a: \(\left(2x-1\right)^2=49\)
=>\(\left[\begin{array}{l}2x-1=7\\ 2x-1=-7\end{array}\right.\Rightarrow\left[\begin{array}{l}2x=8\\ 2x=-6\end{array}\right.\Rightarrow\left[\begin{array}{l}x=4\\ x=-3\end{array}\right.\)
b: (x-2)(3x+5)=(2x-4)(x+1)
=>(x-2)(3x+5)=(x-2)(2x+2)
=>(x-2)(3x+5-2x-2)=0
=>(x-2)(x+3)=0
=>\(\left[\begin{array}{l}x-2=0\\ x+3=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=2\\ x=-3\end{array}\right.\)
c: \(\left(2x+5\right)\left(x-4\right)=\left(x-5\right)\left(4-x\right)\)
=>\(\left(2x+5\right)\left(x-4\right)=\left(-x+5\right)\left(x-4\right)\)
=>\(\left(2x+5\right)\left(x-4\right)-\left(-x+5\right)\left(x-4\right)=0\)
=>(x-4)(2x+5+x-5)=0
=>3x(x-4)=0
=>x(x-4)=0
=>x=0 hoặc x=4
d: \(2\left(9x^2+6x+1\right)=\left(3x+1\right)\left(x-2\right)\)
=>\(2\left(3x+1\right)^2=\left(3x+1\right)\left(x-2\right)\)
=>\(\left(3x+1\right)\left(6x+2-x+2\right)=0\)
=>(3x+1)(5x+4)=0
=>\(\left[\begin{array}{l}3x+1=0\\ 5x+4=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-\frac13\\ x=-\frac45\end{array}\right.\)
e: \(\left(2x+7\right)^2=9\left(x+2\right)^2\)
=>\(\left(3x+6\right)^2=\left(2x+7\right)^2\)
=>\(\left(3x+6\right)^2-\left(2x+7\right)^2=0\)
=>(3x+6+2x+7)(3x+6-2x-7)=0
=>(x-1)(5x+13)=0
=>\(\left[\begin{array}{l}x-1=0\\ 5x+13=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=1\\ x=-\frac{13}{5}\end{array}\right.\)
f: \(27x^2\left(x+3\right)-12\left(x^2+3x\right)=0\)
=>\(27x^2\left(x+3\right)-12x\left(x+3\right)=0\)
=>x(x+3)(27x-12)=0
=>x(x+3)(9x-4)=0
=>\(\left[\begin{array}{l}x=0\\ x+3=0\\ 9x-4=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x=-3\\ x=\frac49\end{array}\right.\)
g: \(16x^2-8x+1=4\left(x+3\right)\left(4x-1\right)\)
=>\(\left(4x-1\right)^2-\left(4x-1\right)\left(4x+12\right)=0\)
=>(4x-1)(4x-1-4x-12)=0
=>-13(4x-1)=0
=>4x-1=0
=>4x=1
=>x=1/4
h: \(\left(5x^2-2x+10\right)^2=\left(3x^2+10x-8\right)^2\)
=>\(\left(5x^2-2x+10-3x^2-10x+8\right)\left(5x^2-2x+10+3x^2+10x-8\right)=0\)
=>\(\left(2x^2-12x+18\right)\left(8x^2+8x+2\right)=0\)
=>\(2\left(x^2-6x+9\right)\cdot2\cdot\left(4x^2+4x+1\right)=0\)
=>\(\left(x-3\right)^2\cdot\left(2x+1\right)^2=0\)
=>\(\left[\begin{array}{l}x-3=0\\ 2x+1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=3\\ x=-\frac12\end{array}\right.\)
