\(1,\\ a,a^3-a^2c+a^2b-abc=a^2\left(a-c\right)+ab\left(a-c\right)=a\left(a+b\right)\left(a-c\right)\\ b,\left(x^2+1\right)^2-4x^2=\left(x^2-2x+1\right)\left(x^2+2x+1\right)=\left(x+1\right)^2\left(x-1\right)^2\\ c,x^2-10x-9y^2+25=\left(x-5\right)^2-9y^2=\left(x-3y-5\right)\left(x+3y-5\right)\\ d,4x^2-36x+56=4\left(x^2-9x+14\right)=4\left(x-7\right)\left(x-2\right)\)
\(2,\\ a,\left(3x+4\right)^2-\left(3x-1\right)\left(3x+1\right)=49\\ \Leftrightarrow9x^2+24x+16-9x^2+1=49\\ \Leftrightarrow24x=32\\ \Leftrightarrow x=\dfrac{4}{3}\\ b,x^2-4x+4=9\left(x-2\right)\\ \Leftrightarrow\left(x-2\right)^2-9\left(x-2\right)=0\\ \Leftrightarrow\left(x-2\right)\left(x-11\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=11\end{matrix}\right.\\ c,x^2-25=3x-15\\ \Leftrightarrow x^2-3x-10=0\\ \Leftrightarrow\left(x-5\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)
Bài 2:
b: ta có: \(x^2-4x+4=9\left(x-2\right)\)
\(\Leftrightarrow\left(x-2\right)\left(x-11\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=11\end{matrix}\right.\)
