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a) ĐKXĐ: \(x\notin\left\{2;-2\right\}\)
Ta có: \(\dfrac{x+1}{x-2}-\dfrac{5}{x+2}=\dfrac{12}{x^2-4}+1\)
\(\Leftrightarrow\dfrac{\left(x+1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{5\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=\dfrac{12}{\left(x-2\right)\left(x+2\right)}+\dfrac{x^2-4}{\left(x-2\right)\left(x+2\right)}\)
Suy ra: \(x^2+3x+2-5x+10=12+x^2-4\)
\(\Leftrightarrow x^2-2x+12-8-x^2=0\)
\(\Leftrightarrow-2x+4=0\)
\(\Leftrightarrow-2x=-4\)
hay x=2(loại)
Vậy: \(S=\varnothing\)
b) Ta có: \(\left|2x+6\right|-x=3\)
\(\Leftrightarrow\left|2x+6\right|=x+3\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+6=x+3\left(x\ge-3\right)\\-2x-6=x+3\left(x< -3\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-x=3-6\\-2x-x=3+6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\left(nhận\right)\\x=-3\left(loại\right)\end{matrix}\right.\)
Vậy: S={-3}
![](https://rs.olm.vn/images/avt/0.png?1311)
bai 1
1 thay k=0 vao pt ta co 4x^2-25+0^2+4*0*x=0
<=>(2x)^2-5^2=0
<=>(2x+5)*(2x-5)=0
<=>2x+5=0 hoăc 2x-5 =0 tiếp tục giải ý 2 tương tự
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(x+1\right)\left(x+2\right)=\left(2-x\right)\left(x+2\right)\)
\(\Leftrightarrow\left(x+1\right)\left(x+2\right)-\left(2-x\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x+1-2+x\right)=0\Leftrightarrow\left(x+2\right)\left(2x-1\right)=0\Leftrightarrow x=-2;x=\dfrac{1}{2}\)
Vậy tập nghiệm của phương trình là S = { -2 ; 1/2 }
Ta có: \(\left(x+1\right)\left(x+2\right)=\left(2-x\right)\left(x+2\right)\)
\(\Leftrightarrow\left(x+1\right)\left(x+2\right)-\left(2-x\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x+1-2+x\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\2x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{1}{2}\end{matrix}\right.\)
Vậy: \(S=\left\{-2;\dfrac{1}{2}\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
ĐKXĐ : \(x\ne\pm1\)
PT : \(\Leftrightarrow\dfrac{x-1-x^2-x+2}{x+1}=\dfrac{x+1-\left(x+2\right)\left(x-1\right)}{x-1}\)
\(\Leftrightarrow\dfrac{1-x^2}{x+1}=1-x=\dfrac{3-x^2}{x-1}\)
\(\Leftrightarrow x^2-3=\left(x-1\right)^2=x^2-2x+1\)
\(\Leftrightarrow-2x=-4\)
\(\Leftrightarrow x=2\left(TM\right)\)
Vậy ...
ĐKXĐ: \(x\notin\left\{1;-1\right\}\)
Ta có: \(\dfrac{x-1}{x+1}-\dfrac{x^2+x-2}{x+1}=\dfrac{x+1}{x-1}-x-2\)
\(\Leftrightarrow\dfrac{x-1-x^2-x+2}{x+1}-\dfrac{x+1}{x-1}+x+2=0\)
\(\Leftrightarrow\dfrac{-x^2+1}{x+1}-\dfrac{x+1}{x-1}+x+2=0\)
\(\Leftrightarrow\dfrac{-\left(x^2-1\right)}{x+1}-\dfrac{x+1}{x-1}+x+2=0\)
\(\Leftrightarrow\dfrac{-\left(x-1\right)\left(x+1\right)}{x+1}-\dfrac{x+1}{x-1}+x+2=0\)
\(\Leftrightarrow-\left(x-1\right)-\dfrac{x+1}{x-1}+x+2=0\)
\(\Leftrightarrow\dfrac{-\left(x-1\right)^2}{x-1}-\dfrac{x+1}{x-1}+\dfrac{\left(x+2\right)\left(x-1\right)}{\left(x+2\right)\left(x-1\right)}=0\)
Suy ra: \(-\left(x^2-2x+1\right)-x-1+x^2-x+2x-2=0\)
\(\Leftrightarrow-x^2+2x-1-x-1+x^2+x-2=0\)
\(\Leftrightarrow2x-4=0\)
\(\Leftrightarrow2x=4\)
hay x=2(nhận)
Vậy: S={2}
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{x-1}{x-2}+\dfrac{5}{x+2}=\dfrac{12}{x^2-4}+1\)
\(\Leftrightarrow x^2+x-2+5x-10=12+x^2-4\)
\(\Leftrightarrow6x-12=8\)
=>6x=20
hay x=10/3(nhận)
x−1x−2+5x+2=12x2−4+1x−1x−2+5x+2=12x2−4+1
⇔x2+x−2+5x−10=12+x2−4⇔x2+x−2+5x−10=12+x2−4
⇔6x−12=8⇔6x−12=8
=>6x=20
hay x=10/3(nhận)
![](https://rs.olm.vn/images/avt/0.png?1311)
Sửa đề: \(\left(x-1\right)^2-\left(3x+2\right)\left(x-12\right)=\left(x^2+1\right)\left(x-2\right)-x^2\)
\(\Leftrightarrow x^3-3x^2+3x-1-\left(3x^2-36x+2x-24\right)=x^3-2x^2+x-2-x^2\)
=>\(x^3-3x^2+3x-1-3x^2+34x+24=x^3-3x^2+x-2\)
=>\(x^3-6x^2+37x+23-x^3+3x^2-x+2=0\)
=>\(-3x^2+36x+25=0\)
=>\(x=\dfrac{18\pm\sqrt{399}}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(x^2+x+1\right)\left(x^2+x+2\right)=2\)
\(x^4+x^3+2x^2+x^3+x^2+2x+x^2+x+2=2\)
\(x^4+2x^3+4x^2+3x=0\)
\(x\left(x^3+2x^2+4x+3\right)=0\)
\(x=0\)( để đó ko quên mất )
\(x^3+2x^2+4x+3=0\)
\(\left(x^2+x+3\right)\left(x+1\right)=0\)
\(x=1\)
Vậy \(x=\left\{0;1\right\}\)
Nháp : \(\Delta=b^2-4ac=1^2-4.1.3=1-12=-11< 0\)
Nên pt vô nghiệm