Bài 3:
a: (x+2)(x-3)=6
=>\(x^2-x-6=6\)
=>\(x^2-x-12=0\)
=>(x-4)(x+3)=0
=>x=4 hoặc x=-3
b: (2x+1)(x+2)=2
=>\(2x^2+4x+x+2=2\)
=>\(2x^2+5x=0\)
=>x(2x+5)=0
=>\(\left[\begin{array}{l}x=0\\ 2x+5=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x=-\frac52\end{array}\right.\)
c:(x+4)(x+3)=2
=>\(x^2+3x+4x+12-2=0\)
=>\(x^2+7x+10=0\)
=>(x+2)(x+5)=0
=>x=-2 hoặc x=-5
d: (2x+5)(3-x)=14
=>\(6x-2x^2+15-5x=14\)
=>\(-2x^2+x+1=0\)
=>\(2x^2-x-1=0\)
=>\(2x^2-2x+x-1=0\)
=>(x-1)(2x+1)=0
=>x=1 hoặc x=-1/2
Bài 4:
a: \(\left(x+5\right)^2-\left(x+5\right)\left(2x+3\right)=0\)
=>(x+5)(x+5-2x-3)=0
=>(x+5)(2-x)=0
=>\(\left[\begin{array}{l}x+5=0\\ 2-x=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-5\\ x=2\end{array}\right.\)
b: \(\left(-3x+5\right)\left(3x-2\right)+\left(3x-2\right)^2=0\)
=>\(\left(3x-2\right)\left(-3x+5+3x-2\right)=0\)
=>3(3x-2)=0
=>3x-2=0
=>3x=2
=>x=2/3
c: \(\left(4-5x\right)\left(3x+\frac12\right)+\left(3x+\frac12\right)^2=0\)
=>\(\left(3x+\frac12\right)\cdot\left(4-5x+3x+\frac12\right)=0\)
=>\(\left(3x+\frac12\right)\left(\frac92-2x\right)=0\)
=>\(\left[\begin{array}{l}3x+\frac12=0\\ \frac92-2x=0\end{array}\right.\Rightarrow\left[\begin{array}{l}3x=-\frac12\\ 2x=\frac92\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-\frac16\\ x=\frac94\end{array}\right.\)
d: \(\left(\frac37+5x\right)\left(-3x-7\right)-\left(-3x-7\right)^2=0\)
=>\(\left(-3x-7\right)\left(5x+\frac37+3x+7\right)=0\)
=>(-3x-7)*(8x+52/7)=0
=>(3x+7)*(8x+52/7)=0
=>\(\left[\begin{array}{l}3x+7=0\\ 8x+\frac{52}{7}=0\end{array}\right.\Rightarrow\left[\begin{array}{l}3x=-7\\ 8x=-\frac{52}{7}\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-\frac73\\ x=-\frac{52}{7}:8=-\frac{52}{56}=-\frac{13}{14}\end{array}\right.\)
