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đơn giản
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a) \(3\left(2x-1\right)-5\left(x-3\right)+6\left(3x-4\right)=24\) (1)
\(\Leftrightarrow6x-3-5x+15+18x-24=24\)
\(\Leftrightarrow19x-12=24\)
\(\Leftrightarrow19x=36\)
\(\Leftrightarrow x=\dfrac{36}{19}\)
Vậy tập nghiệm phương trình (1) là \(S=\left\{\dfrac{36}{19}\right\}\)
b) \(3x\left(x+1\right)-2x\left(x+2\right)=-1-x\)
\(\Leftrightarrow3x^2+3x-2x^2-4x=-1-x\)
\(\Leftrightarrow x^2-x=-1-x\)
\(\Leftrightarrow x^2-x-x=-1\)
\(\Rightarrow x\in\varnothing\)
c) Thiếu đề.
\(a,3\left(2x-1\right)-5\left(x-3\right)+6\left(3x-4\right)=24\) \(\Leftrightarrow6x-3-5x+15+18x-24-24=0\)
\(\Leftrightarrow9x-36=0\)
\(\Leftrightarrow9\left(x-4\right)=0\Rightarrow x-4=0\Rightarrow x=4\)
\(b,3x\left(x-1\right)-2x\left(x+2\right)=-1-x\)
\(\Leftrightarrow3x^2+3x-2x^2-4x+1+x=0\)
\(\Leftrightarrow x^2-1=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)=0\Rightarrow\left[{}\begin{matrix}x-1=0\\x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)\(c,2\left(2x+x^2\right)-x^2\left(x+2\right)+\left(x^3-4x+3\right)\)\(\Leftrightarrow4x+2x^2-x^3-2x^2+x^3-4x+3\)
\(=3\)
Vậy x = mọi giá trị
![](https://rs.olm.vn/images/avt/0.png?1311)
\(=\left(x-y\right)\left(x+y\right)-7\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y-7\right)\)
\(=100^2=10000\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài giải
\(\frac{1}{2}\left(x+1\right)+\frac{1}{4}\left(x+3\right)=3\cdot\frac{1}{3}\cdot\left(x+20\right)\)
\(\frac{1}{2}\left[\left(x+1\right)+\frac{1}{2}\left(x+3\right)\right]=x+20\)
\(\frac{1}{2}\left[x+1+\frac{1}{2}x+\frac{3}{2}\right]=x+20\)
\(\frac{1}{2}\left[x\left(1+\frac{1}{2}\right)+1+\frac{3}{2}\right]=x+20\)
\(\frac{1}{2}\left[\frac{3}{2}x+\frac{5}{2}\right]=x+20\)
\(\frac{3}{4}x+\frac{5}{4}=x+20\)
\(\frac{3}{4}x-x=20-\frac{5}{4}\)
\(\frac{-1}{4}x=\frac{75}{4}\)
\(x=\frac{75}{4}\text{ : }\frac{-1}{4}\)
\(x=-75\)
\(\frac{1}{2}\left(x+1\right)+\frac{1}{4}\left(x+3\right)=3\cdot\frac{1}{3}\cdot\left(x+20\right)\)
\(\frac{1}{2}\left[\left(x+1\right)+\frac{1}{2}\left(x+3\right)\right]=x+20\)
\(\frac{1}{2}\left[x+1+\frac{1}{2}x+\frac{3}{2}\right]=x+20\)
\(\frac{1}{2}\left[x\left(1+\frac{1}{2}\right)+1+\frac{3}{2}\right]=x+20\)
\(\frac{1}{2}\left[\frac{3}{2}x+\frac{5}{2}\right]=x+20\)
\(\frac{3}{4}x+\frac{5}{4}=x+20\)
\(\frac{3}{4}x-x=20-\frac{5}{4}\)
\(\frac{-1}{4}x=\frac{75}{4}\)
\(x=\frac{75}{4}\text{ : }\frac{-1}{4}\)
\(x=-75\)