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LL
1
A
25 tháng 12 2018
\(\left(\frac{1}{x}+1-\frac{3}{x^3+1}-\frac{3}{x^2-x+1}\right)\cdot\frac{3x^2-3x+3}{\left(x+1\right).\left(x+2\right)}-\frac{2x-2}{x^2+2x}\)
\(=\left(\frac{x+1}{x}-\frac{3}{\left(x+1\right).\left(x^2-x+1\right)}+\frac{3.\left(x+1\right)}{\left(x+1\right).\left(x^2-x+1\right)}\right)\cdot\frac{3.\left(x^2-x+1\right)}{\left(x+1\right).\left(x+2\right)}-\frac{2.\left(x-1\right)}{x.\left(x+2\right)}\)
\(=\left[\frac{\left(x+1\right)^2.\left(x^2-x+1\right)-3x+3x^2+3x}{x.\left(x+1\right).\left(x^2-x+1\right)}\right]\cdot\frac{3.\left(x^2-x+1\right)}{\left(x+1\right).\left(x+2\right)}-\frac{2.\left(x-1\right)}{x.\left(x+2\right)}\)
\(=\left[\frac{x^4+x^3+x+1+3x^2}{x.\left(x+1\right).\left(x^2-x+1\right)}\right]\cdot\frac{3.\left(x^2-x+1\right)}{\left(x+1\right).\left(x+2\right)}-\frac{2.\left(x-1\right)}{x.\left(x+2\right)}\)
\(=\frac{3x^4+3x^3+3x+3+9x^2}{x.\left(x+1\right)^2.\left(x+2\right)}-\frac{2.\left(x-1\right)}{x.\left(x+2\right)}=\frac{3x^4+3x^3+3x+3+9x^2}{x.\left(x+1\right)^2.\left(x+2\right)}-\frac{2x^3+2x^2-2x-2}{x.\left(x+1\right)^2.\left(x+2\right)}\)
\(=\frac{3x^4+x^3+7x^2+5x+5}{x.\left(x+1\right)^2.\left(x+2\right)}\)
27 tháng 2 2020
\(\left(\frac{1}{x+1}-\frac{3}{x^3+1}+\frac{3}{x^2-x+1}\right):\frac{3x^2-3x+3}{\left(x+1\right)\left(x+2\right)}-\frac{2x-2}{x^2+2x}\left(x\ne-1;x\ne0;x\ne-2\right)\)
\(=\left(\frac{1}{x+1}-\frac{3}{\left(x+1\right)\left(x^2-x+1\right)}+\frac{3}{x^2-x+1}\right):\frac{3x^3-3x+3}{\left(x+1\right)\left(x+2\right)}-\frac{2\left(x-1\right)}{x\left(x+2\right)}\)
\(=\left(\frac{x^2-x+1}{\left(x+1\right)\left(x^2-x+1\right)}-\frac{3}{\left(x+1\right)\left(x^2-x+1\right)}+\frac{3x+3}{\left(x+1\right)\left(x^2-x+1\right)}\right)\)\(:\frac{3x^2-3x+3}{\left(x+1\right)\left(x+2\right)}-\frac{2\left(x-1\right)}{x\left(x+2\right)}\)
\(=\frac{x^2-x+1-3+3x+3}{\left(x+1\right)\left(x^2-x+1\right)}:\frac{3x^2-3x+3}{\left(x+1\right)\left(x+2\right)}-\frac{2\left(x-1\right)}{x\left(x+2\right)}\)
\(=\frac{x^2+2x+1}{\left(x+1\right)\left(x^2-x+1\right)}:\frac{3\left(x^2-x+1\right)}{\left(x+1\right)\left(x+1\right)}-\frac{2\left(x-1\right)}{x\left(x+2\right)}\)
\(=\frac{\left(x+1\right)\left(x+2\right)}{\left(x+1\right)\left(x^2-x+1\right)}\cdot\frac{\left(x+1\right)\left(x+2\right)}{3\left(x^2-x+1\right)}-\frac{2\left(x-1\right)}{x\left(x+2\right)}\)
\(=\frac{\left(x+2\right)^2\left(x+1\right)}{3\left(x^2-x+1\right)^2}-\frac{2\left(x-1\right)}{x\left(x+2\right)}\)
sắp thi hkì r chỉ jùm vs ik
1 tháng 3 2022
b: \(=\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}\)
\(=\dfrac{\left(x+2\right)\left(x+3\right)+\left(x+1\right)\left(x+3\right)+\left(x+2\right)\left(x+1\right)}{\left(x+2\right)^2\cdot\left(x+1\right)\left(x+3\right)}\)
\(=\dfrac{x^2+5x+6+x^2+4x+3+x^2+3x+2}{\left(x+2\right)^2\cdot\left(x+1\right)\left(x+3\right)}\)
\(=\dfrac{3x^2+12x+11}{\left(x+2\right)^2\cdot\left(x+1\right)\left(x+3\right)}\)
NV
Nguyễn Việt Lâm
Giáo viên
19 tháng 10 2019
a/ Do \(x=0\) không phải nghiệm, pt tương đương:
\(\frac{3}{x+\frac{3}{x}-1}-\frac{2}{x+\frac{3}{x}-3}=-1\)
Đặt \(x+\frac{3}{x}-3=a\) ta được:
\(\frac{3}{a+2}-\frac{2}{a}=-1\)
\(\Leftrightarrow3a-2\left(a+2\right)=-a\left(a+2\right)\)
\(\Leftrightarrow a^2+3a-4=0\Rightarrow\left[{}\begin{matrix}a=1\\a=-4\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x+\frac{3}{x}-3=1\\x+\frac{3}{x}-3=-4\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x^2-4x+3=0\\x^2+x+3=0\end{matrix}\right.\)
b/ Đặt \(x^2+2x+\frac{5}{2}=a>0\)
Phương trình trở thành:
\(\frac{1}{\left(a-\frac{1}{2}\right)^2}+\frac{1}{\left(a+\frac{1}{2}\right)^2}=\frac{5}{4}\)
\(\Leftrightarrow4\left(a+\frac{1}{2}\right)^2+4\left(a-\frac{1}{2}\right)^2=5\left(a^2-\frac{1}{4}\right)^2\)
\(\Leftrightarrow8a^2+2=5\left(a^4-\frac{1}{2}a^2+\frac{1}{16}\right)\)
\(\Leftrightarrow5a^4-\frac{21}{2}a^2-\frac{27}{16}=0\Rightarrow\left[{}\begin{matrix}a^2=\frac{9}{4}\\a^2=-\frac{3}{20}\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x^2+2x+\frac{5}{2}=\frac{3}{2}\\x^2+2x+\frac{5}{2}=-\frac{3}{2}\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
19 tháng 10 2019
c/ ĐKXĐ: \(x\ne\pm1\)
\(\Leftrightarrow\left(\frac{x}{x+1}\right)^2+\left(\frac{x}{x-1}\right)^2+\frac{2x^2}{x^2-1}-\frac{2x^2}{x^2-1}-\frac{10}{9}=0\)
\(\Leftrightarrow\left(\frac{x}{x+1}+\frac{x}{x-1}\right)^2-\frac{2x^2}{x^2-1}-\frac{10}{9}=0\)
\(\Leftrightarrow\left(\frac{2x^2}{x^2-1}\right)^2-\frac{2x^2}{x^2-1}-\frac{10}{9}=0\)
Đặt \(\frac{2x^2}{x^2-1}=a\)
\(\Rightarrow a^2-a-\frac{10}{9}=0\) \(\Rightarrow\left[{}\begin{matrix}a=\frac{5}{3}\\a=-\frac{2}{3}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\frac{2x^2}{x^2-1}=\frac{5}{3}\\\frac{2x^2}{x^2-1}=-\frac{2}{3}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x^2=-5\left(l\right)\\x^2=\frac{1}{4}\end{matrix}\right.\) \(\Rightarrow x=\pm\frac{1}{2}\)
d/ĐKXĐ: ...
\(\Leftrightarrow\left(x^2+\frac{36}{x^2}\right)-13\left(x-\frac{6}{x}\right)=0\)
Đặt \(x-\frac{6}{x}=a\Rightarrow x+\frac{36}{x^2}=a^2+12\)
\(\Rightarrow a^2-13a+12=0\Rightarrow\left[{}\begin{matrix}a=1\\a=12\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x-\frac{6}{x}=1\\x-\frac{6}{x}=12\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x^2-x-6=0\\x^2-12x-6=0\end{matrix}\right.\)
Lời giải:
$\frac{(x-13)^2}{2x^5}.\frac{-3x^2}{x-13}=\frac{-3(x-13)}{2x^3}$