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a: 7x+35=0
=>7x=-35
=>x=-5
b: \(\dfrac{8-x}{x-7}-8=\dfrac{1}{x-7}\)
=>8-x-8(x-7)=1
=>8-x-8x+56=1
=>-9x+64=1
=>-9x=-63
hay x=7(loại)
a, \(7x=-35\Leftrightarrow x=-5\)
b, đk : x khác 7
\(8-x-8x+56=1\Leftrightarrow-9x=-63\Leftrightarrow x=7\left(ktm\right)\)
vậy pt vô nghiệm
2, thiếu đề
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,\dfrac{x-3}{x}=\dfrac{x-3}{x+3}\)\(\left(đk:x\ne0,-3\right)\)
\(\Leftrightarrow\dfrac{x-3}{x}-\dfrac{x-3}{x+3}=0\)
\(\Leftrightarrow\dfrac{\left(x-3\right)\left(x+3\right)-x\left(x-3\right)}{x\left(x+3\right)}=0\)
\(\Leftrightarrow x^2-9-x^2+3x=0\)
\(\Leftrightarrow3x-9=0\)
\(\Leftrightarrow3x=9\)
\(\Leftrightarrow x=3\left(n\right)\)
Vậy \(S=\left\{3\right\}\)
\(b,\dfrac{4x-3}{4}>\dfrac{3x-5}{3}-\dfrac{2x-7}{12}\)
\(\Leftrightarrow\dfrac{4x-3}{4}-\dfrac{3x-5}{3}+\dfrac{2x-7}{12}>0\)
\(\Leftrightarrow\dfrac{3\left(4x-3\right)-4\left(3x-5\right)+2x-7}{12}>0\)
\(\Leftrightarrow12x-9-12x+20+2x-7>0\)
\(\Leftrightarrow2x+4>0\)
\(\Leftrightarrow2x>-4\)
\(\Leftrightarrow x>-2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{3x}{x^2-x+3}-\dfrac{2x}{x^2-3x+3}+1=0\left(a\right)\)
Ta có : \(x^2-x+3=x^2-x+\dfrac{1}{4}+\dfrac{11}{4}=\left(x-\dfrac{1}{2}\right)^2+\dfrac{11}{4}>0\)
\(x^2-3x+3=x^2-3x+\dfrac{9}{4}+\dfrac{3}{4}=\left(x-\dfrac{3}{2}\right)^2+\dfrac{3}{4}>0\)
\(\RightarrowĐKXĐ:x\in R\)
Đặt : \(t=x^2-x+3\)
\(\left(a\right)\Leftrightarrow\dfrac{3x}{t}-\dfrac{2x}{t-2x}+1=0\)
\(\Leftrightarrow3x\left(t-2x\right)-2xt+t\left(t-2x\right)=0\)
\(\Leftrightarrow t^2-xt-6x^2=0\)
\(\Leftrightarrow t^2+2xt-3xt-6x^2=0\)
\(\Leftrightarrow t\left(t+2x\right)-3x\left(t+2x\right)=0\)
\(\Leftrightarrow\left(t-3x\right)\left(t+2x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t-3x=0\\t+2x=0\end{matrix}\right.\left(b\right)\)
Thay \(t=x^2-x+3\) lại vào (b) được :
\(\left[{}\begin{matrix}x^2-x+3-3x=0\\x^2-x+3+2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x^2-4x+3=0\\x^2+x+3=0\end{matrix}\right.\left(c\right)\)
Mà : \(x^2-4x+3=x^2-x-3x+3\)
\(=x\left(x-1\right)-3\left(x-1\right)=\left(x-1\right)\left(x-3\right)\left(c'\right)\)
và : \(x^2+x+3=x^2+x+\dfrac{1}{4}+\dfrac{11}{4}\)
\(=\left(x+\dfrac{1}{2}\right)^2+\dfrac{11}{4}\left(c''\right)\)
Thay (c') và (c'') vào (c) được :
\(\left[{}\begin{matrix}\left(x-1\right)\left(x-3\right)=0\\\left(x+\dfrac{1}{2}\right)^2+\dfrac{11}{4}=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x-1=0\Leftrightarrow x=1\left(tmđk\right)\\x-3=0\Leftrightarrow x=3\left(tmđk\right)\end{matrix}\right.\\\left(x+\dfrac{1}{2}\right)^2=-\dfrac{11}{4}\Leftrightarrow x\in\varnothing\end{matrix}\right.\)
Vậy : Phương trình có tập nghiệm \(S=\left\{1;3\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(x^3-3x^2-4x=0\)
\(\Leftrightarrow x\left(x^2-3x-4\right)=0\)
\(\Leftrightarrow x\left(x-4\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-4=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\\x=-1\end{matrix}\right.\)
Vậy \(S=\left\{0;4;-1\right\}\).
b) \(3x^2-5x-2=0\)
\(\Leftrightarrow3x^2+x-6x-2=0\)
\(\Leftrightarrow x\left(3x+1\right)-2\left(3x+1\right)=0\)
\(\Leftrightarrow\left(3x+1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+1=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{3}\\x=2\end{matrix}\right.\)
Vậy \(S=\left\{-\dfrac{1}{3};2\right\}\).
![](https://rs.olm.vn/images/avt/0.png?1311)
`3x+7=0`
`<=>3x=-7`
`<=>x=-7/3`
Vậy `S={-7/3}`
______________________
`2x(x-2)+2x(5-3x)=0`
`<=>2x(x-2+5-3x)=0`
`<=>2x(3-2x)=0`
`@TH1:2x=0<=>x=0`
`@TH2: 3-2x=0<=>2x=3<=>x=3/2`
Vậy `S={0;3/2}`
3x+7=0
\(\Leftrightarrow3x=-7\Leftrightarrow x=-\dfrac{7}{3}\)
2x(x-2)+2x(5-3x)=0
\(\Leftrightarrow2x\left(x-2+5-3x\right)=0\)
\(\Leftrightarrow2x\left(-2x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=0\\-2x=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{-3}{-2}=\dfrac{3}{2}\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\Leftrightarrow\left(x^2-3x-9-3x+17\right)\left(x^2-3x-9+3x-17\right)=0\)
\(\Leftrightarrow\left(x^2-6x+8\right)\left(x^2-26\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-6x+8=0\\x^2-26=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x_1=4;x_2=2\\x^2=26\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x_1=4;x_2=2\\x=\sqrt{26}\end{matrix}\right.\)
Vậy \(S=\left\{4;2;\sqrt{26}\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
x^2+3x+2=0
<=>x^2+x+2x+2=0
<=>x(x+1)+2(x+1)=0
<=>(x+1)(x+2)=0
<=> x+1=0 hoặc x+2=0
<=> x=-1 hoặc x=-2
Vậy tập nghiệm cỉa pt là S={-1;-2}
pt <=> (x^2+x)+(2x+2)=0
<=> x.(x+1)+2.(x+1)=0
<=>(x+1).(x+2) = 0
<=> x+1 = 0 hoặc x+2 = 0
<=> x=-1 hoặc x=-2
\(\left(x^2-x\right)-\left(2x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-1\right)=0\)
\(\Leftrightarrow x=2;x=1\)
\(x^2-3x+2=0\)
\(\Leftrightarrow x^2-x-2x+2=0\)
\(\Leftrightarrow x\left(x-1\right)-2\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x-2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x=2\end{cases}}}\)
Vậy x=1; x=2