ab(x² + y²) + xy(a² + b²)

= abx² + aby² + xya² + xyb² 

= ax(ay + bx) + by(ay + bx) 

= (ax + by)(ay + bx) 

ab(x²+y²)−xy(a²+b²)ab(x²+y²)−xy(a²+b²)

=abx²+aby²−a²xy−b²xy

=(abx²−b²xy)−(a²xy−aby²)

=bx(ax−by)−ay(ax−by)

=(ax−by)(bx−ay)

\(1,\left(x^3-4x^2\right)-\left(x-4\right)=0\)

=> \(x^2\left(x-4\right)-\left(x-4\right)=0\)

=>  \(\left(x^2-1\right)\left(x-4\right)=0\)

=>  \(\left(x+1\right)\left(x-1\right)\left(x-4\right)=0\)

=>  \(\orbr{\begin{cases}x=\pm1\\x=4\end{cases}}\)

\(2,\left(4x^2-25\right)^2-9\left(2x-5\right)^2=0\)

=>  \(\left[\left(2x-5\right)\left(2x+5\right)\right]^2-\left[3\left(2x-5\right)\right]^2=0\)

=>  \(\left[\left(2x-5\right)\left(2x+5\right)-3\left(2x-5\right)\right]\left[\left(2x-5\right)\left(2x+5\right)+3\left(2x-5\right)\right]=0\)

=>  \(\left[\left(2x-5\right)2\left(x+1\right)\right]\left[\left(2x-5\right)2\left(x+4\right)\right]=0\)

=>   \(4\left(2x-5\right)^2\left(x+1\right)\left(x+4\right)=0\)

=>  \(\orbr{\begin{cases}2x-5=0\\x+1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{2}\\x=-1\end{cases}}}\)

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

\(c,x^4-16x^2=0\)

=>  \(x^2\left(x^2-4^2\right)=0\)

=>  \(x^2\left(x-4\right)\left(x+4\right)=0\)

=>  \(\orbr{\begin{cases}x=0\\x=\pm4\end{cases}}\)

\(d,x^2+x=6\)

=>  \(x^2+x-6=0\)

=>  \(x^2+3x-2x-6=0\)

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

=>  \(\orbr{\begin{cases}x+3=0\\x-2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-3\\x=2\end{cases}}}\)

\(e,x^2-7x=-12\)

=>  \(x^2-7x+12=0\)

=>  \(x^2-3x-4x+12=0\)

=>  \(\left(x-3\right)\left(x-4\right)=0\)

=>  \(\orbr{\begin{cases}x-3=0\\x-4=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\x=4\end{cases}}}\)

\(f,x^3-x^2=-x\)

=>  \(x^3-x^2+x=0\)

=>  \(x\left(x^2-x+1\right)=0\)

=>   \(x=0\)   vì \(\left(x^2-x+1\right)>0\)      

Mình ko biết nhé :)

em mới lớp 4 thôi nên ko bt

lớp bốn thì bài dễ hơn nhiều

$2x\left(x-5\right)-x\left(3+2x\right)=26$

$\iff$ $x\left[2\left(x-5\right)-\left(3+2x\right)\right]=26$

$\iff$ $x\left(2x-10-3-2x\right)=26$

$\iff$ $x\left(-13\right)=26$

$\mathrm{x=26:(-13)}$

$\mathrm{x=2}$

\(\left(2x+5\right)^2=9\left(x-1\right)^2\)

\(\Rightarrow\left(2x+5\right)^2=\left(3x-3\right)^2\)

\(\Rightarrow\left(2x+5\right)^2-\left(3x-3\right)^2=0\)

\(\Rightarrow\left(2x=5-3x+3\right)\left(2x+5+3x-3\right)=0\)

\(\Rightarrow\left(8-x\right)\left(5x+2\right)=0\)

\(\Rightarrow\orbr{\begin{cases}8-x=0\\5x+2=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=8\\x=\frac{-2}{5}\end{cases}}\)