Bài 2:
1: \(\left(2x+5\right)\left(x-4\right)=\left(x-4\right)\left(5-x\right)\)
=>\(\left(2x+5\right)\left(x-4\right)-\left(x-4\right)\left(5-x\right)=0\)
=>\(\left(x-4\right)\left(2x+5-5+x\right)=0\)
=>3x(x-4)=0
=>x(x-4)=0
=>\(\left[{}\begin{matrix}x=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)
2: \(2\left(3x+1\right)^2=\left(3x+1\right)\left(x-2\right)\)
=>\(\left(3x+1\right)\left(6x+2\right)-\left(3x+1\right)\left(x-2\right)=0\)
=>(3x+1)(6x+2-x+2)=0
=>(3x+1)(5x+4)=0
=>\(\left[{}\begin{matrix}3x+1=0\\5x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{3}\\x=-\dfrac{4}{5}\end{matrix}\right.\)
3: \(-5\left(4x-1\right)\left(x-2\right)=2\left(4x-1\right)^2\)
=>\(\left(4x-1\right)\left(8x-2\right)+5\left(4x-1\right)\left(x-2\right)=0\)
=>\(\left(4x-1\right)\left(8x-2+5x-10\right)=0\)
=>(4x-1)(13x-12)=0
=>\(\left[{}\begin{matrix}4x-1=0\\13x-12=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{4}\\x=\dfrac{12}{13}\end{matrix}\right.\)
4: \(\left(x+6\right)\left(5-x\right)=-2\left(5-x\right)\left(7x+8\right)\)
=>\(\left(x+6\right)\left(5-x\right)+2\left(5-x\right)\left(7x+8\right)=0\)
=>\(\left(5-x\right)\left(x+6+14x+16\right)=0\)
=>(5-x)(15x+22)=0
=>\(\left[{}\begin{matrix}5-x=0\\15x+22=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-\dfrac{22}{15}\end{matrix}\right.\)
5: \(\left(2x-1\right)\left(5x-7\right)=\left(2x-1\right)\left(9-7x\right)\)
=>\(\left(2x-1\right)\left(5x-7\right)-\left(2x-1\right)\left(9-7x\right)=0\)
=>(2x-1)(5x-7-9+7x)=0
=>(2x-1)(12x-16)=0
=>(2x-1)(3x-4)=0
=>\(\left[{}\begin{matrix}2x-1=0\\3x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{4}{3}\end{matrix}\right.\)
6: \(\left(x-2\right)\left(7-3x\right)=\left(3x-7\right)\left(8x+32\right)\)
=>(3x-7)(2-x)-(3x-7)(8x+32)=0
=>(3x-7)(2-x-8x-32)=0
=>(3x-7)(-9x-30)=0
=>(3x+10)(3x-7)=0
=>\(\left[{}\begin{matrix}3x+10=0\\3x-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{10}{3}\\x=\dfrac{7}{3}\end{matrix}\right.\)
7: \(\left(2-x\right)\left(x+1\right)=\left(x-2\right)\left(3x+5\right)\)
=>(x-2)(3x+5)-(x+1)(2-x)=0
=>(x-2)(3x+5)+(x+1)(x-2)=0
=>(x-2)(3x+5+x+1)=0
=>(x-2)(4x+6)=0
=>2(2x+3)(x-2)=0
=>(2x+3)(x-2)=0
=>\(\left[{}\begin{matrix}2x+3=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=2\end{matrix}\right.\)
8: \(\left(x-1\right)\left(x+7\right)=\left(1-x\right)\left(3-2x\right)\)
=>(x-1)(2x-3)=(x-1)(x+7)
=>(x-1)(2x-3-x-7)=0
=>(x-1)(x-10)=0
=>\(\left[{}\begin{matrix}x-1=0\\x-10=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=10\end{matrix}\right.\)
9: \(\left(6x-7\right)\left(3x+4\right)=\left(7-6x\right)\left(x-1\right)\)
=>(6x-7)(3x+4)=(6x-7)(1-x)
=>(6x-7)(3x+4-1+x)=0
=>(6x-7)(4x+3)=0
=>\(\left[{}\begin{matrix}6x-7=0\\4x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{6}\\x=-\dfrac{3}{4}\end{matrix}\right.\)
10: \(\left(2x-3\right)\left(5x+1\right)=\left(3-2x\right)\left(x-5\right)\)
=>(2x-3)(5x+1)=(2x-3)(5-x)
=>(2x-3)(5x+1-5+x)=0
=>(2x-3)(6x-4)=0
=>(3x-2)(2x-3)=0
=>\(\left[{}\begin{matrix}3x-2=0\\2x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=\dfrac{3}{2}\end{matrix}\right.\)
