`#3107.101107`
`1)`
`(2x - 1)^2 - (4x^2 - 1) = 0`
`=> (2x - 1)^2 - [(2x)^2 - 1^2] = 0`
`=> (2x - 1)^2 - (2x - 1)(2x + 1) = 0`
`=> (2x - 1)(2x - 1 - 2x - 1) = 0`
`=>`\(\left[{}\begin{matrix}2x-1=0\\2x-1-2x-1=0\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}2x=1\\-2=0\left(\text{vô lý}\right)\end{matrix}\right.\)
`=> x = 1/2`
Vậy, `x = 1/2`
`2)`
\(\left(x+2\right)^2-x\left(x-3\right)=2\\ \Rightarrow x^2+4x+4-x^2+3x=2\\ \Rightarrow7x+4=2\\ \Rightarrow7x=2-4\\ \Rightarrow7x=-2\Rightarrow x=-\dfrac{7}{2}\)
Vậy, `x = -7/2`
`3)`
\(\left(x-5\right)^2-x\left(x+2\right)=5\\ \Rightarrow x^2-10x+25-x^2-2x=5\\ \Rightarrow-12x+25=5\\ \Rightarrow-12x=5-25\\ \Rightarrow-12x=-20\\ \Rightarrow12x=20\\ \Rightarrow x=\dfrac{5}{3}\)
Vậy, `x = 5/3`
`4)`
\(\left(x-1\right)^2+x\left(4-x\right)=11\\ \Rightarrow x^2-2x+1+4x-x^2=11\\ \Rightarrow2x+1=11\\ \Rightarrow2x=11-1\\ \Rightarrow2x=10\\ \Rightarrow x=5\)
Vậy, `x = 5.`
