câu c) mang tính mua vui hay gì hả bn

mếu thật thì x=0,x=số nào cx đc(câu trả lời này mang tính mua vui thôi nhé)

a)Ta có: \(\dfrac{x}{2}-\left(\dfrac{3x}{5}-\dfrac{13}{5}\right)=-\left(\dfrac{7}{5}+\dfrac{7}{10}x\right)\)

\(\Leftrightarrow\dfrac{x}{2}-\dfrac{3x-13}{5}=\dfrac{-7}{5}-\dfrac{7x}{10}\)

\(\Leftrightarrow\dfrac{5x}{10}-\dfrac{2\left(3x-13\right)}{10}=\dfrac{-14}{10}-\dfrac{7x}{10}\)

\(\Leftrightarrow5x-6x+26=-14-7x\)

\(\Leftrightarrow-x+26+14+7x=0\)

\(\Leftrightarrow6x=-40\)

hay \(x=-\dfrac{20}{3}\)

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

\(\Leftrightarrow\dfrac{3\left(2x-3\right)}{6}+\dfrac{-9}{6}=\dfrac{5-3x}{6}-\dfrac{2}{6}\)

\(\Leftrightarrow6x-9-9=5-3x-2\)

\(\Leftrightarrow6x-18-3+3x=0\)

\(\Leftrightarrow x=\dfrac{7}{3}\)

a) \(\dfrac{-x}{4}=\dfrac{-9}{x}\)

\(\Rightarrow-x^2=-36\)

\(\Rightarrow x^2=36\)

\(\Rightarrow\left[{}\begin{matrix}x=6\\x=-6\end{matrix}\right.\)

Vậy: \(x\in\left\{6;-6\right\}\)

b) \(\dfrac{5}{9}+\dfrac{x}{-1}=-\dfrac{1}{3}\)

\(\Rightarrow\dfrac{5}{9}+\dfrac{-9x}{9}=\dfrac{-3}{9}\)

\(\Rightarrow5-9x=-3\)

\(\Rightarrow-9x=-8\)

\(\Rightarrow x=\dfrac{8}{9}\)

Vậy: \(x=\dfrac{8}{9}\)

c) \(x:3\dfrac{1}{5}=1\dfrac{1}{2}\)

\(\Rightarrow x:\dfrac{16}{5}=\dfrac{3}{2}\)

\(\Rightarrow x=\dfrac{3}{2}.\dfrac{16}{5}\)

\(\Rightarrow x=\dfrac{24}{5}\)

Vậy: \(x=\dfrac{24}{5}\)

d) \(\dfrac{3x-1}{-5}=\dfrac{-5}{3x-1}\)

\(\Rightarrow\left(3x-1\right)^2=25\)

\(\Rightarrow\left[{}\begin{matrix}3x-1=5\\3x-1=-5\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}3x=6\\3x=-4\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{4}{3}\end{matrix}\right.\)

Vậy: \(x\in\left\{2;-\dfrac{4}{3}\right\}\)

4 câu đầu hìn như sai đề :v

`m)(3/2-2/(-5)):x-1/2=3/2`

`<=>(3/2+2/5):x=3/2+1/2=2`

`<=>19/10:x=2`

`<=>x=19/10:2=19/20`

`n)(3/2-5/11-3/13)(2x-2)=(-3/4+5/22+3/26)`

`<=>(3/2-5/11-3/13)(2x-2)+3/4-5/22-3/26=0`

`<=>(3/2-5/11-3/13)(2x-2)+1/2(3/2-5/11-3/13)=0`

`<=>(3/2-5/11-3/13)(2x-2+1/2)=0`

Mà `3/2-5/11-3/13>0`

`<=>2x-2+1/2=0`

`<=>2x-3/2=0`

`<=>2x=3/2<=>x=3/4`

h, \(\dfrac{x}{2}-\dfrac{1}{x}=\dfrac{1}{12}\left(x\ne0\right)\)

\(\Leftrightarrow\dfrac{x^2}{2}-1=\dfrac{x}{12}\)

\(\Leftrightarrow x^2-\dfrac{x}{6}-2=0\)

\(\Leftrightarrow x^2-2.x.\dfrac{1}{12}+\dfrac{1}{144}-\dfrac{289}{144}=0\)

\(\Leftrightarrow\left(x-\dfrac{1}{12}\right)^2=\dfrac{289}{144}\)

\(\Leftrightarrow x=\dfrac{1}{12}\pm\dfrac{\sqrt{289}}{12}\)

Vậy ...

i, \(\Leftrightarrow x^2-\dfrac{2.x.7}{12}+\dfrac{49}{144}-\dfrac{1}{144}=0\)

\(\Leftrightarrow\left(x-\dfrac{7}{2}\right)^2=\dfrac{1}{144}\)

\(\Leftrightarrow x=\dfrac{7}{2}\pm\dfrac{1}{12}\)

Vậy ...

h) Ta có: \(\dfrac{x}{2}-\dfrac{1}{x}=\dfrac{1}{12}\)

\(\Leftrightarrow\dfrac{x^2-2}{2x}=\dfrac{1}{12}\)

\(\Leftrightarrow12x^2-24-2x=0\)

\(\Delta=\left(-2\right)^2-4\cdot12\cdot\left(-24\right)=1156\)

Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{2-34}{24}=\dfrac{-8}{3}\\x_2=\dfrac{2+34}{24}=\dfrac{36}{24}=\dfrac{3}{2}\end{matrix}\right.\)

Vậy: \(S=\left\{-\dfrac{8}{3};\dfrac{3}{2}\right\}\)

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

\(\Leftrightarrow\dfrac{19}{10}:x=2\)

hay \(x=\dfrac{19}{20}\)

Vậy: \(S=\left\{\dfrac{19}{20}\right\}\)

\(b,\Rightarrow\dfrac{x}{2}-\dfrac{3x}{5}-\dfrac{13}{5}=-\dfrac{7}{5}-\dfrac{7x}{10}\\ \Rightarrow\dfrac{1}{2}x-\dfrac{3}{5}x+\dfrac{7}{10}x=\dfrac{6}{5}\\ \Rightarrow\dfrac{3}{5}x=\dfrac{6}{5}\Rightarrow x=2\\ c,\Rightarrow\dfrac{2x-3}{3}-\dfrac{5-3x}{6}=-\dfrac{1}{3}+\dfrac{3}{2}=\dfrac{7}{6}\\ \Rightarrow\dfrac{4x-6-5+3x}{6}=\dfrac{7}{6}\\ \Rightarrow7x-11=7\Rightarrow x=\dfrac{18}{7}\\ d,\Rightarrow\dfrac{2}{3x}+\dfrac{7}{x}=\dfrac{4}{5}+2+\dfrac{3}{12}=\dfrac{61}{20}\\ \Rightarrow\dfrac{23}{3x}=\dfrac{61}{20}\\ \Rightarrow183x=460\\ \Rightarrow x=\dfrac{460}{183}\\ e,\Rightarrow2\left(x-1\right)-\left(x-1\right)^2=0\\ \Rightarrow\left(x-1\right)\left(2-x+1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)

e: Ta có: \(\left(x-1\right)^2=2\left(x-1\right)\)

\(\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)

Bài 1: 

b) ĐKXĐ: \(x\ne3\)

Ta có: \(\dfrac{3-x}{20}=\dfrac{-5}{x-3}\)

\(\Leftrightarrow\dfrac{x-3}{-20}=\dfrac{-5}{x-3}\)

\(\Leftrightarrow\left(x-3\right)^2=100\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=10\\x-3=-10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=13\left(nhận\right)\\x=-7\left(nhận\right)\end{matrix}\right.\)

Vậy: \(x\in\left\{13;-7\right\}\)

a: x=4/27-2/3=4/27-18/27=-14/27

b: =>3/4x-1/4x=1/6+7/3

=>1/2x=1/6+14/6=5/2

hay x=5

c: =>13/10x=7/2+5/2=6

=>x=13/10:6=13/60

d: (3x+2)(-2/5x-7)=0

=>3x+2=0 hoặc 2/5x+7=0

=>x=-2/3 hoặc x=-35/2

a) x = 4/27 - 2/3

    x = -14/27

a: x=4/27-2/3=4/27-18/27=-14/27

b: =>3/4x-1/4x=1/6+7/3

=>1/2x=1/6+14/6=5/2

hay x=5

c: =>13/10x=7/2+5/2=6

=>x=13/10:6=13/60

d: (3x+2)(-2/5x-7)=0

=>3x+2=0 hoặc 2/5x+7=0

=>x=-2/3 hoặc x=-35/2

`h)x/2-1/x=1/12(x ne 0)`

`<=>6x^2-12=x`

`<=>6x^2-x-12=0`

`<=>6x^2-9x+8x-12=0`

`<=>3x(2x-3)+4(2x-3)=0`

`<=>(2x-3)(3x+4)=0`

`<=>` \(\left[ \begin{array}{l}x=\dfrac32\\x=-\dfrac43\end{array} \right.\) 

`i)x^2-7/6x+1/3=0`

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

`<=>6x^2-3x-4x+2=0`

`<=>3x(2x-1)-2(2x-1)=0`

`<=>(2x-1)(3x-2)=0`

`<=>` \(\left[ \begin{array}{l}x=\dfrac12\\x=\dfrac23\end{array} \right.\) 

Câu cuối không có dấu "=" nên không tìm được x :v

- Hai câu h, i bấm nốt đáp án để đẹp nha ;-; câu k thiếu đề :v

h) Ta có: \(\dfrac{x}{2}-\dfrac{1}{x}=\dfrac{1}{12}\)

\(\Leftrightarrow\dfrac{x^2-2}{2x}=\dfrac{1}{12}\)

\(\Leftrightarrow12\left(x^2-2\right)-2x=0\)

\(\Leftrightarrow12x^2-2x-24=0\)

\(\Delta=\left(-2\right)^2-4\cdot12\cdot\left(-24\right)=1156\)

Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{2+34}{12}=\dfrac{36}{12}=3\\x_2=\dfrac{2-34}{12}=\dfrac{-32}{12}=-\dfrac{8}{3}\end{matrix}\right.\)

Vậy: \(S=\left\{3;-\dfrac{8}{3}\right\}\)

i) Ta có: \(x^2-\dfrac{7}{6}x+\dfrac{1}{3}=0\)

\(\Leftrightarrow6x^2-7x+2=0\)

\(\Leftrightarrow6x^2-3x-4x+2=0\)

\(\Leftrightarrow3x\left(2x-1\right)-2\left(2x-1\right)=0\)

\(\Leftrightarrow\left(2x-1\right)\left(3x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\3x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{2}{3}\end{matrix}\right.\)

Vậy: \(S=\left\{\dfrac{1}{2};\dfrac{2}{3}\right\}\)

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