+1-x^2\right)\)
\(=\frac{x^2-2}{x^2+1}\).
11 tháng 1 2022
\(A=\left(\dfrac{1}{x^2-4x}+\dfrac{2}{16-x^2}+\dfrac{4}{4x+16}\right):\dfrac{1}{4x}\left(x\ne4;x\ne-4;x\ne0\right).\)
\(A=\left(\dfrac{1}{x\left(x-4\right)}+\dfrac{-2}{\left(x+4\right)\left(x-4\right)}+\dfrac{1}{x+4}\right).4x\).
\(A=\dfrac{x+4-2x+x^2-4x}{x\left(x-4\right)\left(x+4\right)}.4x.\)
\(A=\dfrac{x^2-5x+4}{\left(x-4\right)\left(x+4\right)}.4.\)
\(A=\dfrac{\left(x-4\right)\left(x-1\right)}{\left(x-4\right)\left(x+4\right)}.4.\)
\(A=\dfrac{4\left(x-1\right)}{x+4}.\)
13 tháng 3 2020
\(A=\left(\frac{2}{x+2}-\frac{4}{x^2+4x+4}\right):\left(\frac{2}{x^2-4}+\frac{1}{2-x}\right)\)
\(ĐKXĐ:x\ne\pm2\)
\(A=\left(\frac{2\left(x+2\right)}{\left(x+2\right)^2}-\frac{4}{\left(x+2\right)^2}\right):\left(\frac{2}{\left(x-2\right)\left(x+2\right)}-\frac{x+2}{\left(x-2\right)\left(x+2\right)}\right)\)
\(=\frac{2x+4-4}{\left(x+2\right)^2}:\frac{2-x-2}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{2x}{\left(x+2\right)^2}\frac{\left(x-2\right)\left(x+2\right)}{-x}\)
\(=\frac{-2\left(x-2\right)}{x+2}=\frac{4-2x}{x+2}\)
HH
13 tháng 3 2020
\(ĐKXĐ:x\ne\pm2;x\ne0\)
\(A=\left(\frac{2}{2+x}-\frac{4}{x^2+4x +4}\right):\left(\frac{2}{x^2-4}+\frac{1}{2-x}\right)\)
\(A=\left(\frac{2\left(x+2\right)}{\left(x+2\right)^2}-\frac{4}{\left(x+2\right)^2}\right):\left(\frac{2}{\left(x-2\right)\left(x+2\right)}-\frac{x+2}{\left(x-2\right)\left(x+2\right)}\right)\)
\(A=\frac{2x+4-4}{\left(x+2\right)^2}:\frac{2-x-2}{\left(x-2\right)\left(x+2\right)}\)
\(A=\frac{2x}{\left(x+2\right)^2}.\frac{\left(x-2\right)\left(x+2\right)}{-x}\)
\(A=\frac{4-2x}{x+2}\)
ND
1
1 tháng 12 2018
ĐK: \(\hept{\begin{cases}x^2+4x+4\ne0\\4-x^2\ne0\end{cases}}\Rightarrow\hept{\begin{cases}\left(x+2\right)^2\ne0\\\left(2-x\right)\left(2+x\right)\ne0\end{cases}\Rightarrow\hept{\begin{cases}x\ne-2\\x\ne2\end{cases}}}\)
\(P=\frac{x^3-4x}{x^2+4}.\left(\frac{1}{x^2+4x+4}+\frac{1}{4-x^2}\right)\)
\(=\frac{x\left(x^2-4\right)}{x^2+4}.\left(\frac{1}{\left(x+2\right)^2}+\frac{-1}{\left(x-2\right)\left(x+2\right)}\right)\)
\(=\frac{x\left(x^2-4\right)}{x^2+4}.\left(\frac{x-2-\left(x+2\right)}{\left(x+2\right)^2\left(x-2\right)}\right)\)
\(=\frac{x\left(x-2\right)\left(x+2\right)}{x^2+4}.\frac{-4}{\left(x+2\right)^2\left(x-2\right)}=\frac{-4x}{\left(x^2+4\right)\left(x+2\right)}\)
30 tháng 1 2022
\(A=\left(\dfrac{1}{x-2}+\dfrac{2x}{\left(x-2\right)\left(x+2\right)}+\dfrac{1}{x+2}\right)\cdot\dfrac{2-x}{x}\)
\(=\dfrac{x+2+2x+x-2}{-\left(2-x\right)\left(x+2\right)}\cdot\dfrac{2-x}{x}\)
\(=\dfrac{4x}{-\left(x+2\right)\cdot x}=\dfrac{-4}{x+2}\)
=\(\left(1+\frac{4}{x-2}\right):\left(\frac{x^2-4}{2}\right)\)
=\(\left(\frac{x-2}{x-2}+\frac{4}{x-2}\right):\left(\frac{\left(x-2\right)\left(x+2\right)}{2}\right)\)
=\(\frac{x-2+4}{x-2}\cdot\frac{2}{\left(x-2\right)\left(x+2\right)}\)
=\(\frac{x+2}{x-2}.\frac{2}{\left(x-2\right).\left(x+2\right)}\)
=\(\frac{2}{\left(x-2\right)^2}\)
Mình quên ĐKXĐ
Sorry bạn nha