a: \(x+6\dfrac{1}{8}=8\)

=>\(x+\dfrac{49}{8}=\dfrac{64}{8}\)

=>\(x=\dfrac{64}{8}-\dfrac{49}{8}=\dfrac{15}{8}\)

b: \(\dfrac{11}{2}\cdot x=\dfrac{1}{5}:\dfrac{1}{3}\)

=>\(x\cdot\dfrac{11}{2}=\dfrac{1}{5}\cdot3=\dfrac{3}{5}\)

=>\(x=\dfrac{3}{5}:\dfrac{11}{2}=\dfrac{3}{5}\cdot\dfrac{2}{11}=\dfrac{6}{55}\)

c: \(x\cdot\dfrac{3}{5}+\dfrac{2}{5}\cdot x=\dfrac{4}{9}+\dfrac{1}{3}\)

=>\(x\left(\dfrac{3}{5}+\dfrac{2}{5}\right)=\dfrac{4}{9}+\dfrac{3}{9}\)

=>\(x\cdot1=\dfrac{7}{9}\)

=>\(x=\dfrac{7}{9}\)

`4.(3-x)^10=9(3-x)^8`

`=>(3-x)^{8}(4(3-x)^2-9)=0`

`=>` $\left[ \begin{array}{l}3-x=0\\(3-x)^2=\dfrac{9}{4}\end{array} \right.$

`=>` $\left[ \begin{array}{l}x=3\\3-x=\dfrac{3}{2}\\3-x=-\dfrac{3}{2}\end{array} \right.$

`=>` $\left[ \begin{array}{l}x=3\\x=\dfrac{3}{2}\\x=\dfrac{9}{2}\end{array} \right.$

Vậy `x=3` hoặc `x=3/2` hoặc `x=9/2`

`b,(x-1)^{x+3}=(x-1)^{x+1}`

`=>(x-1)^{x+1}[(x-1)^2-1]=0`

`=>` $\left[ \begin{array}{l}x-=1\\x-=-1\end{array} \right.$

`=>` $\left[ \begin{array}{l}x=2\\x=1\\x=0\end{array} \right.$

Vậy `x=0` hoặc `x=1` hoặc `x=2`

a) 4(3 - x)¹⁰ = 9(3 - x)⁸

=> (x - 3)⁸.[4(x - 3)² - 9) = 0

<=> (x - 3)⁸.(2x - 6 - 3)(2x - 6 + 3) = 0

<=> (x - 3)⁸(2x - 9)(2x - 3) = 0

<=> \(\left[{}\begin{matrix}x-3=0\\2x-9=0\\2x-3=0\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=3\\x=\dfrac{9}{2}\\x=\dfrac{3}{2}\end{matrix}\right.\)

Vậy S = {3;9/2; 3/2}

b) (x - 1)^{x + 3} = (x - 1)^{x  + 1}

<=> (x - 1)^{x + 1}.[(x - 3)² - 1) = 0

<=>(x - 1)^{x + 1}.(x - 3 - 1)(x - 3 + 1) = 0

<=> (x - 1)^{x + 1}.(x - 4)(x - 2) = 0

<=> \(\left[{}\begin{matrix}x-1=0\\x-4=0\\x-2=0\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=1\\x=4\\x=2\end{matrix}\right.\)

b) Ta có: \(\left(x-1\right)^{x+3}=\left(x-1\right)^{x+1}\)

\(\Leftrightarrow\left(x-1\right)^{x+3}-\left(x-1\right)^{x+1}=0\)

\(\Leftrightarrow\left(x-1\right)^{x+1}\left[\left(x-1\right)^2-1\right]=0\)

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

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

Vậy: S={0;1;2}

\(a,x+\dfrac{1}{4}=\dfrac{5}{8}\)

\(\Leftrightarrow x=\dfrac{5}{8}-\dfrac{1}{4}\)

\(\Leftrightarrow x=\dfrac{5}{8}-\dfrac{2}{8}\)

\(\Leftrightarrow x=\dfrac{3}{8}\)

\(b,x-\dfrac{3}{5}=\dfrac{1}{10}\)

\(\Leftrightarrow x=\dfrac{1}{10}+\dfrac{3}{5}\)

\(\Leftrightarrow x=\dfrac{1}{10}+\dfrac{6}{10}\)

\(\Leftrightarrow x=\dfrac{7}{10}\)

\(c,x\times\dfrac{2}{7}=\dfrac{6}{11}\)

\(\Leftrightarrow x=\dfrac{6}{11}:\dfrac{2}{7}\)

\(\Leftrightarrow x=\dfrac{6}{11}\times\dfrac{7}{2}\)

\(\Leftrightarrow x=\dfrac{42}{22}\)

\(\Leftrightarrow x=\dfrac{21}{11}\)

\(d,x:\dfrac{3}{2}=\dfrac{1}{4}\)

\(\Leftrightarrow x=\dfrac{1}{4}\times\dfrac{3}{2}\)

\(\Leftrightarrow x=\dfrac{3}{8}\)

\(đ,\dfrac{7}{4}-x=\dfrac{5}{7}\)

\(\Leftrightarrow x=\dfrac{7}{4}-\dfrac{5}{7}\)

\(\Leftrightarrow x=\dfrac{49}{28}-\dfrac{20}{28}\)

\(\Leftrightarrow x=\dfrac{29}{28}\)

\(e,\dfrac{5}{4}:x=\dfrac{1}{8}\)

\(x=\dfrac{5}{4}:\dfrac{1}{8}\)

\(\Leftrightarrow x=\dfrac{5}{4}\times8\)

\(\Leftrightarrow x=\dfrac{40}{4}\)

\(\Leftrightarrow x=10\)

a.

\(\left|5x\right|=3x+8\Leftrightarrow\left[{}\begin{matrix}-5x=3x+8\\5x=3x+8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=4\end{matrix}\right.\)

b.

\(\left|-4x\right|=-2x+11\Leftrightarrow\left[{}\begin{matrix}-4x=-2x+11\\4x=-2x+11\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{11}{2}\\x=\dfrac{11}{6}\end{matrix}\right.\)

c.

\(\left|3x-1\right|=4x+1\Leftrightarrow\left[{}\begin{matrix}-3x+1=4x+1\\3x-1=4x+1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)

d.

\(\left|3-2x\right|=3x-7\Leftrightarrow\left[{}\begin{matrix}-3+2x=3x-7\\3-2x=3x-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=2\end{matrix}\right.\)

e.

\(9-\left|-5x\right|+2x=0\Leftrightarrow\left[{}\begin{matrix}9-5x+2x=0\\9+5x+2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{9}{7}\end{matrix}\right.\)

f.

\(\left(x+1\right)^2+\left|x+10\right|-x^2-12=0\Leftrightarrow\left[{}\begin{matrix}x^2+2x+1-x-10-x^2-12=0\\x^2+2x+1+x+10-x^2-12=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=21\\x=\dfrac{1}{3}\end{matrix}\right.\)

Câu 1 : 

a, \(\frac{3\left(2x+1\right)}{4}-\frac{5x+3}{6}=\frac{2x-1}{3}-\frac{3-x}{4}\)

\(\Leftrightarrow\frac{6x+3}{4}+\frac{3-x}{4}=\frac{2x-1}{3}+\frac{5x+3}{6}\)

\(\Leftrightarrow\frac{5x+6}{4}=\frac{9x+1}{6}\Leftrightarrow\frac{30x+36}{24}=\frac{36x+4}{24}\)

Khử mẫu : \(30x+36=36x+4\Leftrightarrow-6x=-32\Leftrightarrow x=\frac{32}{6}=\frac{16}{3}\)

tương tự 

\(\frac{19}{4}-\frac{2\left(3x-5\right)}{5}=\frac{3-2x}{10}-\frac{3x-1}{4}\)

\(< =>\frac{19.5}{20}-\frac{8\left(3x-5\right)}{20}=\frac{2\left(3-2x\right)}{20}-\frac{5\left(3x-1\right)}{20}\)

\(< =>95-24x+40=6-4x-15x+5\)

\(< =>-24x+135=-19x+11\)

\(< =>5x=135-11=124\)

\(< =>x=\frac{124}{5}\)

\(\frac{\left(x-2\right).3}{2}+3+\frac{\left(x-3\right).5}{3}+5+\frac{\left(x-5\right).2}{5}+2=10\)

\(< =>\frac{\left(x-2\right).3.15}{30}+\frac{\left(x-3\right).5.10}{30}+\frac{\left(x-5\right).2.6}{30}=10-2-3-5\)

\(< =>\frac{\left(x-2\right).45+\left(x-3\right).50+\left(x-5\right).12}{30}=0\)

\(< =>45x-90+50x-150+12x-60=0\)

\(< =>107x-300=0< =>x=\frac{300}{107}\)

\(1,\\ x+\dfrac{1}{2}=-\dfrac{5}{3}\\ x=-\dfrac{5}{3}-\dfrac{1}{2}\\ x=-\dfrac{13}{6}\\ Vậyx=-\dfrac{13}{6}\)

\(2,\\ \dfrac{1}{3}-x=\dfrac{3}{5}\\ x=\dfrac{1}{3}-\dfrac{3}{5}\\ x=-\dfrac{4}{15}\\ Vậyx=-\dfrac{4}{15}\)

\(3,\\ 3-4+x=\dfrac{7}{2}\\ -1+x=\dfrac{7}{2}\\ x=\dfrac{7}{2}+1\\ x=\dfrac{9}{2}\\ Vậyx=\dfrac{9}{2}\)

\(4,\\ x-\dfrac{4}{3}=-\dfrac{7}{9}\\ x=-\dfrac{7}{9}+\dfrac{4}{3}\\ x=\dfrac{15}{27}\\ Vậyx=\dfrac{15}{27}\)

\(5,\\ x-\left(-\dfrac{7}{3}\right)=\dfrac{5}{6}\\ x=\dfrac{5}{6}-\dfrac{7}{3}\\ x=-\dfrac{27}{18}\\ Vậyx=-\dfrac{27}{18}\)

\(6,\\ x-\dfrac{1}{5}=\dfrac{9}{10}\\ x=\dfrac{9}{10}+\dfrac{1}{5}\\ x=\dfrac{11}{10}\\ Vậyx=\dfrac{11}{10}\)

\(7,\\ x+\dfrac{5}{12}=\dfrac{3}{8}\\ x=\dfrac{3}{8}-\dfrac{5}{12}\\ x=-\dfrac{1}{24}\\ Vậyx=-\dfrac{1}{24}\)

\(8,\\ x+\dfrac{5}{4}=\dfrac{7}{6}\\ x=\dfrac{7}{6}-\dfrac{5}{4}\\ x=-\dfrac{9}{24}\\ Vậyx=-\dfrac{9}{24}\)

\(9,\\ x-\dfrac{2}{7}=\dfrac{1}{35}\\ x=\dfrac{1}{35}+\dfrac{2}{7}\\ x=\dfrac{11}{35}\\ Vậyx=\dfrac{11}{35}\\ 10,\\ x-\dfrac{1}{5}=-\dfrac{7}{10}\\ x=-\dfrac{7}{10}+\dfrac{1}{5}\\ x=-\dfrac{1}{2}\\ Vậyx=-\dfrac{1}{2}\)

ahiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii hãy ấn a

Giúp mình nhanh nha! Mình sẽ thick người đó

`x/8 = 3/4 +(-5/8)`

`=>x/8 = 6/8 +(-5/8)`

`=>x/8 = 1/8`

`=>x=1`

`-----`

`x/12 =3/4 +(-2/3)`

`=>x/12 = 9/12 + (-8/12)`

`=> x/12=1/12`

`=>x=1`

`----`

`1+11/13=24/x`

`=> 13/13 +11/13=24/x`

`=> 24/13 =24/x`

`=>x=13`

`----`

`x/6 -3/4=1/12`

`=>x/6 = 1/12 +3/4`

`=>x/6 = 1/12 + 9/12`

`=>x/6 = 10/12`

`=>x/6= 5/6`

`=>x=5`

\(a)\)

\(\dfrac{x}{8}=\dfrac{3}{4}+\dfrac{-5}{8}\)

\(\Rightarrow\dfrac{x}{8}=\dfrac{1}{8}\)

\(\Rightarrow x.8=1.8\)

\(\Rightarrow x.8=8\)

\(\Rightarrow x=8:8\)

\(\Rightarrow x=1\)

\(b)\)

\(\dfrac{x}{12}=\dfrac{3}{4}+\dfrac{-2}{3}\)

\(\Rightarrow\dfrac{x}{12}=\dfrac{1}{12}\)

\(\Rightarrow x.12=1.12\)

\(\Rightarrow x.12=12\)

\(\Rightarrow x=12:12\)

\(\Rightarrow x=1\)

\(c)\)

\(1+\dfrac{11}{13}=\dfrac{24}{x}\)

\(\Rightarrow\dfrac{24}{13}=\dfrac{24}{x}\)

\(\Rightarrow x.24=24.13\)

\(\Rightarrow x.24=312\)

\(\Rightarrow x=312:24\)

\(\Rightarrow x=13\)

\(d)\)

\(\dfrac{x}{6}-\dfrac{3}{4}=\dfrac{1}{12}\)

\(\Rightarrow\dfrac{x}{6}=\dfrac{1}{12}+\dfrac{3}{4}\)

\(\Rightarrow\dfrac{x}{6}=\dfrac{5}{6}\)

\(\Rightarrow x.6=5.6\)

\(\Rightarrow x.6=30\)

\(\Rightarrow x=30:6\)

\(\Rightarrow x=5\)

`a)1/3x-1/4x=1`

`(1/3-1/4)x=1`

`1/12x=1`

`x=1:1/12=12`

______________________________

`b)4/5+5/7:x=1/6`

`5/7:x=1/6-4/5`

`5/7:x=-19/30`

`x=5/7:(-19/30)`

`x=-150/133`

__________________________________

`c)[2x]/3-x/4=5/6`

`x(2/3-1/4)=5/6`

`x. 5/12=5/6`

`x=5/6:5/12=2`

a) 1/3x - 1/4x = 1

1/12x = 1

x = 1 . 12

x = 12

b) 4/5 + 5/7: x = 1/6

5/7 : x = 1/6 - 4/5

5/7 : x = -19/30

x = 5/7 : -19/30

x = -150/133

c) 2x/3 - x/4 = 5/6

4 . 2x - 3x = 10

8x - 3x = 10

5x = 10

x = 10 : 5

x = 2.