giúp

(2x+1)(x+1)^2 (2x+3)=18

(2x+1)(2x+3)(x^2+2x+1)=18

(2x+1)(2x+3)(x^2+2x+1)-18=0

(4x^2+8x+3)(x^2+2x+1)-18=0

[4(x^2+2x)+3](x^2+2x+1)-18=0

dat x^2+2x=y

=>(4y+3)(y+1)-18=0

4y^2+7y-15=0

4y(y+3)-5(y+3)=0

(y+3)(4y-5)=0

y+3=0 hoac 4y-5=0

y=-3,y=5/4

th1 x^2+2x=-3

x^2+2x+3=0

=>x vo nghiem vi x^2+2x+3>0 voi moi x

th2 x^2+2x=5/4

x^2+2x-5/4=0

4x^2+8x-5=0

2x(2x-1)+5(2x-1)=0

(2x-1)(2x+5)=0

2x-1=0 hoac 2x+5=0

x=1/2,x=-5/2

S={1/2;-5/2}

em mới học lớp 6 thui

ĐKXĐ:\(x\ne\pm1\)

\(\dfrac{4x+5}{x-1}+\dfrac{2x-1}{x+1}=6\\ \Leftrightarrow\dfrac{\left(x+1\right)\left(4x+5\right)}{\left(x+1\right)\left(x-1\right)}+\dfrac{\left(x-1\right)\left(2x-1\right)}{\left(x+1\right)\left(x-1\right)}=6\\ \Leftrightarrow\dfrac{\left(x+1\right)\left(4x+5\right)+\left(x-1\right)\left(2x-1\right)}{\left(x+1\right)\left(x-1\right)}=6\)

\(\Leftrightarrow4x^2+4x+5x+5+2x^2-2x-x+1=6\left(x^2-1\right)\\ \Leftrightarrow6x^2+6x+6=6x^2-6\\ \Leftrightarrow6x=-12\\ \Leftrightarrow x=-2\left(tm\right)\)

\(\dfrac{4x+5}{x-1}+\dfrac{2x-1}{x+1}=6\)

\(\dfrac{\left(4x+5\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}+\dfrac{\left(2x-1\right)\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{\left(4x+5\right)\left(x+1\right)+\left(2x-1\right)\left(x-1\right)}{x^2-1}\)

\(\dfrac{4x^2+9x+5+2x^2-3x+1}{x^2-1}=\dfrac{6x^2+6x+6}{x^2-1}=6\)

\(\Rightarrow6x^2+6x+6=6\left(x^2-1\right)=6x^2-6\)

\(\Rightarrow6x+12=0\Rightarrow x=-2\)

\(\frac{x+2}{x-2}-\frac{1}{x}=\frac{x^2+3}{x^2-2x}\)

<=> \(\frac{x+2}{x-2}-\frac{1}{x}=\frac{x^2+3}{x\left(x-2\right)}\)

<=> \(\frac{x\left(x+2\right)-x+2}{x\left(x-2\right)}=\frac{x^2+3}{x\left(x-2\right)}\)

=> x²+2x-x+2=x²+3

<=>x=3