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Lỗi sai: Khi chuyển vế hạng tử -x từ vế phải sang vế trái và hạng tử -6 từ vế trái sang vế phải không đổi dấu của hạng tử đó.
Sửa lại:
3x – 6 + x = 9 – x
⇔ 3x + x + x = 9 + 6
⇔ 5x = 15
⇔ x = 3.
Vậy phương trình có nghiệm duy nhất x = 3.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(1,\\ a,=3x^3-2x^2+5x\\ b,=2x^3y^2+\dfrac{2}{9}x^4y^2-\dfrac{1}{3}x^2y^3\\ c,=x^2-2x+6x-12=x^2+4x-12\\ 2,\\ a,\Rightarrow6x-9+4-2x=-3\\ \Rightarrow4x=2\Rightarrow x=\dfrac{1}{2}\\ b,\Rightarrow5x-2x^2+2x^2-2x=13\\ \Rightarrow3x=13\Rightarrow x=\dfrac{13}{3}\\ c,\Rightarrow5x^2-5x-5x^2+7x-10x+14=6\\ \Rightarrow-8x=-8\Rightarrow x=1\\ d,\Rightarrow6x^2+9x-6x^2+4x-15x+10=8\\ \Rightarrow-2x=-2\Rightarrow x=1\)
\(3,\\ A=2x^2+x-x^3-2x^2+x^3-x+3=3\\ B=6x^2-10x+33x-55-6x^2-14x-9x-21=-76\)
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c: \(=\dfrac{2\left(x+3\right)}{x\left(3x-1\right)}\cdot\dfrac{-\left(3x-1\right)}{x\left(x+3\right)}=\dfrac{-2}{x^2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(x-1\right)\left(x+1\right)-3x-6=6\)
\(x^2-1^2-3x-6-6=0\)
\(x^2-1-3x-12=0\)
\(x^2-3x-13=0\)
\(\orbr{\begin{cases}x=\frac{3-\sqrt{61}}{2}\\x=\frac{3+\sqrt{61}}{2}\end{cases}}\)
\(\left(x-1\right)\left(x+1\right)-3x-6=6\)
\(\left(x-1\right)\left(x+1\right)-3x=12\)
\(\left(x-1\right)x-\left(x-1\right)1-\left(1+2\right)x=12\)
\(\left(x-1-1+2\right)x-x-1=12\)
\(\left(x-1-1+2-1\right)x=11\)
\(\left(x-1\right)x=11\)
\(x^2-x=11\)
Đk : x > 4
\(x=4\Rightarrow16-4=11\left(\varnothing\right)\)
\(x\in\varnothing\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(b,3\left(x-2\right)+2\left(3x-5\right)=10\\ \Leftrightarrow3x-6+6x-10=10\\ \Leftrightarrow3x+6x=10+10+6\\ \Leftrightarrow9x=26\\ \Leftrightarrow x=\dfrac{26}{9}\\ c,2x-\left(3x+1\right)=5x-2\\ \Leftrightarrow2x-3x-1=5x-2\\ \Leftrightarrow2x-3x-5x=-2+1\\ \Leftrightarrow-6x=-1\\ \Leftrightarrow x=\dfrac{1}{6}\\ d,3x+2=-5+6 \\ \Leftrightarrow3x=-5+6-2\\ \Leftrightarrow3x=-2\\ \Leftrightarrow x=-\dfrac{1}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, 3x ( 3x + 5) - (3x - 1) ( 3x + 1) = 2
<=>9x2+15x-9x2+1
<=>15x+1=2
<=>15x=1
<=>x=1/15
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,\Rightarrow5x+3x^2-3x^2-x+2=6\\ \Rightarrow4x=4\Rightarrow x=1\\ b,\Rightarrow\left(2x+\dfrac{1}{2}-1+2x\right)\left(2x+\dfrac{1}{2}+1-2x\right)=2\\ \Rightarrow\dfrac{3}{2}\left(4x-\dfrac{1}{2}\right)=2\\ \Rightarrow6x-\dfrac{3}{4}=2\\ \Rightarrow6x=\dfrac{11}{4}\\ \Rightarrow x=\dfrac{11}{24}\\ c,\Rightarrow\left(x+3\right)\left(x-2\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)
\(\left(3x-6\right)^{5021}=\left(3x-6\right)^{5020}\\ \Leftrightarrow\left(3x-6\right)^{5020}.\left(3x-6\right)=\left(3x-6\right)^{5020}\\ \Leftrightarrow3x-6=0\\ \Leftrightarrow x=2\)
\(\left(3x-6\right)^{5021}=\left(3x-6\right)^{5020}\)
\(\left(3x-6\right)^{5021}-\left(3x-6\right)^{5020}=0\)
\(\left(3x-6\right)^{5020}\left(3x-6-1\right)=0\)
\(\left(3x-6\right)^{5020}\left(3x-7\right)=0\)
⇔ \(\left[{}\begin{matrix}3x-6=0\\3x-7=0\end{matrix}\right.\)
⇔ \(\left[{}\begin{matrix}3x=6\\3x=7\end{matrix}\right.\)
⇔ \(\left[{}\begin{matrix}x=2\\x=\dfrac{7}{3}\end{matrix}\right.\)