a) \(15.3-x=2\left(x-4\right)\)
\(\Leftrightarrow15-x=2x-8\)
\(\Leftrightarrow15+8=2x+x\)
\(\Leftrightarrow23=3x\)
\(\Leftrightarrow x=\dfrac{23}{3}\)
b) \(10-\left|3-x\right|=6.\left(-7\right)\)
\(\Leftrightarrow10-\left|3-x\right|=-42\)
\(\Leftrightarrow\left|3-x\right|=10+42\)
\(\Leftrightarrow\left|3-x\right|=52\)
\(\Leftrightarrow\left[{}\begin{matrix}3-x=52\\3-x=-52\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-49\\x=55\end{matrix}\right.\)
c) \(\left(x-2\right)\left(2y-1\right)=-6\)
\(\Leftrightarrow\left(x-2\right)\left(2y-1\right)=\left(-1\right).6=6.\left(-1\right)=\left(-2\right).3=3.\left(-2\right)=\left(-6\right).1=1.\left(-6\right)=\left(-3\right).2=2.\left(-3\right)\)
Ta có bảng sau:
| \(x-2\) | \(-1\) | \(6\) | \(-6\) | \(1\) | \(-2\) | \(3\) | \(-3\) | \(2\) |
| \(2y+1\) | \(6\) | \(-1\) | \(1\) | \(-6\) | \(3\) | \(-2\) | \(2\) | \(-3\) |
| \(x\) | \(1\) | \(8\) | \(-4\) | \(3\) | \(0\) | \(5\) | \(-1\) | \(4\) |
| \(y\) | \(\dfrac{5}{2}\) | \(-1\) | \(0\) | \(\dfrac{-7}{2}\) | \(1\) | \(\dfrac{-3}{2}\) | \(\dfrac{1}{2}\) | \(-2\) |
KL: Vậy...
