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bài 2 giải các phương trình sau

b,\(\dfrac{2\left(3-7x\right)}{1+x}=\dfrac{1}{2}\)               m,\(\dfrac{3x-1}{x+1}=\dfrac{2x+1}{x-1}\)

d,\(\dfrac{3x-14}{x+5}=\dfrac{2}{3}\)                   p,\(\dfrac{4x+7}{x-1}=\dfrac{12x+5}{3x+4}\)

f,\(\dfrac{6}{x}-1=\dfrac{2x-3}{3}\)               r,\(\dfrac{1}{x+3}+\dfrac{1}{x-1}=\dfrac{10}{\left(x+3\right)\left(x-1\right)}\)

h,\(\dfrac{1}{x-2}+3=\dfrac{x-3}{2-x}\)         t,\(\dfrac{3x}{x-2}-\dfrac{x}{x-5}=\dfrac{3x}{\left(x-2\right)\left(5-x\right)}\)

j,\(\dfrac{5}{3x+2}=2x-1\)              u,\(\dfrac{x+3}{x+1}+\dfrac{x-2}{x}=\dfrac{2\left(x^2+x-1\right)}{x\left(x+1\right)}\)

w,\(\dfrac{5x}{2x+2}+1=-\dfrac{6}{x+1}\)         s, \(\dfrac{6}{x-1}-\dfrac{4}{x-3}=\dfrac{2x}{\left(x-1\right)\left(x-3\right)}\)

ơ,\(\dfrac{1}{x-1}+\dfrac{2}{x+1}=\dfrac{x}{x^2-1}\)          v,\(\dfrac{x}{2\left(x-3\right)}+\dfrac{x}{2\left(x+1\right)}=\dfrac{2x}{\left(x+1\right)\left(x-3\right)}\)

z,\(\dfrac{1}{x-1}-\dfrac{3x^2}{x^3-1}=\dfrac{2x}{x^2+x+1}\)             ư,\(\dfrac{x+2}{x-2}-\dfrac{-2}{x^2-2x}=\dfrac{1}{x}\)

o,\(x+\dfrac{1}{x}=x^2+\dfrac{1}{x^2}\)          ô,\(1-\dfrac{1}{1-x}=\dfrac{x^2}{x^2-1}\)       zz,\(\dfrac{12}{8+x^3}=1+\dfrac{1}{x+2}\)

H24
13 tháng 1 2023 lúc 19:40

Bạn chia nhỏ các phần ra nhé.

Bình luận (1)
NT
13 tháng 1 2023 lúc 19:43

b: =>\(4\left(3-7x\right)=x+1\)

=>12-28x=x+1

=>-29x=-11

=>x=11/29

m:=>(3x-1)(x-1)=(2x+1)(x+1)

=>3x^2-4x+1=2x^2+3x+1

=>x^2-7x=0

=>x=0 hoặcx=7

d: =>9x-42=2x+10

=>7x=52

=>x=52/7

p: \(\Leftrightarrow\left(4x+7\right)\left(3x+4\right)=\left(12x+5\right)\left(x-1\right)\)

=>12x^2+16x+21x+28=12x^2-12x+5x-5

=>37x+28=7x-5

=>30x=-33

=>x=-11/10

j: =>(2x-1)(3x+2)=5

=>6x^2+4x-3x-2-5=0

=>6x^2-x-7=0

=>6x^2-7x+6x-7=0

=>(6x-7)(x+1)=0

=>x=7/6 hoặc x=-1

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