giải phương trình
1/x +1/x+6 = 1/4
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![](https://rs.olm.vn/images/avt/0.png?1311)
\(1,\Leftrightarrow7-2x-4=-x-4\)
\(\Leftrightarrow x-2x=-4-7+4\)
\(\Leftrightarrow-x=-7\)
\(\Leftrightarrow x=7\)
Vậy \(S=\left\{7\right\}\)
\(2,\Leftrightarrow x-1-2x+1=9-x\)
\(\Leftrightarrow x+x-2x=9-1+1\)
\(\Leftrightarrow0x=9\)
\(\Rightarrow x\in\varnothing\)
Vậy \(S=\left\{\varnothing\right\}\)
\(3,\Leftrightarrow2x^2+3x-2x+3=2x^2+10x-x-5\)
\(\Leftrightarrow2x^2-2x^2+3x-2x-10x+x=-5-3\)
\(-8x=-8\)
\(\Rightarrow x=1\)
Vậy \(S=\left\{1\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
1) \(\left(x-2\right)\left(3+2x\right)-2x\left(x+5\right)=6\)
\(3x+2x^2-6-4x-2x^2-10x-6=0\)
\(-11x=12\)
\(x=-\dfrac{12}{11}\)
2) \(x^2-4-\left(x-5\right)\left(x-2\right)=0\)
\(\left(x-2\right)\left(x+2\right)-\left(x-5\right)\left(x-2\right)=0\)
\(\left(x-2\right)\left(x+2-x+5\right)=0\)
\(7\left(x-2\right)=0\)
\(\Leftrightarrow x=2\)
1, \(3x+2x^2-6-4x-2x^2-10x=0\Leftrightarrow-11x-6=0\Leftrightarrow x=-\dfrac{6}{11}\)
2, \(\left(x-2\right)\left(x+2\right)-\left(x-5\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2-x+5\right)=0\Leftrightarrow x=2\)
3, bạn xem lại đề
5, đk x khác -4 ; 4
\(96=\left(2x-1\right)\left(x-4\right)+\left(3x-1\right)\left(x+4\right)-6\left(x^2-16\right)\)
\(\Leftrightarrow96=2x^2-9x+4+3x^2+11x-4-6x^2+96\)
\(\Leftrightarrow-x^2+2x=0\Leftrightarrow-x\left(x-2\right)=0\Leftrightarrow x=0;x=2\)(tm)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(1,2\left(x-3\right)+1=2\left(x+1\right)-9\\ \Rightarrow2x-6+1=2x+2-9\\ \Rightarrow2x-5=2x-7\\ \Rightarrow-2=0\left(vô.lí\right)\)
\(2,\dfrac{5-x}{2}=\dfrac{3x-4}{6}\\ \Rightarrow30-6x=6x-8\\ \Rightarrow12x=38\\ \Rightarrow x=\dfrac{19}{6}\)
\(3,\left(x-1\right)^2+\left(x+2\right)\left(x-2\right)=\left(2x+1\right)\left(x-3\right)\\ \Rightarrow x^2-2x+1+x^2-4=2x^2-6x+x-3\\ \Rightarrow2x^2-2x-3=2x^2-5x-3\\ \Rightarrow3x=0\\ \Rightarrow x=0\)
\(4,\left(x+5\right)\left(x-1\right)-\left(x+1\right)\left(x+2\right)=1\\ \Rightarrow x^2+5x-x-5-x^2-2x-x-2=1\\ \\ \Rightarrow x-7=1\\ \Rightarrow x=8\)
\(5,\dfrac{6x-1}{15}-\dfrac{x}{5}=\dfrac{2x}{3}\\ \Rightarrow\dfrac{6x-1}{15}-\dfrac{3x}{15}=\dfrac{10x}{15}\\ \Rightarrow6x-1-3x=10x\\ \Rightarrow3x-1=10x\\ \Rightarrow7x=-1\\ \Rightarrow x=\dfrac{-1}{7}\)
\(6,\dfrac{5\left(x-2\right)}{2}-\dfrac{x+5}{3}=1-\dfrac{4\left(x-3\right)}{5}\\ \Rightarrow\dfrac{75\left(x-2\right)}{30}-\dfrac{10\left(x+5\right)}{30}=\dfrac{30}{30}-\dfrac{24\left(x-3\right)}{30}\\ \Rightarrow75\left(x-2\right)-10\left(x+5\right)=30-24\left(x-3\right)\\ \Rightarrow75x-150-10x-50=30-24x+72\\ \Rightarrow65x-200=102-24x\\ \Rightarrow89x=302\\ \Rightarrow x=\dfrac{320}{89}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left\{{}\begin{matrix}\dfrac{3}{x+1}-2x=-1\left(ĐK:x\ne-1\right)\\\dfrac{5}{x+1}+3y=11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{9}{x+1}-6y=-3\\\dfrac{10}{x+1}+6y=22\end{matrix}\right.\)
<=> \(\left\{{}\begin{matrix}\dfrac{9}{x+1}+\dfrac{10}{x+1}=19\\\dfrac{3}{x+1}-2y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=\dfrac{5}{4}\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1)\(\sqrt{4x^2+12x+9}=2-x\)
\(\Leftrightarrow\sqrt{\left(2x+3\right)^2}=2-x\)
\(\Leftrightarrow\left|2x+3\right|=2-x\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+3=2-x\\2x+3=x-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=-1\\x=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{3}\\x=-5\end{matrix}\right.\)
\(\)
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1, x(x-1)=2(x-1)
<=> x(x-1)-2(x-1)=0
<=> (x-2)(x-1)=0
<=>x=2 hoặc x=1
vậy ...
2, (x+2)(2x-3)=x^2 -4
<=>(x+2)(2x-3)=(x-2)(x+2)
<=> (x+2)(2x-3)-(x-2)(x+2)=0
<=> (x+2)(2x-3-x+2)=0
<=> x=-2 hoặc x=1
vây...
3,x^2 +3x +2=0
<=> x^2 +x+2x+2=0
<=>(x+2)(x+1)=0
<=> x=-2 hoặc x=-1
vậy ...
5, x^3+x^2-12x =0
<=> x(x^2+x-12)=0
<=>x(x^2-3x+4x-12)=0
<=>x(x+4)(x-3)=0
<=> x=0 hoặc x=-4 hoặc x=3
vậy ...
\(\Leftrightarrow4\left(x+6\right)+4x=x\left(x+6\right)\)
\(\Leftrightarrow x^2+6x=8x+24\)
\(\Leftrightarrow x^2-2x-24=0\)
=>(x-6)(x+4)=0
=>x=6(nhận) hoặc x=-4(nhận)