\(4\left(x+1\right)^2+\left(2x-1\right)-8\left(x-1\right)\left(x+1\right)=11\)

\(\Leftrightarrow10x-4x^2=0\)

\(\Leftrightarrow2x\left(5-2x\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\5-2x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{5}{2}\end{cases}}}\)

ái tick đúng thang Tran thế xem lại thần kinh

4(x+1)²+(2x-1)²-8(x-1)(x+1)=11

<=>4.(x²+2x+1)+(4x²-4x+1)-8.(x²-1)=11

<=>4x²+8x+4+4x²-4x+1-8x²+8=11

<=>4x+13=11

<=>4x=-2

<=>x=-1/2

  4( x +1)^2 + (2x - 1)^2 - 8( x - 1)( x + 1) = 11 

=>4 ( x^2 + 2x + 1) + 4x^2 - 4x + 4 - 8( x^2 - 1) = 11

=>4x^2 + 8x  + 4 + 4x^2 - 4x + 4 - 8x^2 + 8    = 11 

=> 4x = 11 - 4 - 4 - 8 

=> 4x = -5 

=> x = -5/4

\(4.\left(x+1\right)^2+\left(2x-1\right)^2-8\left(x-1\right)\left(x+1\right)=11\)

\(\Leftrightarrow4\left(x^2+2x+1\right)+\left(4x^2-4x+1\right)-8\left(x^2-1\right)-11=0\)

\(\Leftrightarrow4x^2+8x+4+4x^2-4x+1-8x^2+8-11=0\)

\(\Leftrightarrow4x+1=0\Leftrightarrow4x=-1\Leftrightarrow x=-\frac{1}{4}\)

a: Ta có: \(4\left(x+1\right)^2+\left(2x+1\right)^2-8\left(x-1\right)\left(x+1\right)-11=0\)

\(\Leftrightarrow4x^2+8x+4+4x^2+4x+1-8x^2+8-11=0\)

\(\Leftrightarrow12x=-2\)

hay \(x=-\dfrac{1}{6}\)

b: Ta có: \(\left(x+3\right)^2-\left(x-4\right)\left(x+8\right)-1=0\)

\(\Leftrightarrow x^2+6x+9-x^2-4x+32-1=0\)

\(\Leftrightarrow2x=-40\)

hay x=-20

\(\Leftrightarrow\)4.(x²+2x+1)+4x+4x²+1-8(x²+10x-x-10)=11

\(\Leftrightarrow\)4x²+8x+4+4x+4x²+1-8x²-80x+8x+80=11

\(\Leftrightarrow\)-60x=-74

\(\Leftrightarrow\)x=\(\frac{37}{30}\)

4 ( x + 1 )^2 + ( 2x - 1 )^2 - 8 ( x - 1 ).( x + 1 ) = 11

 <=>4.(x²+2x+1)+(4x²-4x+1)-8.(x²-1)=11

<=>4x²+8x+4+4x²-4x+1-8x²+8=11

<=>4x+13=11

<=>4x=-2

<=>x=-1/2

Nhân ra rút gọn hết x^2 chỉ còn x thôi rồi gải bt