Giải pt
(2x +1)2(x+1)x=105
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(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}
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
(2x + 1)2(x + 1)x = 105
<=> 4x4 + 8x3 + 5x2 + x -105 = 0
<=> (4x4 + 4x3 - 20x2) + (4x3 + 4x2 - 20x) + ( 21x2 + 21x - 105)
<=> (x2 + x - 5)(4x2 + 4x + 21) = 0
Tới đâu thì đơn giản rồi nên tự tiếp nhé