-2X^2 cộng 5X cộng 3=0
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\(\Leftrightarrow-2x^2+6x-x+3=0\)
\(\Leftrightarrow-2x\left(x-3\right)-\left(x-3\right)=0\)
\(\Leftrightarrow\left(2x+1\right)\left(x-3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-\frac{1}{2}\\x=3\end{matrix}\right.\)
\(-2x^2+5x+3=0\)
\(\Leftrightarrow-2x^2+6x-x+3=0\)
\(\Leftrightarrow\left(-2x^2+6x\right)-\left(x-3\right)=0\)
\(\Leftrightarrow-2x.\left(x-3\right)-\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right).\left(-2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\-2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\-2x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\frac{1}{2}\end{matrix}\right.\)
Vậy phương trình có tập hợp nghiệm là: \(S=\left\{3;-\frac{1}{2}\right\}.\)
Chúc bạn học tốt!
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có : \(x^4+2x^3+5x^2+4x-12=0\)
\(\Leftrightarrow x^4+2x^3+5x^2+10x-6x-12=0\)
\(\Leftrightarrow x^3\left(x+2\right)+5x\left(x+2\right)-6\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x^3+5x-6\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x^3-x^2+x^2-x+6x-6\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left[x^2\left(x-1\right)+x\left(x-1\right)+6\left(x-1\right)\right]=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-1\right)\left(x^2+x+6\right)=0\)
\(\Leftrightarrow\)\(x+2=0\)
hoặc \(x-1=0\)
hoặc \(x^2+x+6=0\)
\(\Leftrightarrow\) \(x=-2\)(tm)
hoặc \(x=1\)(tm)
hoặc \(\left(x+\frac{1}{2}\right)^2+\frac{23}{4}=0\)(ktm)
Vậy tập nghiệm của phương trình là \(S=\left\{-2;1\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: 3x-2=2x-3
=>x=-1
b: 2x+3=5x+9
=>-3x=6
=>x=-2
c: 5-2x=7
=>2x=-2
=>x=-2
d: 10x+3-5x=4x+12
=>5x+3=4x+12
=>x=9
e: 11x+42-2x=100-9x-22
=>9x+42=78-9x
=>18x=36
=>x=2
f: 2x-(3-5x)=4(x+3)
=>2x-3+5x=4x+12
=>7x-3=4x+12
=>3x=15
=>x=5
![](https://rs.olm.vn/images/avt/0.png?1311)
A+B+C
=2x^3-5x^2+x-7+x^2-2x+6+C
=2x^3-4x^2-x-1-x^3+4x^2-1
=x^3-x-2
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có : \(\frac{\left(5x+3\right)\left(3x+11\right)}{4}-\frac{x-7}{12}=0\)
=> \(\frac{3\left(5x+3\right)\left(3x+11\right)}{12}-\frac{x-7}{12}=0\)
=> \(3\left(5x+3\right)\left(3x+11\right)-\left(x-7\right)=0\)
=> \(3\left(15x^2+9x+55x+33\right)-x+7=0\)
=> \(45x^2+27x+165x+99-x+7=0\)
=> \(45x^2+191x+106=0\)
=> \(45x^2+2.\sqrt{45}x.\frac{191}{2\sqrt{45}}+\frac{191^2}{\left(2\sqrt{45}\right)^2}-\frac{17401}{180}=0\)
=> \(\left(x\sqrt{45}+\frac{191}{2\sqrt{45}}\right)^2-\left(\sqrt{\frac{17401}{180}}\right)^2=0\)
=> \(\left(x\sqrt{45}+\frac{191}{2\sqrt{45}}-\sqrt{\frac{17401}{180}}\right)\left(x\sqrt{45}+\frac{191}{2\sqrt{45}}+\sqrt{\frac{17401}{180}}\right)=0\)
=> \(\left[{}\begin{matrix}x\sqrt{45}+\frac{191}{2\sqrt{45}}-\sqrt{\frac{17401}{180}}=0\\x\sqrt{45}+\frac{191}{2\sqrt{45}}+\sqrt{\frac{17401}{180}}=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x\sqrt{45}=-\frac{191}{2\sqrt{45}}+\sqrt{\frac{17401}{180}}\\x\sqrt{45}=-\frac{191}{2\sqrt{45}}-\sqrt{\frac{17401}{180}}\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=\frac{-\frac{191}{2\sqrt{45}}+\sqrt{\frac{17401}{180}}}{\sqrt{45}}\\x=\frac{-\frac{191}{2\sqrt{45}}-\sqrt{\frac{17401}{180}}}{\sqrt{45}}\end{matrix}\right.\)
Vậy phương trình trên có nghiệm là \(\left[{}\begin{matrix}x=\frac{-\frac{191}{2\sqrt{45}}+\sqrt{\frac{17401}{180}}}{\sqrt{45}}\\x=\frac{-\frac{191}{2\sqrt{45}}-\sqrt{\frac{17401}{180}}}{\sqrt{45}}\end{matrix}\right.\) .
![](https://rs.olm.vn/images/avt/0.png?1311)
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b, TH1 2.3X-2+5X-6=0
=>6x-4+5x-6=0
=>11x+-10=0=>11x=0+10==>x=\(\frac{10}{11}\)
c,
\(-2x^2+5x+3=0\)
Ta có \(\Delta=5^2+4.2.3=49,\sqrt{\Delta}=7\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{-5+7}{4}=\frac{1}{2}\\x=\frac{-5-7}{4}=-3\end{cases}}\)