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Ta có : 6x2 - 11x + 3
= 6x2 - 2x - 9x + 3
= (6x2 - 2x) - (9x - 3)
= 2x(3x - 1) - 3(3x - 1)
= (2x - 3)(3x - 1)
\(\left|2x-3\right|=3-2x\)
\(ĐK:x\le\dfrac{3}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=3-2x\\3-2x=3-2x\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\0=0\left(đúng\right)\end{matrix}\right.\)
Vậy \(S=\left\{x\in R;x=\dfrac{3}{2}\right\}\)
d) \(2x^2+5x-7=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{7}{2}\end{matrix}\right.\) \(\left(a+b+c=1\right)\)
a: \(=\dfrac{2\left(x+2\right)\left(x-1\right)}{x+2}=2x-2\)
b: \(=\dfrac{2x^3+x^2-6x^2-3x+2x+1}{2x+1}=x^2-3x+1\)
c: \(=\dfrac{x^3+2x^2-2x^2-4x+2x+4}{x+2}=x^2-2x+2\)
d: \(=\dfrac{x^2\left(x-3\right)}{x-3}=x^2\)
\(\left(2x-3\right)\left(2x+3\right)=2\left(2x-3\right)\left(2x-3\right)\)
\(\Leftrightarrow\left(2x-3\right)\left(2x+3\right)-2\left(2x-3\right)\left(2x-3\right)=0\)
\(\Leftrightarrow\left(2x-3\right)\left(2x+3-4x+6\right)=0\)
\(\Leftrightarrow\left(2x-3\right)\left(-2x+9\right)=0\)
\(\Leftrightarrow2x-3=0\) hay \(-2x+9=0\)
\(\Leftrightarrow x=\dfrac{3}{2}\) hay \(x=\dfrac{9}{2}\)
-Vậy \(S=\left\{\dfrac{3}{2};\dfrac{9}{2}\right\}\)