x^5+x^3+x^2+1/x^3+x^2+1
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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)
Bạn cần viết đề bài bằng công thức toán để được hỗ trợ tốt hơn.
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a: \(\dfrac{x+5}{x-1}+\dfrac{8}{x^2-4x+3}=\dfrac{x+1}{x-3}\)
=>(x+5)(x-3)+8=x^2-1
=>x^2+2x-15+8=x^2-1
=>2x-7=-1
=>x=3(loại)
b: \(\dfrac{x-4}{x-1}-\dfrac{x^2+3}{1-x^2}+\dfrac{5}{x+1}=0\)
=>(x-4)(x+1)+x^2+3+5(x-1)=0
=>x^2-3x-4+x^2+3+5x-5=0
=>2x^2+2x-6=0
=>x^2+x-3=0
=>\(x=\dfrac{-1\pm\sqrt{13}}{2}\)
e: =>x^2-2x+1+2x+2=5x+5
=>x^2+3=5x+5
=>x^2-5x-2=0
=>\(x=\dfrac{5\pm\sqrt{33}}{2}\)
g: (x-3)(x+4)*x=0
=>x=0 hoặc x-3=0 hoặc x+4=0
=>x=0;x=3;x=-4
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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)
1.
$(x-2)(x-5)=(x-3)(x-4)$
$\Leftrightarrow x^2-7x+10=x^2-7x+12$
$\Leftrightarrow 10=12$ (vô lý)
Vậy pt vô nghiệm.
2.
$(x-7)(x+7)+x^2-2=2(x^2+5)$
$\Leftrightarrow x^2-49+x^2-2=2x^2+10$
$\Leftrightarrow 2x^2-51=2x^2+10$
$\Leftrightarrow -51=10$ (vô lý)
Vậy pt vô nghiệm.
3.
$(x-1)^2+(x+3)^2=2(x-2)(x+2)$
$\Leftrightarrow (x^2-2x+1)+(x^2+6x+9)=2(x^2-4)$
$\Leftrightarrow 2x^2+4x+10=2x^2-8$
$\Leftrightarrow 4x+10=-8$
$\Leftrightarrow 4x=-18$
$\Leftrightarrow x=-4,5$
4.
$(x+1)^2=(x+3)(x-2)$
$\Leftrightarrow x^2+2x+1=x^2+x-6$
$\Leftrightarrow x=-7$