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TN
13 tháng 3 2020
\(a.\frac{x-6}{x-4}=\frac{x}{x-2}\\\Leftrightarrow \frac{\left(x-6\right)\left(x-2\right)}{\left(x-4\right)\left(x-2\right)}=\frac{x\left(x-4\right)}{\left(x-4\right)\left(x-2\right)}\\\Leftrightarrow \left(x-6\right)\left(x-2\right)=x\left(x-4\right)\\\Leftrightarrow \left(x-6\right)\left(x-2\right)-x\left(x-4\right)=0\\ \Leftrightarrow x^2-2x-6x+12-x^2+4x=0\\\Leftrightarrow -4x+12=0\\\Leftrightarrow -4x=-12\\ \Leftrightarrow x=3\)
\(b.1+\frac{2x-5}{x-2}-\frac{3x-5}{x-1}=0\\ \Leftrightarrow\frac{\left(x-2\right)\left(x-1\right)}{\left(x-2\right)\left(x-1\right)}+\frac{\left(2x-5\right)\left(x-1\right)}{\left(x-2\right)\left(x-1\right)}-\frac{\left(3x-5\right)\left(x-2\right)}{\left(x-2\right)\left(x-1\right)}=0\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)+\left(2x-5\right)\left(x-1\right)-\left(3x-5\right)\left(x-2\right)=0\\ \Leftrightarrow x^2-x-2x+3+2x^2-2x-5x+5-3x^2+6x+5x-10=0\\ \Leftrightarrow x-2=0\\ \Leftrightarrow x=2\\ \)
AO
21 tháng 1 2018
\(\frac{x+1}{x-1}-\frac{x-1}{x+1}=\frac{16}{x^2-1}\)
\(\frac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}-\frac{\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}=\frac{16}{x^2-1}\)
\(\Rightarrow\left(x+1\right)^2-\left(x-1\right)^2=16\)
\(\Rightarrow\left(x+1-x+1\right)\left(x+1+x-1\right)=16\)
\(\Rightarrow2\left(2x\right)=16\)
\(\Rightarrow4x=16\)
\(\Rightarrow x=4\)
vậy \(x=4\)
\(\frac{6x+1}{x^2-7x+10}+\frac{5}{x-2}=\frac{3}{x-5}\)
\(\frac{6x+1}{\left(x-2\right)\left(x-5\right)}+\frac{5\left(x-5\right)}{\left(x-2\right)\left(x-5\right)}=\frac{3\left(x-2\right)}{\left(x-2\right)\left(x-5\right)}\)
\(\Rightarrow6x+1+5x-5=3x-6\)
\(\Rightarrow11x-3x=-6+4\)
\(\Rightarrow8x=-2\)
\(\Rightarrow x=\frac{-1}{4}\)
3) \(\frac{1}{x-1}+\frac{2x^2-5}{x^3-1}=\frac{4}{x^2+x+1}\)
\(\frac{x^2+x+1}{x^3-1}+\frac{\left(2x^2-5\right)}{x^3-1}=\frac{4\left(x-1\right)}{x^3-1}\)
\(\Rightarrow x^2+x+1+2x^2-5=4x-4\)
\(\Rightarrow3x^2-3x=-4+4\)
\(\Rightarrow3x\left(x-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3x=0\\x-1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)
16 tháng 4 2020
a, \(\frac{6x+1}{x^2+7x+10}+\frac{5}{x-2}=\frac{3}{x-5}\)
\(11x^3-31x^2-72x-240=3\left(x+2\right)\left(x+5\right)\left(x-2\right)\)
\(11x^3-31x^2-72x-240-3\left(x+2\right)\left(x+5\right)\left(x-2\right)=0\)
\(8x^3-46x^2-60x-180=0\)
=> vô nghiệm
16 tháng 4 2020
b) \(\frac{2}{x^2-4}-\frac{x-1}{x\left(x-2\right)}+\frac{x-4}{x\left(x+2\right)}=0\left(x\ne0;x\ne\pm2\right)\)
\(\Leftrightarrow\frac{2x}{\left(x-2\right)\left(x+2\right)x}-\frac{\left(x+2\right)\left(x-1\right)}{x\left(x-2\right)\left(x+2\right)}+\frac{\left(x+4\right)\left(x-2\right)}{x\left(x-2\right)\left(x+2\right)}=0\)
\(\Leftrightarrow\frac{2x}{x\left(x-2\right)\left(x+2\right)}-\frac{x^2+x-2}{x\left(x-2\right)\left(x+2\right)}+\frac{x^2+2x-8}{x\left(x-2\right)\left(x+2\right)}=0\)
\(\Leftrightarrow\frac{2x-x^2-x+2+x^2+2x-8}{x\left(x-2\right)\left(x+2\right)}=0\)
\(\Leftrightarrow\frac{3x-6}{x\left(x-2\right)\left(x+2\right)}=0\)
=> 3x-6=0
<=> x=2 (ktm)
Vậy pt vô nghiệm
21 tháng 1 2017
2. \(\frac{1}{x-1}-\frac{7}{x-2}=\frac{1}{\left(x-1\right)\left(2-x\right)}\) (ĐKXĐ:\(x\ne1,x\ne2\))
\(\Leftrightarrow\frac{1}{x-1}+\frac{7}{2-x}=\frac{1}{\left(x-1\right)\left(2-x\right)}\)
\(\Leftrightarrow\frac{2-x+7\left(x-1\right)}{\left(x-1\right)\left(2-x\right)}=\frac{1}{\left(x-1\right)\left(2-x\right)}\)
\(\Rightarrow2-x+7\left(x-1\right)=1\)
\(\Leftrightarrow2-x+7x-7=1\)
\(\Leftrightarrow-x+7x=1-2+7\)
\(\Leftrightarrow6x=6\)
\(\Leftrightarrow x=1\) (Không thỏa mãn ĐKXĐ)
Vậy phương trình trên vô nghiệm
22 tháng 1 2017
ko phan tich duoc nha bn
chuc bn hoc gioi
happy new year
20 tháng 2 2020
a) \(\frac{4x-8}{2x^2+1}=0\)
\(\Rightarrow4x-8=0\left(2x^2+1\ne0\right)\)
\(\Leftrightarrow4x=8\)
\(\Leftrightarrow x=2\)
Vậy x=2
b)
\(\frac{x^2-x-6}{x-3}=0\)
\(\Leftrightarrow\frac{\left(x-3\right)\left(x+2\right)}{x-3}=0\)
\(\Rightarrow x+2=0\)
\(\Leftrightarrow x=-2\)
Vậy x=-2
12 tháng 2 2020
a) ĐKXĐ: x≠-5
Ta có: \(\frac{2x-5}{x+5}=3\)
\(\Leftrightarrow\frac{2x-5}{x+5}-3=0\)
\(\Leftrightarrow\frac{2x-5}{x+5}-\frac{3\left(x+5\right)}{x+5}=0\)
\(\Leftrightarrow2x-5-3\left(x+5\right)=0\)
\(\Leftrightarrow2x-5-3x-15=0\)
\(\Leftrightarrow-x-20=0\)
\(\Leftrightarrow-x=20\)
\(\Leftrightarrow x=-20\)(tmđk)
Vậy: x=-20
b) ĐKXĐ: x≠1;x≠-1
Ta có: \(\frac{2}{x-1}=\frac{6}{x+1}\)
\(\Leftrightarrow\frac{2}{x-1}-\frac{6}{x+1}=0\)
\(\Leftrightarrow\frac{2\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}-\frac{6\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}=0\)
\(\Leftrightarrow2\left(x+1\right)-6\left(x-1\right)=0\)
\(\Leftrightarrow2x+2-6x+6=0\)
\(\Leftrightarrow-4x+8=0\)
\(\Leftrightarrow-4x=-8\)
\(\Leftrightarrow x=2\)(tmđk)
Vậy: x=2
c) ĐKXĐ: x≠1;x≠-1
Ta có: \(\frac{2x+1}{x-1}=\frac{5\left(x-1\right)}{x+1}\)
\(\Leftrightarrow\frac{2x+1}{x-1}-\frac{5\left(x-1\right)}{x+1}=0\)
\(\Leftrightarrow\frac{\left(2x+1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}-\frac{5\left(x-1\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}=0\)
\(\Leftrightarrow\left(2x+1\right)\left(x+1\right)-5\left(x^2-2x+1\right)=0\)
\(\Leftrightarrow2x^2+2x+x+1-5x^2+10x-5=0\)
\(\Leftrightarrow-3x^2+13x-4=0\)
\(\Leftrightarrow-3x^2+x+12x-4=0\)
\(\Leftrightarrow x\left(-3x+1\right)+4\left(3x-1\right)=0\)
\(\Leftrightarrow x\left(1-3x\right)-4\left(1-3x\right)=0\)
\(\Leftrightarrow\left(1-3x\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}1-3x=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=1\\x=4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{3}\\x=4\end{matrix}\right.\)(thỏa mãn điều kiện)
Vậy: \(x\in\left\{\frac{1}{3};4\right\}\)
d) ĐKXĐ: x≠1;x≠-1
Ta có: \(\frac{x}{x-1}-\frac{2x}{x^2-1}=0\)
\(\Leftrightarrow\frac{x\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}-\frac{2x}{\left(x-1\right)\left(x+1\right)}=0\)
\(\Leftrightarrow x\left(x+1\right)-2x=0\)
\(\Leftrightarrow x^2+x-2x=0\)
\(\Leftrightarrow x^2-x=0\)
\(\Leftrightarrow x\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=1\left(ktm\right)\end{matrix}\right.\)
Vậy: x=0
e) ĐKXĐ: x≠2
Ta có: \(\frac{1}{x-2}+3=\frac{x-3}{2-x}\)
⇔\(\frac{1}{x-2}+3-\frac{x-3}{2-x}=0\)
⇔\(\frac{1}{x-2}+3+\frac{x-3}{x-2}=0\)
⇔\(\frac{1}{x-2}+\frac{3\left(x-2\right)}{x-2}+\frac{x-3}{x-2}=0\)
\(\Leftrightarrow1+3\left(x-2\right)+x-3=0\)
\(\Leftrightarrow1+3x-6+x-3=0\)
\(\Leftrightarrow4x-8=0\)
\(\Leftrightarrow4x=8\)
\(\Leftrightarrow x=2\)(không thỏa mãn)
Vậy: x∈∅
f) ĐKXĐ: \(x\ne\pm2\)
Ta có: \(\frac{x+1}{x-2}+\frac{x-1}{x+2}=\frac{2\left(x^2+2\right)}{x^2-4}\)
⇔\(\frac{x+1}{x-2}+\frac{x-1}{x+2}-\frac{2\left(x^2+2\right)}{x^2-4}=0\)
⇔\(\frac{\left(x+1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{\left(x-1\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}-\frac{2x^2+4}{\left(x+2\right)\left(x-2\right)}=0\)
\(\Leftrightarrow\left(x+1\right)\left(x+2\right)+\left(x-1\right)\left(x-2\right)-2x^2-4=0\)
\(\Leftrightarrow x^2+2x+x+2+x^2-2x-x+2-2x^2-4=0\)
\(\Leftrightarrow0=0\)
Vậy: x∈R
g) ĐKXĐ: \(x\ne\pm2\)
Ta có: \(\frac{x+2}{x-2}+\frac{1}{x+2}=\frac{x\left(x-5\right)}{x^2-4}\)
⇔\(\frac{x+2}{x-2}+\frac{1}{x+2}-\frac{x\left(x-5\right)}{\left(x-2\right)\left(x+2\right)}=0\)
⇔\(\frac{\left(x+2\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{x-2}{\left(x+2\right)\left(x-2\right)}-\frac{x^2-5x}{\left(x-2\right)\left(x+2\right)}=0\)
⇔\(\left(x+2\right)^2+x-2-x^2+5x=0\)
\(\Leftrightarrow x^2+4x+4+x-2-x^2+5x=0\)
\(\Leftrightarrow10x-2=0\)
\(\Leftrightarrow10x=2\)
\(\Leftrightarrow x=\frac{2}{10}=\frac{1}{5}\)(thỏa mãn)
Vậy: \(x=\frac{1}{5}\)
\(x-\frac{1}{x}=5\Rightarrow\left(x-\frac{1}{x}\right)^2=5^2\)
\(x^2-2+\frac{1}{x^2}=25\)
\(x^2+\frac{1}{x^2}=25+2\)
\(x^2+\frac{1}{x^2}=27\)
vậy A = 27