\(\left(2x-1\right)^6=\left(2x-1\right)^8\)

\(\Rightarrow\left(2x-1\right)^8-\left(2x-1\right)^6=0\)

\(\Rightarrow\left(2x-1\right)^6\left[\left(2x-1\right)^2-1\right]=0\)

\(\Rightarrow\left(2x-1\right)^6\left(2x-1+1\right)\left(2x-1-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}\left(2x-1\right)^6=0\\2x=0\\2x-2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=0\\x=1\end{matrix}\right.\)

Vậy......................

a, \(x^3-5x=0\)

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

\(\Rightarrow\left[{}\begin{matrix}x=0\\x^2-5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=\pm\sqrt{5}\end{matrix}\right.\)

b, \(4x^3-9x=0\)

\(\Rightarrow x\left(4x^2-9\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\4x^2-9=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=\pm\sqrt{\dfrac{9}{4}}\end{matrix}\right.\)

c, \(2x^3-72x=0\)

\(\Rightarrow2x\left(x^2-36\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x^2-36=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=\pm6\end{matrix}\right.\)

d, \(4\left(3x+1\right)^2+16=25\)

\(\Rightarrow4\left(3x+1\right)^2-9=0\)

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

\(\Rightarrow\left[{}\begin{matrix}2\left(3x+1\right)-3=0\\2\left(3x+1\right)+3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}3x+1=\dfrac{3}{2}\\3x+1=-\dfrac{3}{2}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{6}\\x=-\dfrac{5}{6}\end{matrix}\right.\)

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

\(\Rightarrow5x^2-4x^2+8x-4-5=0\)

\(\Rightarrow x^2-x+9x-9=0\)

\(\Rightarrow x\left(x-1\right)+9\left(x-1\right)=0\)

\(\Rightarrow\left(x-1\right)\left(x+9\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-1=0\\x+9=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-9\end{matrix}\right.\)

b, \(\left(x^2-9\right)^2-\left(x-3\right)^2=0\)

\(\Rightarrow\left(x^2-9-x+3\right)\left(x^2-9+x-3\right)=0\)

\(\Rightarrow\left(x^2-x-6\right)\left(x^2+x-12\right)=0\)

\(\Rightarrow\left(x^2-3x+2x-6\right)\left(x^2+4x-3x-12\right)=0\)

\(\Rightarrow\left[x\left(x-3\right)+2\left(x-3\right)\right]\left[x\left(x+4\right)-3\left(x+4\right)\right]=0\)

\(\Rightarrow\left(x-3\right)\left(x+2\right)\left(x+4\right)\left(x-3\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-3=0\\x+2=0\\x+4=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x=-2\\x=-4\end{matrix}\right.\)

c, \(x^3-3x+2=0\)

\(\Rightarrow x^3+2x^2-2x^2-4x+x+2=0\)

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

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

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

\(\Rightarrow\left[{}\begin{matrix}x+2=0\\\left(x-1\right)^2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-2\\x=1\end{matrix}\right.\)

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

\(\Rightarrow5x^2-4x^2+8x-4-5=0\)

\(\Rightarrow x^2+8x-9=0\)

\(\Rightarrow x^2-x+9x-9=0\)

\(\Rightarrow x\left(x-1\right)+9\left(x-1\right)=0\)

\(\Rightarrow\left(x-1\right)\left(x+9\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-1=0\\x+9=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-9\end{matrix}\right.\)

b, \(x^3-3x+2=0\)

\(\Rightarrow x^3+2x^2-2x^2-4x+x+2=0\)

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

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

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

\(\Rightarrow\left[{}\begin{matrix}x+2=0\\\left(x-1\right)^2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-2\\x=1\end{matrix}\right.\)

Bài 1:

a, \(x^2-y^2-2yz-z^2=x^2-\left(y^2+2yz+z^2\right)\)

\(=x^2-\left(y+z\right)^2=\left(x-y-z\right)\left(x+y+z\right)\)

b, \(2xy-x^2-y^2+16=16-\left(x^2-2xy+y^2\right)\)

\(=16-\left(x-y\right)^2=\left(4-x+y\right)\left(4+x-y\right)\)

c, \(3a^2-6ab+3b^2-12c^2\)

\(=3\left(a-b\right)^2-12c^2=3\left[\left(a-b\right)^2-4c^2\right]\)

\(=3\left(a-b+2c\right)\left(a-b-2c\right)\)

Bài 2:

a, \(37,5.8,5-7,5.3,4+6,6.7,5+1,5.37,5\)

\(=37,5.\left(8,5+1,5\right)-7,5.\left(3,4+6,6\right)\)

\(=37,5.10-7,5.10=10.\left(37,5-7,5\right)\)

\(=10.30=300\)

b, \(35^2+40^2-25^2+80.35\)

\(=\left(40^2+2.40.35+35^2\right)-25^2\)

\(=\left(40+35\right)^2-25^2=75^2-25^2\)

\(=\left(75-25\right)\left(75+25\right)=50.100=5000\)