\(2,\\ a,\Leftrightarrow9x^2+24x+16-9x^2+1=49\\ \Leftrightarrow24x=32\\ \Leftrightarrow x=\dfrac{4}{3}\\ b,\Leftrightarrow\left(x-2\right)^2-9\left(x-2\right)=0\\ \Leftrightarrow\left(x-2-9\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=11\\x=2\end{matrix}\right.\\ c,x^2-25=3x-15\\ \Leftrightarrow\left(x-5\right)\left(x+5\right)-3\left(x-5\right)=0\\ \Leftrightarrow\left(x-5\right)\left(x+5-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)
\(1,\\ a,=a^2\left(a-c\right)+ab\left(a-c\right)=a\left(a+b\right)\left(a-c\right)\\ b,\left(x^2+1\right)^2-4x^2=\left(x^2+2x+1\right)\left(x^2-2x+1\right)=\left(x+1\right)^2\left(x-1\right)^2\\ c,=x^2-10x+25-9y^2=\left(x-5\right)^2-9y^2=\left(x-3y-5\right)\left(x+3y-5\right)\\ d,4x^2-36x+56=4x^2-8x-28x+56=4\left(x-2\right)\left(x-7\right)\)
Câu 5:
a: Ta có: \(x\left(x-2\right)+x-2=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
b: Ta có: \(5x\left(x-3\right)-x+3=0\)
\(\Leftrightarrow\left(x-3\right)\left(5x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{5}\end{matrix}\right.\)
Câu 3:
a: \(y^2+2y+1=\left(y+1\right)^2\)
b: \(9x^2-6xy+y^2=\left(3x-y\right)^2\)
c: \(25a^2+20ab+4b^2=\left(5a+2b\right)^2\)
