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\(=\dfrac{1}{4}-1+\dfrac{2}{4}x-\dfrac{1}{6}+\dfrac{5}{4}=\dfrac{1}{2}x+\dfrac{1}{3}\)
* Trả lời:
\(\left(1\right)\) \(-3\left(1-2x\right)-4\left(1+3x\right)=-5x+5\)
\(\Leftrightarrow-3+6x-4-12x=-5x+5\)
\(\Leftrightarrow6x-12x+5x=3+4+5\)
\(\Leftrightarrow x=12\)
\(\left(2\right)\) \(3\left(2x-5\right)-6\left(1-4x\right)=-3x+7\)
\(\Leftrightarrow6x-15-6+24x=-3x+7\)
\(\Leftrightarrow6x+24x+3x=15+6+7\)
\(\Leftrightarrow33x=28\)
\(\Leftrightarrow x=\dfrac{28}{33}\)
\(\left(3\right)\) \(\left(1-3x\right)-2\left(3x-6\right)=-4x-5\)
\(\Leftrightarrow1-3x-6x+12=-4x-5\)
\(\Leftrightarrow-3x-6x+4x=-1-12-5\)
\(\Leftrightarrow-5x=-18\)
\(\Leftrightarrow x=\dfrac{18}{5}\)
\(\left(4\right)\) \(x\left(4x-3\right)-2x\left(2x-1\right)=5x-7\)
\(\Leftrightarrow4x^2-3x-4x^2+2x=5x-7\)
\(\Leftrightarrow-x-5x=-7\)
\(\Leftrightarrow-6x=-7\)
\(\Leftrightarrow x=\dfrac{7}{6}\)
\(\left(5\right)\) \(3x\left(2x-1\right)-6x\left(x+2\right)=-3x+4\)
\(\Leftrightarrow6x^2-3x-6x^2-12x=-3x+4\)
\(\Leftrightarrow-15x+3x=4\)
\(\Leftrightarrow-12x=4\)
\(\Leftrightarrow x=-\dfrac{1}{3}\)
b) (4x -1)2 = (1-4x)4 (1)
Vì (1 - 4x) = (4x - 1)
\(\Rightarrow\)(1 - 4x)4 = [ -( 4x -1)4 ]
Vì (1-4x)4 = ( 4x - 1)4
Do đó (1) có dạng :
(4x - 1)2 = (4x - 1)2
Đặt 4x - 1 = x, ta có :
x2 = x4
x2 ( 1 - x2 ) = 0
\(\Rightarrow\)\(\orbr{\begin{cases}x=0\\x^2=\orbr{\begin{cases}1^2\\\left(-1\right)^2\end{cases}}\end{cases}}\)\(\Leftrightarrow\)\(\orbr{\begin{cases}x=0\\x=\orbr{\begin{cases}x=1\\x=-1\end{cases}}\end{cases}}\)
Thay x = 4x -1 = 0
x = \(\frac{1}{4}\)
- x = 1 \(\Leftrightarrow\) 4x - 1 = 1
x = \(\frac{1}{2}\)
- x = -1 \(\Leftrightarrow\) 4x -1 = -1
x = 0
Vậy x = \(\frac{1}{2}\) hoặc x = 0