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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 \(S=\left\{-2;1\right\}\)
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\(\frac{\left(x+6\right)\left(x+6\right)}{2}-\frac{4x}{3}=0\)
\(\Leftrightarrow\frac{\left(x+6\right)^2}{2}=\frac{4x}{3}\)
\(\Leftrightarrow\frac{3\left(x+6\right)^2}{6}=\frac{8x}{6}\)
\(\Leftrightarrow3\left(x+6\right)^2=8x\)
\(\Leftrightarrow3\left(x^2+12x+36\right)-8x=0\)
\(\Leftrightarrow3x^2+36x+108-8x=0\)
\(\Leftrightarrow3x^2+28x+108=0\)
=> pt vô ngiệp
\(\frac{\left(x+6\right)^2}{2}-\frac{4x}{3}=0\)
\(\Rightarrow\frac{x^2+12x+36}{2}-\frac{4x}{3}=0\)
\(\Rightarrow\frac{3x^2+36x+108}{6}-\frac{8x}{6}=0\)
\(\Rightarrow\frac{3x^2+28x+108}{6}=0\)
\(\Rightarrow3x^2+28x+108=0\)
Ta có: \(\Delta=28^2-4.3.108=-512< 0\)
Vậy pt vô nghiệm
![](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 \(\left[{}\begin{matrix}x=\frac{-\frac{191}{2\sqrt{45}}+\sqrt{\frac{17401}{180}}}{\sqrt{45}}\\x=\frac{-\frac{191}{2\sqrt{45}}-\sqrt{\frac{17401}{180}}}{\sqrt{45}}\end{matrix}\right.\) .
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Đề hai có nhân 9 nha bạn làm mình hoang mang cái đề quá
\(9\left(2x+1\right)^2-4\left(x+1\right)^2=0\\\Leftrightarrow \left[3\left(2x+1\right)\right]^2-\left[2\left(x+1\right)\right]^2=0\\ \Leftrightarrow\left[3\left(2x+1\right)-2\left(x+1\right)\right]\left[3\left(2x+1\right)+2\left(x+1\right)\right]=0\\\Leftrightarrow \left[6x+3-2x-2\right]\left[6x+3+2x+2\right]=0\\\Leftrightarrow \left(4x+1\right)\left(8x+5\right)=0\\ \Rightarrow\left[{}\begin{matrix}4x+1=0\\8x+5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\frac{1}{4}\\x=-\frac{5}{8}\end{matrix}\right.\)
Vậy tập nghiệm của phương trình trên là \(S=\left\{-\frac{1}{4};-\frac{5}{8}\right\}\)
\(\left(2x+1\right)^2-4.\left(x+1\right)^2=0\\ \Leftrightarrow4x^2+4x+1-4.\left(x^2+2x+1\right)=0\\ \Leftrightarrow4x^2+4x+1-4x^2-8x-4=0\\ \Leftrightarrow-4x=3\\ \Leftrightarrow x=-\frac{3}{4}\)
Ta có
\(2\left(x+3\right)\left(x-4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2\left(x+3\right)=0\\x-4=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x+3=0\\x=4\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-3\\x=4\end{cases}}\)
Vậy \(x\in\left\{-3;4\right\}\)
@@ Học tốt @@
## Chiyuki Fujito
2.(x+3)(x-4)=0
* 2(x+3)=0 * x-4=0
x+3=0:2 x=0+4
x+3=0 x=4
x=0-3
x=-3
vậy x=-3 hoặc x=4