giúp mình giải bài toán trên với. Mình cảm ơn rất nhiều

a: =>1/2x-3/4x=-5/6+7/3

=>-1/4x=14/6-5/6=3/2

=>x=-3/2*4=-6

b: =>4/5x-3/2x=1/2+6/5

=>-7/10x=17/10

=>x=-17/7

c: =>6/5x+6/20=6/5-1/3x

=>6/5x+1/3x=6/5-3/10=12/10-3/10=9/10

=>x=27/46

d: =>6x+3/2+4/5=1/2-2x

=>8x=1/2-3/2-4/5=-1-4/5=-9/5

=>x=-9/40

bn đăng bên toán nhé

a) Ta có: \(\dfrac{2x-1}{5}-\dfrac{x-2}{3}=\dfrac{x+7}{15}\)

\(\Leftrightarrow\dfrac{3\left(2x-1\right)}{15}-\dfrac{5\left(x-2\right)}{15}=\dfrac{x+7}{15}\)

\(\Leftrightarrow6x-3-5x+10-x-7=0\)

\(\Leftrightarrow0x=0\)(luôn đúng)

Vậy: S={x|\(x\in R\)}

Ai giúp vs

a,2/5 = 2/5 ; 3/8=6/16 ; 1/9=3/27

b, 4/3=8/6 ; -1=-1 ; -4/-2=-8/4

tick cho mik nhé 

a) x= 2, x= 8.(6 : 3) = 16, x= 1. (27 : 9)= 3

b) x= 6 : (8 : 4) = 3, x= -1, x= -2 . -8 = x.x => 16 = x² => 4² = x² => x=4

        Tick cho mình đi 

\(\dfrac{x+1}{3}+\dfrac{x+1}{4}+\dfrac{x+1}{5}=\dfrac{x+1}{6}\)

\(\dfrac{x+1}{3}+\dfrac{x+1}{4}+\dfrac{x+1}{5}-\dfrac{x+1}{6}=0\)

\(\left(x+1\right)\left(\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}\right)=0\)

\(\)vì \(\dfrac{1}{3}>\dfrac{1}{6};\dfrac{1}{4}>\dfrac{1}{6};\dfrac{1}{5}>\dfrac{1}{6}=>\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}>0\)

\(=>x+1=0\)

\(=>x=-1\)

b,

\(\dfrac{x+1}{2020}+\dfrac{x+2}{2019}=\dfrac{x+3}{2018}+\dfrac{x+4}{2017}\)

\(\left(\dfrac{x+1}{2020}+1\right)+\left(\dfrac{x+2}{2019}+1\right)=\left(\dfrac{x+3}{2018}+1\right)+\left(\dfrac{x+4}{2017}+1\right)\)

\(\dfrac{x+2021}{2020}+\dfrac{x+2021}{2019}=\dfrac{x+2021}{2018}+\dfrac{x+2021}{2017}\)

\(=>\dfrac{x+2021}{2020}+\dfrac{x+2021}{2019}-\dfrac{x+2021}{2018}-\dfrac{x+2021}{2017}=0\)

\(=>\left(x+2021\right)\left(\dfrac{1}{2020}+\dfrac{1}{2019}-\dfrac{1}{2018}-\dfrac{1}{2017}\right)=0\)

Vì \(\dfrac{1}{2020}< \dfrac{1}{2018};\dfrac{1}{2019}< \dfrac{1}{2017}=>\dfrac{1}{2020}+\dfrac{1}{2019}-\dfrac{1}{2018}-\dfrac{1}{2017}< 0\)

\(=>x+2021=0\)

\(=>x=-2021\)

c,

\(\dfrac{x+2}{327}+\dfrac{x+3}{326}+\dfrac{x+4}{325}+\dfrac{x+5}{324}+\dfrac{x+349}{5}=0\)

\(\left(\dfrac{x+2}{327}+1\right)+\left(\dfrac{x+3}{326}+1\right)+\left(\dfrac{x+4}{325}+1\right)+\left(\dfrac{x+5}{324}+1\right)+\left(\dfrac{x+349}{5}-4\right)=0\)

\(\dfrac{x+329}{327}+\dfrac{x+329}{326}+\dfrac{x+329}{325}+\dfrac{x+329}{324}+\dfrac{x+329}{5}=0\)

\(=>\left(x+329\right)\left(\dfrac{1}{327}+\dfrac{1}{326}+\dfrac{1}{325}+\dfrac{1}{324}+\dfrac{1}{5}\right)=0\)

Vì \(\dfrac{1}{327}+\dfrac{1}{326}+\dfrac{1}{325}+\dfrac{1}{324}+\dfrac{1}{5}>0\)

\(=>x+329=0\)

\(=>x=-329\)

\(\begin{array}{l} n) \Leftrightarrow \dfrac{{x + 1}}{7} + 1 + \dfrac{{x + 2}}{6} + 1 = \dfrac{{x + 3}}{5} + 1 + \dfrac{{x + 4}}{4} + 1\\ \Leftrightarrow \dfrac{{x + 8}}{7} + \dfrac{{x + 8}}{6} - \dfrac{{x + 8}}{5} - \dfrac{{x + 8}}{4} = 0\\ \Leftrightarrow \left( {x + 8} \right)\underbrace {\left( {\dfrac{1}{7} + \dfrac{1}{8} - \dfrac{1}{5} - \dfrac{1}{6}} \right)}_{ < 0} = 0\\ \Leftrightarrow x + 8 = 0\\ \Leftrightarrow x = - 8 \end{array}\)

k/

\(8-\dfrac{x-2}{3}=\dfrac{x}{4}\)

\(\Leftrightarrow\dfrac{96}{12}-\dfrac{4\left(x-2\right)}{12}=\dfrac{3x}{12}\)

\(\Leftrightarrow96-4x+8=3x\)

\(\Leftrightarrow96-4x+8-3x=0\)

\(\Leftrightarrow104-7x=0\)

\(\Leftrightarrow7x=104\)

\(\Leftrightarrow x=104:7\)

\(\Leftrightarrow x=\dfrac{104}{7}\)

Vậy tập nghiệm của phương trình là \(S=\left\{\dfrac{104}{7}\right\}\)

m/ 

\(\dfrac{3x+2}{2}-\dfrac{3x+1}{6}=2x+\dfrac{5}{3}\)

\(\Leftrightarrow\dfrac{3\left(3x+2\right)}{6}-\dfrac{3x+1}{6}=\dfrac{12x}{6}+\dfrac{10}{6}\)

\(\Leftrightarrow9x+6-3x-1-12x-10=0\)

\(\Leftrightarrow-6x-5=0\)

\(\Leftrightarrow-6x=5\)

\(\Leftrightarrow x=-\dfrac{5}{6}\)

Vậy tập nghiệm của phương trình là \(S=\left\{-\dfrac{5}{6}\right\}\)

k) Ta có: \(8-\dfrac{x-2}{2}=\dfrac{x}{4}\)

\(\Leftrightarrow\dfrac{32}{4}-\dfrac{2\left(x-2\right)}{4}=\dfrac{x}{4}\)

\(\Leftrightarrow32-2x+4-x=0\)

\(\Leftrightarrow28-x=0\)

hay x=28

Vậy: S={28}

m) Ta có: \(\dfrac{3x+2}{2}-\dfrac{3x+1}{6}=2x+\dfrac{5}{3}\)

\(\Leftrightarrow\dfrac{3\left(3x+2\right)}{6}-\dfrac{3x+1}{6}=\dfrac{12x}{6}+\dfrac{10}{6}\)

\(\Leftrightarrow9x+6-3x-1=12x+10\)

\(\Leftrightarrow6x+5-12x-10=0\)

\(\Leftrightarrow-6x=5\)

hay \(x=-\dfrac{5}{6}\)

Vậy: \(S=\left\{-\dfrac{5}{6}\right\}\)

n) Ta có: \(\dfrac{x+1}{7}+\dfrac{x+2}{6}=\dfrac{x+3}{5}+\dfrac{x+4}{4}\)

\(\Leftrightarrow\dfrac{x+1}{7}+1+\dfrac{x+2}{6}+1=\dfrac{x+3}{5}+1+\dfrac{x+4}{4}+1\)

\(\Leftrightarrow\dfrac{x+8}{7}+\dfrac{x+8}{6}=\dfrac{x+8}{5}+\dfrac{x+8}{4}\)

\(\Leftrightarrow\dfrac{x+8}{7}+\dfrac{x+8}{6}-\dfrac{x+8}{5}-\dfrac{x+8}{4}=0\)

\(\Leftrightarrow\left(x+8\right)\left(\dfrac{1}{7}+\dfrac{1}{6}-\dfrac{1}{5}-\dfrac{1}{4}\right)=0\)

mà \(\dfrac{1}{7}+\dfrac{1}{6}-\dfrac{1}{5}-\dfrac{1}{4}\ne0\)

nên x+8=0

hay x=-8

Vậy: S={-8}

bạn tách một câu vài câu hỏi chứ đừng gộp như thế này ko ai trả lời đâu

a: =>\(4x-5=2x-2+x=3x-2\)

=>x=3

b: \(\Leftrightarrow7x-35=3x+6\)

=>4x=41

=>x=41/4

c: =>(2x+5)(x+5)-2x^2=0

=>2x^2+10x+5x+25-2x^2=0

=>15x=-25

=>x=-5/3

e: \(\Leftrightarrow\dfrac{11}{x}=\dfrac{9x-36+2x+2}{\left(x+1\right)\left(x-4\right)}\)

=>11(x^2-3x-4)=x(11x-34)

=>11x^2-33x-44=11x^2-34x

=>x=44

a: =>4x-5=2x-2+x

=>4x-5=3x-2

=>x=3(nhận)

b: =>7x-35=3x+6

=>4x=41

=>x=41/4

d: =>(2x+5)(x+5)-2x^2=0

=>2x^2+10x+5x+25+2x^2=0

=>15x=-25

=>x=-5/3

a)

\(\dfrac{4x-5}{x-1}=2+\dfrac{x}{x-1}\left(x\ne1\right)\)

suy ra

`4x-5=2(x-1)+x`

`<=>4x-5=2x-2+x`

`<=>4x-2x-x=-2+5`

`<=>x=3(tm)`

b)

\(\dfrac{7}{x+2}=\dfrac{3}{x-5}\left(x\ne-2;x\ne5\right)\)

suy ra

`7(x-5)=3(x+2)`

`<=>7x-35=3x+6`

`<=>7x-3x=6+35`

`<=>4x=41`

`<=>x=41/4(tm)`

c)

\(\dfrac{2x+5}{2x}-\dfrac{x}{x+5}=0\left(x\ne0;x\ne-5\right)\)

suy ra

`(2x+5)(x+5)-2x^2=0`

`<=>2x^2+10x+5x+25-2x^2=0`

`<=>15x=-25`

`<=>x=-5/3(tm)`

\(=\dfrac{4}{7}\)

a) Đk: \(x>0;x\ne9;x\ne25\)

Đặt \(A=\left(\dfrac{\sqrt{x}}{3+\sqrt{x}}+\dfrac{2x}{9-x}\right):\left(\dfrac{\sqrt{x}-1}{x-3\sqrt{x}}-\dfrac{2}{\sqrt{x}}\right)\)

\(=\left[\dfrac{\sqrt{x}\left(3-\sqrt{x}\right)}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}+\dfrac{2x}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}\right]\)\(:\left[\dfrac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-3\right)}-\dfrac{2\left(\sqrt{x}-3\right)}{\sqrt{x}\left(\sqrt{x}-3\right)}\right]\)

\(=\dfrac{\sqrt{x}\left(3-\sqrt{x}\right)+2x}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}:\dfrac{\sqrt{x}-1-2\left(\sqrt{x}-3\right)}{\sqrt{x}\left(\sqrt{x}-3\right)}\)

\(=\dfrac{3\sqrt{x}+x}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}:\dfrac{-\sqrt{x}+5}{\sqrt{x}\left(\sqrt{x}-3\right)}\)

\(=\dfrac{\sqrt{x}\left(3+\sqrt{x}\right)}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}.\dfrac{\sqrt{x}\left(3-\sqrt{x}\right)}{\sqrt{x}-5}\)

\(=\dfrac{x}{\sqrt{x}-5}\)

b) Đk: \(x\ge0;x\ne1;x\ne25\)

Biểu thức

\(=\left[\dfrac{\sqrt{x}-2}{\sqrt{x}+5}+\dfrac{\sqrt{x}}{\sqrt{x}-5}-\dfrac{x+9}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\right]:\dfrac{1-\sqrt{x}}{5+\sqrt{x}}\)

\(=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-5\right)+\sqrt{x}\left(\sqrt{x}+5\right)-x-9}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}.\dfrac{\sqrt{x}+5}{1-\sqrt{x}}\)

\(=\dfrac{x-7\sqrt{x}+10+x+5\sqrt{x}-x-9}{\left(\sqrt{x}-5\right)\left(1-\sqrt{x}\right)}\)

\(=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-5\right)\left(1-\sqrt{x}\right)}\)\(=\dfrac{\left(1-\sqrt{x}\right)^2}{\left(\sqrt{x}-5\right)\left(1-\sqrt{x}\right)}=\dfrac{1-\sqrt{x}}{\sqrt{x}-5}\)

Đk: \(x>0;x\ne4\)

Biểu thức:

\(=\left[\dfrac{1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\dfrac{1}{\left(\sqrt{x}-2\right)^2}\right].\dfrac{\sqrt{x}\left(2-\sqrt{x}\right)}{\sqrt{x}}\)

\(=\left[\dfrac{-1}{\left(2-\sqrt{x}\right)\left(\sqrt{x}+2\right)}-\dfrac{1}{\left(2-\sqrt{x}\right)^2}\right].\left(2-\sqrt{x}\right)\)

\(=-\dfrac{1}{\sqrt{x}+2}-\dfrac{1}{2-\sqrt{x}}\)

\(=\dfrac{-\left(2-\sqrt{x}\right)-\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(2-\sqrt{x}\right)}=\dfrac{-4}{4-x}\)\(=\dfrac{4}{x-4}\)

Khoảng cách hai ta là 100 bước, em bước 100, anh lùi 1