\(4x+2x\left(x+1\right)=\left(2x-1\right)^2\)
\(=4x+2x^2+2x=4x^2-4x+1\)
\(=6x+2x^2=4x^2-4x+1\)
\(=2x^2-4x^2+6x+4x=1\)
\(=-2x^2+10x-1=0\)
\(\Delta=10^2-4\cdot\left(-2\right)\cdot\left(-1\right)=92\)
\(\Leftrightarrow\left\{{}\begin{matrix}x_1=\dfrac{-10+\sqrt{92}}{2\cdot\left(-2\right)}=\dfrac{5-\sqrt{23}}{2}\\x_2=\dfrac{-10-\sqrt{92}}{2\cdot\left(-2\right)}=\dfrac{5+\sqrt{23}}{2}\end{matrix}\right.\)
Vậy ....
=>4x^2-4x+1=4x+2x^2+2x=2x^2+6x
=>2x^2-10x+1=0
Δ=(-10)^2-4*2*1=100-8=92>0
=>Phương trình có hai nghiệm pb là:
\(\left\{{}\begin{matrix}x_1=\dfrac{10-2\sqrt{23}}{4}=\dfrac{5-\sqrt{23}}{2}\\x_2=\dfrac{5+\sqrt{23}}{2}\end{matrix}\right.\)
`4x+2x(x+1)=(2x-1)^2`
`<=>4x+2x^2 +2x=4x^2 -4x+1`
`<=>4x+2x^2 +2x-4x^2 +4x-1=0`
`<=>-2x^2 +10x-1=0`
`<=>2x^2 -10x+1=0`
Ta có: `Δ=(-10)^2 -4*1*2=92>0`
\(=>\left[{}\begin{matrix}x_1=\dfrac{-\left(-10\right)+\sqrt{92}}{2\cdot2}=\dfrac{5+\sqrt{23}}{2}\\x_2=\dfrac{-\left(-10\right)-\sqrt{92}}{2\cdot2}=\dfrac{5-\sqrt{23}}{2}\end{matrix}\right.\)
