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Ta có : |2x - 5| + |4 + x| = 0
Mà : |2x - 5| \(\ge0\forall x\)
|4 + x| \(\ge0\forall x\)
Nên \(\orbr{\begin{cases}\left|2x-5\right|=0\\\left|4+x\right|=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x-5=0\\4+x=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x=5\\x=-4\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{2}\\x=-4\end{cases}}\)
a)\(x\left(x-3\right)-2x+6=0\)
\(\Leftrightarrow x\left(x-3\right)-2\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x-3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\\x=3\end{cases}}\)
b)\(\left(3x-5\right)\left(5x-7\right)+\left(5x+1\right)\left(2-3x\right)=4\)
\(\Leftrightarrow15x^2-46x+35-15x^2+7x+2-4=0\)
\(\Leftrightarrow33-39x=0\Leftrightarrow33=39x\Leftrightarrow x=\frac{33}{39}\)
a) \(x\left(x-3\right)-2x+6=0\)
\(x\left(x-3\right)-2\left(x-3\right)=0\)
\(\left(x-3\right)\left(x-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-3=0\\x-2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=3\\x=2\end{cases}}}\)
b) \((3x-5)(5x-7)+(5x+1)(2-3x)=4\)
\(15x^2-46x+35+10x-15x^2+2-3x-4=0\)
\(33-39x=0\)
\(3\left(11-13x\right)=0\)
\(11-13x=0\)
\(13x=11\)
\(x=\frac{11}{13}\)
\(\dfrac{2x-7}{3}\ge3x-7\)
\(\Leftrightarrow2x-7\ge3\left(3x-7\right)\)
\(\Leftrightarrow2x-7\ge9x-21\)
\(\Leftrightarrow7x\le14\)
\(\Leftrightarrow x\le2\)
(2x-7)/3≥3x-7
<=>(2x-7)/3≥3*(3x-7)/3
<=>2x-7≥3*(3x-7)
<=>2x-7≥9x-21
<=>2x-9x≥-21+7
<=>-7x≥-14
<=>7x≤14
<=>x≤2