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a:=>3x=15
=>x=5
b: =>8-11x<52
=>-11x<44
=>x>-4
c: \(VT=\left(\dfrac{x^2-\left(x-6\right)^2}{x\left(x+6\right)\left(x-6\right)}\right)\cdot\dfrac{x\left(x+6\right)}{2x-6}+\dfrac{x}{6-x}\)
\(=\dfrac{12x-36}{2x-6}\cdot\dfrac{1}{x-6}-\dfrac{x}{x-6}=\dfrac{6}{x-6}-\dfrac{x}{x-6}=-1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 3:
b: \(\Leftrightarrow x^2\left(x+1\right)^2=0\)
hay \(x\in\left\{0;-1\right\}\)
c: \(\Leftrightarrow\left(x-1\right)\left(x^2+x+1\right)=0\)
=>x-1=0
hay x=1
d: \(\Leftrightarrow6x^2-3x-4x+2=0\)
\(\Leftrightarrow\left(2x-1\right)\left(3x-2\right)=0\)
hay \(x\in\left\{\dfrac{1}{2};\dfrac{2}{3}\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{1}{x+2}+\dfrac{6x+12}{x^3+8}-\dfrac{7}{x^2-2x+4}=0\) \(\left(đk:x\ne-2\right)\)
\(\Leftrightarrow\dfrac{x^2-2x+4+6x+12-7\left(x+2\right)}{x^3+8}=0\)
\(\Leftrightarrow\dfrac{x^2-3x+2}{x^3+8}=0\)
\(\Leftrightarrow x^2-3x+2=0\)
\(\Leftrightarrow\left(x^2-2x\right)-\left(x-2\right)=0\)
\(\Leftrightarrow x\left(x-2\right)-\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)(TM)
Vậy ...
dk : x khac -2
\(\Rightarrow x^2-2x+4+6x+12-7\left(x+2\right)=0\)
\(\Leftrightarrow x^2+4x+16-7x-14=0\Leftrightarrow x^2-3x+2=0\)
\(\Leftrightarrow x^2-2x-x+2=0\Leftrightarrow x\left(x-2\right)-\left(x-2\right)=0\Leftrightarrow\left(x-1\right)\left(x-2\right)=0\Leftrightarrow x=1;x=2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(ĐKXĐ:\)\(x\ne-2;\)\(x\ne-3;\)\(x\ne-4\)
\(x+\frac{x}{x+2}+\frac{x+3}{x^2+5x+6}+\frac{x+4}{x^2+6x+8}=1\)
\(\Leftrightarrow\)\(x+\frac{x}{x+2}+\frac{x+3}{\left(x+2\right)\left(x+3\right)}+\frac{x+4}{\left(x+2\right)\left(x+4\right)}=1\)
\(\Leftrightarrow\)\(x+\frac{x}{x+2}+\frac{1}{x+2}+\frac{1}{x+2}=1\)
\(\Leftrightarrow\)\(\frac{x\left(x+2\right)+x+1+1}{x+2}=1\)
\(\Leftrightarrow\)\(\frac{x^2+3x+2}{x+2}=1\)
\(\Leftrightarrow\)\(x^2+3x+2=x+2\)
\(\Leftrightarrow\)\(x\left(x+2\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=0\\x+2=0\end{cases}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=0\\x=-2\left(L\right)\end{cases}}\)
Vậy pt có nghiệm \(x=0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(ĐKXĐ:\hept{\begin{cases}x\ne2\\x\ne4\end{cases}}\)
\(\frac{x+3}{x-4}+\frac{x-1}{x-2}=\frac{2}{6x-8-x^2}\)
\(\Leftrightarrow\frac{x+3}{x-4}+\frac{x-1}{x-2}+\frac{2}{\left(x-2\right)\left(x-4\right)}=0\)
\(\Leftrightarrow\frac{\left(x+3\right)\left(x-2\right)+\left(x-1\right)\left(x-4\right)+2}{\left(x-2\right)\left(x-4\right)}=0\)
\(\Leftrightarrow x^2+x-6+x^2-5x+4+2=0\)
\(\Leftrightarrow2x^2-4x=0\)
\(\Leftrightarrow2x\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-2=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\left(tm\right)\\x=2\left(ktm\right)\end{cases}}\)
Vậy tập nghiệm của phương trình là \(S=\left\{0\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
(x^2)^2+(x^2+6x+3^2)-1=0
(x^2)^2-1^2+(x+3)^2=0
(x^2-1)(x^2+1)+(x+3)^2=0
(x+3)^2 luôn lớn hơn 0
nên x^2-1=0 => x=1
x^2+1=0 => x vô nghiệm
\(x^4+x^2+6x-8=0\)
\(\Leftrightarrow x^4-x^3+4x^2+x^3-x^2+4x-2x^2+2x-8=0\)
\(\Leftrightarrow x^2\left(x^2-x+4\right)+x\left(x^2-x+4\right)-2\left(x^2-x+4\right)=0\)
\(\Leftrightarrow\left(x^2+x-2\right)\left(x^2-x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+x-2=0\\x^2-x+4=0\left(vô-nghiệm\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)