1/3(x-1)+2/x+1
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
1) ĐKXĐ: \(x\notin\left\{1;-1\right\}\)
Ta có: \(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}=\dfrac{4}{x^2-1}\)
\(\Leftrightarrow\dfrac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}-\dfrac{\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}=\dfrac{4}{\left(x-1\right)\left(x+1\right)}\)
Suy ra: \(x^2+2x+1-\left(x^2-2x+1\right)=4\)
\(\Leftrightarrow x^2+2x+1-x^2+2x-1=4\)
\(\Leftrightarrow4x=4\)
hay x=1(loại)
Vậy: \(S=\varnothing\)
2) ĐKXĐ: \(x\notin\left\{2;-2\right\}\)
Ta có: \(\dfrac{x+2}{x-2}+\dfrac{x}{x+2}=2\)
\(\Leftrightarrow\dfrac{\left(x+2\right)^2}{\left(x-2\right)\left(x+2\right)}+\dfrac{x\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{2\left(x^2-4\right)}{\left(x-2\right)\left(x+2\right)}\)
Suy ra: \(x^2+4x+4+x^2-2x=2x^2-8\)
\(\Leftrightarrow2x^2+2x+4-2x^2-8=0\)
\(\Leftrightarrow2x-4=0\)
\(\Leftrightarrow2x=4\)
hay x=2(loại)
Vậy: \(S=\varnothing\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: =>9x^2+12x+4-9x^2+12x-4=5x+38
=>24x=5x+38
=>19x=38
=>x=2
e: =>x^3+1-2x=x^3-x
=>-2x+1=-x
=>-x=-1
=>x=1
f: =>x^3-6x^2+12x-8+9x^2-1=x^3+3x^2+3x+1
=>12x-9=3x+1
=>9x=10
=>x=10/9
b: \(\Leftrightarrow3x^2-12x+12+9x-9=3x^2+3x-9\)
=>-3x+3=3x-9
=>-6x=-12
=>x=2
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{1-x}{1+x}+3=\frac{2x+3}{x+1}\left(ĐKXĐ:x\ne-1\right)\)
\(\Leftrightarrow\frac{1-x}{x+1}+\frac{3\left(x+1\right)}{x+1}=\frac{2x+3}{x+1}\)
\(\Leftrightarrow\frac{1-x+3\left(x+1\right)}{x+1}=\frac{2x+3}{x+1}\)
\(\Rightarrow1-x+3\left(x+1\right)=2x+3\)
\(\Leftrightarrow1-x+3x+3=2x+3\)
\(\Leftrightarrow2x+4=2x+3\)
\(\Leftrightarrow0x=-1\)(vô nghiệm)
Vậy phương trình vô nghiệm.
\(\frac{\left(x+2\right)^2}{2x-3}-1=\frac{x^2-10}{2x-3}\left(ĐKXĐ:x\ne\frac{3}{2}\right)\)
\(\Leftrightarrow\frac{x^2+4x+4}{2x-3}-\frac{2x-3}{2x-3}=\frac{x^2-10}{2x-3}\)
\(\Leftrightarrow\frac{x^2+4x+4-2x+3}{2x-3}=\frac{x^2-10}{2x-3}\)
\(\Rightarrow x^2+4x+4-2x+3=x^2-10\)
\(\Leftrightarrow2x+7=-10\)
\(\Leftrightarrow2x=-17\)
\(\Leftrightarrow x=\frac{-17}{2}\)(thỏa mãn ĐKXĐ)
Vậy phương trình có nghiệm duy nhất : \(x=\frac{-17}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: Ta có: \(\left(x+1\right)^3-\left(x+2\right)\left(x-1\right)^2-3\left(x-3\right)\left(x+3\right)=5\)
\(\Leftrightarrow x^3+3x^2+3x+1-\left(x+2\right)\left(x^2-2x+1\right)-3\left(x^2-9\right)=5\)
\(\Leftrightarrow x^3+3x^2+3x+1-\left(x^3-2x^2+x+2x^2-4x+2\right)-3\left(x^2-9\right)=5\)
\(\Leftrightarrow x^3+3x^2+3x+1-x^3+3x-2-3x^2+9=5\)
\(\Leftrightarrow6x=-3\)
hay \(x=-\dfrac{1}{2}\)
b: Ta có: \(\left(x+1\right)^3+\left(x-1\right)^3=\left(x+2\right)^3+\left(x-2\right)^3\)
\(\Leftrightarrow x^3+3x^2+3x+1+x^3-3x^2+3x-1=x^3+6x^2+12x+8+x^3-6x^2+12x-8\)
\(\Leftrightarrow2x^3+6x=2x^3+24x\)
\(\Leftrightarrow x=0\)
c: Ta có: \(\left(x+1\right)^3-\left(x-1\right)^3-6\left(x-1\right)^2=-10\)
\(\Leftrightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1-6x^2+12x-1=-10\)
\(\Leftrightarrow12x=-11\)
hay \(x=-\dfrac{11}{12}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: =>x-2/5=3/4:1/3=3/4*3=9/4
=>x=9/4+2/5=45/20+8/20=53/20
b: =>x-2/3=7/3:4/5=7/3*5/4=35/12
=>x=35/12+2/3=43/12
c: 1/3(x-2/5)=4/5
=>x-2/5=4/5*3=12/5
=>x=12/5+2/5=14/5
d: =>2/3x-1/3-1/4x+1/10=7/3
=>5/12x-7/30=7/3
=>5/12x=7/3+7/30=77/30
=>x=77/30:5/12=154/25
e: \(\Leftrightarrow x\cdot\dfrac{3}{7}-\dfrac{2}{7}+\dfrac{1}{2}-\dfrac{5}{4}x+\dfrac{5}{2}=0\)
=>\(x\cdot\dfrac{-23}{28}=\dfrac{2}{7}-3=\dfrac{-19}{7}\)
=>x=19/7:23/28=76/23
f: =>1/2x-3/2+1/3x-4/3+1/4x-5/4=1/5
=>13/12x=1/5+3/2+4/3+5/4=257/60
=>x=257/65
i: =>x^2-2/5x-x^2-2x+11/4=4/3
=>-12/5x=4/3-11/4=-17/12
=>x=17/12:12/5=85/144
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(2\dfrac{3}{4}-x=\dfrac{3}{4}\)
\(\Rightarrow\dfrac{11}{4}-x=\dfrac{3}{4}\)
\(\Rightarrow x=\dfrac{11}{4}-\dfrac{3}{4}=\dfrac{8}{4}=2\)
b) \(x:\dfrac{5}{6}=-\dfrac{3}{5}\)
\(\Rightarrow x=-\dfrac{3}{5}.\dfrac{5}{6}=-\dfrac{15}{30}=-\dfrac{1}{2}\)
c) \(1\dfrac{1}{3}+\dfrac{2}{3}:x=1\)
\(\Rightarrow\dfrac{2}{3}:x=1-1\dfrac{1}{3}\)
\(\Rightarrow\dfrac{2}{3}:x=-\dfrac{1}{3}\)
\(\Rightarrow x=\dfrac{2}{3}:-\dfrac{1}{3}\)
\(\Rightarrow x=-2\)
d) \(x-\dfrac{1}{9}=\dfrac{8}{3}\)
\(\Rightarrow x=\dfrac{8}{3}+\dfrac{1}{9}\)
\(\Rightarrow x=\dfrac{25}{9}\)
e) \(\dfrac{1}{2}x+650\%x-x=-6\)
\(\Rightarrow\dfrac{1}{2}x+\dfrac{13}{2}x-x=-6\)
\(\Rightarrow x\left(\dfrac{1}{2}+\dfrac{13}{2}-1\right)-6\)
\(\Rightarrow6x=-6\)
\(\Rightarrow x=\dfrac{-6}{6}=-1\)
g) \(2\left(x-\dfrac{1}{2}\right)+3\left(-1+\dfrac{x}{3}\right)=x\left(\dfrac{2}{x}-1\right)\) \(\text{Đ}K:x\ne0\)
\(\Rightarrow2x-1-3+x=2-x\)
\(\Rightarrow3x-4=2-x\)
\(\Rightarrow3x+x=2+4\)
\(\Rightarrow4x=6\)
\(\Rightarrow x=\dfrac{6}{4}=\dfrac{3}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
a. $x(3x+1)+(x-1)^2-(2x+1)(2x-1)=0$
$\Leftrightarrow (3x^2+x)+(x^2-2x+1)-(4x^2-1)=0$
$\Leftrightarrow 3x^2+x+x^2-2x+1-4x^2+1=0$
$\Leftrightarrow (3x^2+x^2-4x^2)+(x-2x)+(1+1)=0$
$\Leftrightarrow -x+2=0$
$\Leftrightarrow x=2$
b.
$(x+1)^3+(2-x)^3-9(x-3)(x+3)=0$
$\Leftrightarrow [(x+1)+(2-x)][(x+1)^2-(x+1)(2-x)+(2-x)^2]-9(x-3)(x+3)=0$
$\Leftrightarrow 3[x^2+2x+1-(x-x^2+2)+(x^2-4x+4)]-9(x-3)(x+3)=0$
$\Leftrightarrow 3(3x^2-3x+3)-9(x^2-9)=0$
$\Leftrightarrow 9(x^2-x+1)-9(x^2-9)=0$
$\Leftrightarrow 9(x^2-x+1-x^2+9)=0$
$\Leftrightarrow 9(-x+10)=0$
$\Leftrightarrow -x+10=0\Leftrightarrow x=10$
c.
$(x-1)^3-(x+3)(x^2-3x+9)+3x^2=25$
$\Leftrightarrow (x^3-3x^2+3x-1)-(x^3+3^3)+3x^2=25$
$\Leftrightarrow x^3-3x^2+3x-1-x^3-27+3x^2=25$
$\Leftrightarrow (x^3-x^3)+(-3x^2+3x^2)+3x-28=25$
$\Leftrightarrow 3x-28=25$
$\Leftrightarrow x=\frac{53}{3}$
d.
$(x+2)^3-(x+1)(x^2-x+1)-6(x-1)^2=23$
$\Leftrightarrow (x^3+6x^2+12x+8)-(x^3+1)-6(x^2-2x+1)=23$
$\Leftrightarrow x^3+6x^2+12x+8-x^3-1-6x^2+12x-6=23$
$\Leftrightarrow (x^3-x^3)+(6x^2-6x^2)+(12x+12x)+(8-1-6)=23$
$\Leftrightarrow 24x+1=23$
$\Leftrgihtarrow 24x=22$
$\Leftrightarrow x=\frac{11}{12}$