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a)\(\Leftrightarrow-\frac{x}{x+1}+\frac{1}{x+1}+\frac{x}{x-1}+\frac{1}{x-1}=-\frac{3x^2}{x+1}+\frac{3x}{x+1}+3x\)
\(\Rightarrow\frac{3x^2}{x+1}-\frac{4x}{x+1}+\frac{1}{x+1}+\frac{x}{x-1}-3x+\frac{1}{x-1}=0\)
\(\Leftrightarrow-\frac{2x\left(3x-5\right)}{\left(x-1\right)\left(x+1\right)}=0\)
\(\Rightarrow\int^{\frac{x-1}{1}=0}_{\frac{x+1}{1}=0}\Rightarrow x=0\)
=>3x=5
\(\Rightarrow x=\frac{3}{5}\)
vậy \(x=\frac{3}{5}\) hoặc 0
b)x = -(20309916*i+23555105)/9277755;
x = -(985155752*i-35635815)/916564140;
x = (985155752*i+35635815)/916564140;
x = (20309916*i-23555105)/9277755;
c)\(\Leftrightarrow\frac{x+2}{x-1}=\frac{1}{1}\Rightarrow\left(x+2\right)1=\left(x-1\right)1\)
vì \(\left(x+2\right)1\ne\left(x-1\right)1\)
=>x vô nghiệm hoặc đề sai
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\(\left(\frac{1}{x-1}+\frac{1}{x-4}\right)-\left(\frac{1}{x-2}+\frac{1}{x-3}\right)=0\)
\(\Leftrightarrow\frac{x-4+x-1}{\left(x-1\right).\left(x-4\right)}-\frac{x-3-x-2}{\left(x-2\right).\left(x-3\right)}=0\)
\(\Leftrightarrow\frac{2x-5}{x^2-5x+4}-\frac{2x-5}{x^2-5x+6}=0\)
\(\Leftrightarrow\left(2x-5\right).\left(\frac{1}{x^2-5x+4}-\frac{1}{x^2-5x+6}=0\right)\)
\(\Leftrightarrow\orbr{\begin{cases}2x-5=0\\\frac{1}{x^2-5x+4}-\frac{1}{x^2-5x+6}=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{2}\\x^2-5x+4=x^2-5x+6\left(loai\right)\end{cases}}}\)
Vậy..
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a,\(\frac{2x-5}{3}-\frac{3x-1}{2}< \frac{3-x}{5}-\frac{2x-1}{4}\)
\(\Leftrightarrow\frac{\left(2x-5\right)20}{60}-\frac{\left(3x-1\right)30}{60}< \frac{\left(3-x\right)12}{60}-\frac{\left(2x-1\right)15}{60}\)
\(\Leftrightarrow40x-100-90x+30< 36-12x-30x+15\)
\(\Leftrightarrow40x-90x+12x+30x< 36+15+100-30\)
\(\Leftrightarrow-8x< 121\)
\(\Leftrightarrow x>-\frac{378}{25}\)
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\(a,\frac{1}{2x-3}-\frac{3}{x\left(2x-3\right)}=\frac{5}{x}\) ĐKXĐ : \(x\ne0;x\ne\frac{3}{2}\)
\(\Leftrightarrow\frac{x}{x\left(2x-3\right)}-\frac{3}{x\left(2x-3\right)}=\frac{5\left(2x-3\right)}{x\left(2x-3\right)}\)
\(\Leftrightarrow x-3=10x-15\)
\(\Leftrightarrow x-10x=3-15\)
\(\Leftrightarrow-9x=-12\)
\(\Leftrightarrow x=\frac{-12}{-9}=\frac{4}{3}\)(TMĐKXĐ)
KL :....
\(b,\frac{x+2}{x-2}-\frac{1}{x}=\frac{2}{x\left(x-2\right)}\) ĐKXĐ : \(x\ne0;2\)
\(\Leftrightarrow\frac{x\left(x+2\right)}{x\left(x-2\right)}-\frac{x-2}{x\left(x-2\right)}=\frac{2}{x\left(x-2\right)}\)
\(\Leftrightarrow x^2+2x-x+2=2\)
\(\Leftrightarrow x^2+x=2-2\)
\(\Leftrightarrow x\left(x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=-1\end{cases}}\)
KL ::
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điều kiền x # 0
đặt \(t=x+\frac{1}{x};đk:t\ge2\)=>\(x^2+\frac{1}{x^2}=t^2-2\)
Ta được phương trình mới ẩn t : \(t^2-2t-5=0\)
tự giải phương trình nhé. lấy nghiệm t>= 2
ĐK: x khác -1 và x khác 1.
\(PT\Leftrightarrow\frac{7x.\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}-\frac{5x.\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}+\frac{x+21}{\left(x-1\right)\left(x+1\right)}=0\)
<=> 7x2 + 7x - 5x2 + 5x + x + 21 = 0
<=> 2x2 + 13x + 21 = 0
<=> 2x2 + 6x + 7x + 21 = 0
<=> 2x.(x + 3) + 7.(x + 3) = 0
<=> (x + 3).(2x + 7) = 0
<=> x + 3 = 0 hoặc 2x + 7 = 0
<=> x = -3 hoặc x = -7/2
Vậy S = {-7/2; -3}.