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1. \(\dfrac{1}{x}-\dfrac{2}{x+1}=\dfrac{3}{x^2+x}\)
\(\Leftrightarrow\dfrac{x+1}{x^2+x}-\dfrac{2x}{x^2+x}=\dfrac{3}{x^2+x}\)
\(\Rightarrow x+1-2x=3\)
\(\Leftrightarrow1-x=3\)
\(\Leftrightarrow-x=2\\ \Leftrightarrow x=-2\)
Vậy phương trình có nghiệm duy nhất \(x=-2\)
2. \(\dfrac{x+2}{x-2}-\dfrac{1}{x}=\dfrac{2}{x\left(x-2\right)}\)
\(\Leftrightarrow\dfrac{x^2+2x}{x\left(x-2\right)}-\dfrac{x-2}{x\left(x-2\right)}=\dfrac{2}{x\left(x-2\right)}\)
\(\Rightarrow x^2+2x-x+2=2\)
\(\Leftrightarrow x^2+x+2=2\\ \Leftrightarrow x^2+x=0\)
\(\Leftrightarrow x\left(x+1\right)=0 \)
\(\Leftrightarrow x=0\) hoặc x + 1= 0
⇔ x = 0 hoặc x= -1
Vậy phương trình có tập nghiệm là S={0;-1}
1) ĐKXĐ: \(x\notin\left\{0;-1\right\}\)
Ta có: \(\dfrac{1}{x}-\dfrac{2}{x+1}=\dfrac{3}{x^2+x}\)
\(\Leftrightarrow\dfrac{x+1}{x\left(x+1\right)}-\dfrac{2x}{x\left(x+1\right)}=\dfrac{3}{x\left(x+1\right)}\)
Suy ra: \(x+1-2x=3\)
\(\Leftrightarrow-x+1=3\)
\(\Leftrightarrow-x=2\)
hay x=-2(thỏa ĐK)
Vậy: S={-2}
=>(x^2+1)^2+x^2/x*(x^2+1)=5/2
=>\(\dfrac{\left(x^2+1\right)^2+x^2}{x\left(x^2+1\right)}=\dfrac{5}{2}\)
=>\(2\left(x^4+2x^2+1+x^2\right)=5\left(x^3+x\right)\)
=>2x^4+6x^2+2-5x^3-5x=0
=>2x^4-5x^3+6x^2-5x+2=0
=>2x^4-2x^3-3x^3+3x^2+3x^2-3x-2x+2=0
=>(x-1)(2x^3-3x^2+3x-2)=0
=>(x-1)(2x^3-2x^2-x^2+x+2x-2)=0
=>(x-1)^2*(2x^2-x+2)=0
=>x-1=0
=>x=1
1: Ta có: \(\dfrac{x+2}{x-2}+\dfrac{2}{x+2}=\dfrac{x^2}{x^2-4}\)
Suy ra: \(x^2+4x+4+2x-4=x^2\)
\(\Leftrightarrow6x=0\)
hay \(x=0\left(nhận\right)\)
2: Ta có: \(\dfrac{1}{x-6}-\dfrac{2}{x+6}=\dfrac{3x+6}{x^2-36}\)
Suy ra: \(x+6-2x+12=3x+6\)
\(\Leftrightarrow-x-3x=6-18=-12\)
hay \(x=3\left(nhận\right)\)
Lời giải:
1. ĐKXĐ: $x\neq \pm 2$
PT \(\Leftrightarrow \frac{(x+2)^2+2(x-2)}{(x-2)(x+2)}=\frac{x^2}{x^2-4}\)
\(\Leftrightarrow \frac{x^2+6x}{x^2-4}=\frac{x^2}{x^2-4}\)
\(\Rightarrow x^2+6x=x^2\Leftrightarrow x=0\) (tm)
2. ĐKXĐ: $x\neq \pm 6$
PT \(\Leftrightarrow \frac{6+x-2(x-6)}{(x-6)(6+x)}=\frac{3x+6}{x^2-36}\)
\(\Leftrightarrow \frac{18-x}{x^2-36}=\frac{3x+6}{x^2-36}\)
\(\Rightarrow 18-x=3x+6\Leftrightarrow 12=4x\Leftrightarrow x=3\) (tm)
Sửa đề: \(\dfrac{2x-1}{x+2}+\dfrac{3x+2}{x^2+2x}=\dfrac{x+1}{x}\)
ĐKXĐ: \(x\notin\left\{0;-2\right\}\)
\(\dfrac{2x-1}{x+2}+\dfrac{3x+2}{x^2+2x}=\dfrac{x+1}{x}\)
=>\(\dfrac{2x-1}{x+2}+\dfrac{3x+2}{x\left(x+2\right)}=\dfrac{x+1}{x}\)
=>\(x\left(2x-1\right)+3x+2=\left(x+1\right)\left(x+2\right)\)
=>\(2x^2-x+3x+2=x^2+3x+2\)
=>\(2x^2+2x-x^2-3x=0\)
=>\(x^2-x=0\)
=>x(x-1)=0
=>\(\left[{}\begin{matrix}x=0\left(loại\right)\\x=1\left(nhận\right)\end{matrix}\right.\)
\(\dfrac{1}{x^2+2x}+\dfrac{1}{x^2+6x+8}+\dfrac{1}{x^2+10x+24}+\dfrac{1}{x^2+14x+48}=\dfrac{4}{105}\)
\(\Leftrightarrow\dfrac{2}{x\left(x+2\right)}+\dfrac{2}{\left(x+2\right)\left(x+4\right)}+\dfrac{2}{\left(x+4\right)\left(x+6\right)}+\dfrac{2}{\left(x+6\right)\left(x+8\right)}=\dfrac{8}{105}\)
\(\Leftrightarrow\left(\dfrac{1}{x}-\dfrac{1}{x+2}\right)+\left(\dfrac{1}{x+2}-\dfrac{1}{x+4}\right)+\left(\dfrac{1}{x+4}-\dfrac{1}{x+6}\right)+\left(\dfrac{1}{x+6}-\dfrac{1}{x+8}\right)=\dfrac{8}{105}\)
\(\Leftrightarrow\dfrac{1}{x}-\dfrac{1}{x+8}=\dfrac{8}{105}\)
\(\Leftrightarrow\dfrac{8}{x\left(x+8\right)}=\dfrac{8}{105}\)
\(\Leftrightarrow x\left(x+8\right)=105\)
\(\Leftrightarrow x^2+8x-105=0\)
\(\Leftrightarrow x^2-7x+15x-105=0\)
\(\Leftrightarrow x\left(x-7\right)+15\left(x-7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-15\end{matrix}\right.\)
Thử lại ta có nghiệm của phương trình trên là \(x=7\text{v}à\text{x}=15\)
a: =>10x=3(5-3x)
=>10x=15-9x
=>19x=15
=>x=15/19
b: =>\(\dfrac{x\left(x-4\right)+x^2-1}{x\left(x+1\right)}=2\)
=>2x^2+2x=x^2-4x+x^2-1=2x^2-4x-1
=>2x=-4x-1
=>6x=-1
=>x=-1/6
c:=>x(x+2)-x+2=2
=>x^2+2x-x=0
=>x(x+1)=0
=>x=0(loại) hoặc x=-1(nhận)
d: =>x+1+3x=2
=>4x=1
=>x=1/4
e: =>x(x+1)+x(x-3)=2x
=>x^2+x+x^2-3x=2x
=>2x^2-4x=0
=>x=0(nhận) hoặc x=2(nhận)
f: =>2x+6-4x+12=5
=>-2x=-13
=>x=13/2
\(ĐK:x\ne-1;x\ne1\\ PT\Leftrightarrow\dfrac{\dfrac{2x^2+4x+2-x^2+2x-1}{2\left(x+1\right)\left(x-1\right)}}{\dfrac{x-1+x+1}{x-1}}=\dfrac{x-1}{2\left(x+1\right)}\\ \Leftrightarrow\dfrac{x^2+6x+1}{2\left(x-1\right)\left(x+1\right)}\cdot\dfrac{x-1}{2x}=\dfrac{x-1}{2\left(x+1\right)}\\ \Leftrightarrow\dfrac{x^2+6x+1}{4x\left(x+1\right)}=\dfrac{x-1}{2\left(x+1\right)}\\ \Leftrightarrow x^2+6x+1=2x\left(x-1\right)\\ \Leftrightarrow x^2+6x+1=2x^2-2x\\ \Leftrightarrow x^2-8x-1=0\\ \Leftrightarrow\left[{}\begin{matrix}x=4+\sqrt{17}\left(tm\right)\\x=4-\sqrt{17}\left(tm\right)\end{matrix}\right.\)
d: ĐKXĐ: \(x\notin\left\{2;-3\right\}\)
\(\dfrac{1}{x-2}-\dfrac{6}{x+3}=\dfrac{5}{6-x^2-x}\)
=>\(\dfrac{1}{x-2}-\dfrac{6}{x+3}=\dfrac{-5}{\left(x+3\right)\left(x-2\right)}\)
=>\(x+3-6\left(x-2\right)=-5\)
=>x+3-6x+12=-5
=>-5x+15=-5
=>-5x=-20
=>x=4(nhận)
e: ĐKXĐ: x<>-2
\(\dfrac{2}{x+2}-\dfrac{2x^2+16}{x^3+8}=\dfrac{5}{x^2-2x+4}\)
=>\(\dfrac{2}{x+2}-\dfrac{2x^2+16}{\left(x+2\right)\left(x^2-2x+4\right)}=\dfrac{5}{x^2-2x+4}\)
=>\(2\left(x^2-2x+4\right)-2x^2-16=5\left(x+2\right)\)
=>\(2x^2-4x+8-2x^2-16=5x+10\)
=>5x+10=-4x-8
=>9x=-18
=>x=-2(loại)
f: ĐKXĐ: \(x\in\left\{1;-1\right\}\)
\(\dfrac{x+1}{x^2+x+1}-\dfrac{x-1}{x^2-x+1}=\dfrac{2\left(x+2\right)^2}{x^6-1}\)
\(\Leftrightarrow\dfrac{x+1}{x^2+x+1}-\dfrac{x-1}{x^2-x+1}=\dfrac{2\left(x+2\right)^2}{\left(x-1\right)\left(x+1\right)\left(x^2+x+1\right)\left(x^2-x+1\right)}\)
=>\(\dfrac{\left(x+1\right)\left(x^2-x+1\right)\left(x^2-1\right)-\left(x-1\right)\left(x^2+x+1\right)\left(x^2-1\right)}{\left(x-1\right)\left(x+1\right)\left(x^2+x+1\right)\left(x^2-x+1\right)}=\dfrac{2\left(x+2\right)^2}{\left(x-1\right)\left(x+1\right)\left(x^2+x+1\right)\left(x^2-x+1\right)}\)
=>\(\left(x^3+1\right)\left(x^2-1\right)-\left(x^3-1\right)\left(x^2-1\right)=2\left(x^2+4x+4\right)\)
=>\(\left(x^2-1\right)\cdot\left(x^3+1-x^3+1\right)=2\left(x^2+4x+4\right)\)
=>\(2x^2+8x+8=\left(x^2-1\right)\cdot2=2x^2-2\)
=>8x=-10
=>x=-5/4(nhận)
1: Ta có: \(\dfrac{3}{x+2}-\dfrac{x-1}{x^2-4}=\dfrac{2}{x-2}\)
Suy ra: \(3x-6-x+1=2x+4\)
\(\Leftrightarrow2x-5=2x+4\left(vôlý\right)\)
2: Ta có: \(\dfrac{x-5}{2x-3}-\dfrac{x}{2x+3}=\dfrac{1-6x}{4x^2-9}\)
Suy ra: \(\left(x-5\right)\left(2x+3\right)-x\left(2x-3\right)=1-6x\)
\(\Leftrightarrow2x^2-7x-15-2x^2+6x+6x-1=0\)
\(\Leftrightarrow5x=16\)
hay \(x=\dfrac{16}{5}\)
đk : x khác 1 ; -1
<=> \(-x\left(x+1\right)+x^2+2=2\left(x-1\right)\)
\(\Leftrightarrow-x+2=2x-2\Leftrightarrow x=\dfrac{4}{3}\)(tm)
\(\Leftrightarrow-x\left(x+1\right)+x^2+2=2x-2\)
\(\Leftrightarrow-x^2-x+x^2+2-2x+2=0\)
=>-3x+4=0
hay x=4/3(nhận)