\(x+\frac{16}{49}+x+\frac{18}{47}=x+\frac{20}{45}-1\)

\(2x+\frac{16}{49}+\frac{18}{47}-x-\frac{20}{45}=-1\)

\(x+\frac{5494}{20727}=-1\)

\(x=-1-\frac{5494}{20727}=-\frac{26221}{20707}\)

(x+16)/49.... cơ Dung à 

\(=\left(\left(x+2\right)\left(x+5\right)\right)\left(\left(x+3\right)\left(x+4\right)\right)-24\)

\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)

Đặt \(x^2+7x+10=y\)ta có:

    \(y\left(y+2\right)-24\)

\(=y^2+2y-24\)

\(=y^2-4y+6y-24\)

\(=y\left(y-4\right)+6\left(y-4\right)\)

\(=\left(y-4\right)\left(y+6\right)\)

\(=\left(x^2+7x+10-4\right)\left(x^2+7x+10+6\right)\)

\(=\left(x^2+7x+6\right)\left(x^2+7x+16\right)\)

\(=\left(x^2+x+6x+6\right)\left(x^2+7x+16\right)\)

\(=\left(x\left(x+1\right)+6\left(x+1\right)\right)\left(x^2+7x+16\right)\)

\(=\left(x+1\right)\left(x+6\right)\left(x^2+7x+16\right)\)

a) \(\Leftrightarrow\left(2x-2\right)^2-\left(3x+6\right)^2=0\)

    \(\Leftrightarrow\left(\left(2x-2\right)+\left(3x+6\right)\right)\left(\left(2x-2\right)-\left(3x+6\right)\right)=0\)

     \(\Leftrightarrow\left(5x+4\right)\left(-x-8\right)=0\)

    \(\Leftrightarrow\orbr{\begin{cases}5x+4=0\\-x-8=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-\frac{4}{5}\\x=-8\end{cases}}}\)

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

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

  \(\Leftrightarrow4x+13=11\)

 \(\Leftrightarrow x=-\frac{1}{2}\)

a) \(4\left(x-1\right)^2-9\left(x+2\right)^2=0\)

\(\Leftrightarrow\left[2\left(x-1\right)\right]^2-\left[3\left(x+2\right)\right]^2=0\)

\(\Leftrightarrow\left(2x-2\right)^2-\left(3x+6\right)^2=0\)

\(\Leftrightarrow\left(2x-2+3x+6\right)\left(2x-2-3x-6\right)=0\)

\(\Leftrightarrow\left(5x+4\right)\left(-x-8\right)=0\)

\(\Leftrightarrow\hept{\begin{cases}5x+4=0\\-x-8=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=-\frac{4}{5}\\x=-8\end{cases}}}\)

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

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

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

\(\Leftrightarrow4x+13=11\)

\(\Leftrightarrow4x=-2\)

\(\Leftrightarrow x=-\frac{2}{4}=-\frac{1}{2}\)

(Nhớ k cho mình với nhé!)

\(4x^4+4x^3-x^2-x\)

\(=4x^3\left(x+1\right)-x\left(x+1\right)\)

\(=\left(x+1\right)\left(4x^3-x\right)\)

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

\(=x\left(x+1\right)\left[\left(2x\right)^2-1\right]\)

\(=x\left(x+1\right)\left(2x+1\right)\left(2x-1\right)\)

(Nhớ k cho mình với nhá!)

vậy x= 0 ,25

=>x=0,25

nhớ cho mik nha