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\(\frac{3}{x-5}=-\frac{4}{x+2}\)
\(\Leftrightarrow3\left(x+2\right)=-4\left(x-5\right)\)
\(\Leftrightarrow3x+6=-4x+20\)
\(\Leftrightarrow7x=14\)
\(\Leftrightarrow x=2\)
\(\frac{x}{-2}=-\frac{8}{x}\)
\(\Leftrightarrow x^2=16\)
\(\Leftrightarrow x=\pm4\)
\(-\frac{2}{x}=\frac{y}{3}\)
\(\Leftrightarrow xy=-6\)
\(\Leftrightarrow x;y\inƯ\left(-6\right)=\left\{\pm1;\pm2;\pm3;\pm6\right\}\)
Xét bảng
| x | 1 | -1 | 2 | -2 | 3 | -3 | 6 | -1 |
| y | -6 | 6 | -3 | 3 | -2 | 2 | -1 | 6 |
Vậy.................
\(\frac{2x-9}{240}=\frac{39}{80}\)
\(\Leftrightarrow2x-9=\frac{240.39}{80}\)
\(\Leftrightarrow2x-9=117\)
\(\Leftrightarrow2x=126\)
\(\Leftrightarrow x=63\)
1)=>3(x-5)(2x+9)+3(x-5)=0=>(x-5)(6x+30)
=>x-5=0=>x=5
6x+30=0=>x=-5
2)=>x^2-16=0=>x=+-4
12-4x=0=>x=3
3)=>9-x^2=0=>x=+-3
4x-8=0=>x=2
4)=>8-x^3=0=>x=3
5^x-125=0=>x=2
5)=>2^x.2^x=8=>2^2x=8=>2x=3=>x=1,5
g) \(\left(x+\frac{1}{2}\right)\left(\frac{2}{3}-2x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+\frac{1}{2}=0\\\frac{2}{3}-2x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{-1}{2}\\x=\frac{1}{3}\end{cases}}\)
Vây \(x\in\left\{\frac{-1}{2};\frac{1}{3}\right\}\)
\(a,x=3x^2\Rightarrow x-3x^2=0\Rightarrow x\left(1-3x\right)=0\Rightarrow\orbr{\begin{cases}x=0\\1-3x=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{1}{3}\end{cases}}\)
\(b,\left(2x-6\right)\left(x+4\right)+2\left(2x-6\right)=0\)
\(\Rightarrow\left(2x-6\right)\left(x+4+2\right)=0\)
\(\Rightarrow\left(2x-6\right)\left(x+6\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x-6=0\\x+6=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=3\\x=-6\end{cases}}\)
\(c,\left(2x-5\right)\left(x+9\right)+6x-15=0\)
\(\Rightarrow\left(2x-5\right)\left(x+9\right)+3\left(2x-5\right)=0\)
\(\Rightarrow\left(2x-5\right)\left(x+9+3\right)=0\)
\(\Rightarrow\left(2x-5\right)\left(x+12\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x-5=0\\x+12=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{5}{2}\\x=-12\end{cases}}\)
a) \(15+2\left|x\right|=-3\\ \\ < =>2\left|x\right|=15-\left(-3\right)\\ < =>2\left|x\right|=18\\ =>\left|x\right|=\frac{18}{2}=9\\ =>x=9hoặcx=-9\)
b) \(\left|x-2\right|=7\\ < =>x-2=7hoặcx-2=-7\\ =>x=9hoặcx=-5\)
c) \(100-4.x^2=224\\ < =>4.x^2=100-224=-124\\ < =>x^2=-\frac{124}{4}=-31\\ Mà:x^2\ge0\\ =>xkhôngcógiátrịnàothoảmãn\)
d)\(2x-\frac{9}{240}=\frac{39}{80}\\ < =>2x-\frac{3}{80}=\frac{39}{80}\\ =>2x=\frac{39}{80}+\frac{3}{80}=\frac{21}{40}\\ =>x=\frac{\frac{21}{40}}{2}=\frac{21}{80}\)
\(15+2\left|x\right|=-3\)
\(\Rightarrow2\left|x\right|=15+3\)
\(\Rightarrow2\left|x\right|=18\)
\(\Rightarrow\left|x\right|=\frac{18}{2}\)
\(\Rightarrow\left|x\right|=9\)
\(\Rightarrow\left[\begin{matrix}x=9\\x=-9\end{matrix}\right.\)
Vậy, x = 9 hoặc x = -9.