a) Ta có: \(\dfrac{-3}{5}x+\dfrac{-7}{4}=\dfrac{3}{10}\)

\(\Leftrightarrow\dfrac{-3}{5}x=\dfrac{3}{10}+\dfrac{7}{4}=\dfrac{41}{20}\)

\(\Leftrightarrow x=\dfrac{41}{20}:\dfrac{-3}{5}=\dfrac{41}{20}\cdot\dfrac{-5}{3}\)

hay \(x=-\dfrac{41}{12}\)

Vậy: \(x=-\dfrac{41}{12}\)

2: 

a: =>x-1/5=2/15

=>x=2/15+3/15=5/15=1/3

b: =>x+7/12=-5/6-2/6=-7/6

=>x=-14/12-7/12=-21/12=-7/4

c: =>x+2/3=-10/3

=>x=-4

d: =>1/4:x=-11/4

=>x=-1/4:11/4=-1/11

e: =>8:x=1,6

=>x=5

a)

\(\frac{x-3}{10}=\frac{4}{x-3}\)

=> ( x - 3 )² = 4 . 10.

     ( x - 3 )² = 40

Mà x - 3 thuộc Z ( vì x thuộc Z ) nên ( x - 3 )² là số chính phương.

Do 40 không là số chính phương.

=> Ko tìm được x thuộc Z thỏa mãn đề bài.

b) 

\(\frac{x+5}{9}=\frac{4}{x+5}\)

=> ( x + 5 )² = 4 . 9

     ( x + 5 )² = 36

=> x + 5 = 6 hoặc x + 5 = -6.

+) x + 5 = 6

           x = 1.

+) x + 5 = -6

          x = -11.

Vậy x = 1; x = -11.

a/ \(2x+\frac{1}{7}=\frac{1}{3}\)

=> \(2x=\frac{1}{3}-\frac{1}{7}=\frac{7}{21}-\frac{3}{21}\)

=> \(2x=\frac{4}{21}\)

=> \(x=\frac{4}{21}:2=\frac{4}{21}.\frac{1}{2}=\frac{2}{21}\)

b/ \(3\left(x-\frac{1}{2}\right)=\frac{4}{9}\)

=> \(x-\frac{1}{2}=\frac{4}{9}:3=\frac{4}{9}.\frac{1}{3}\)

=> \(x-\frac{1}{2}=\frac{4}{27}\)

=> \(x=\frac{4}{27}+\frac{1}{2}=\frac{8}{54}+\frac{27}{54}=\frac{35}{54}\)

c/ \(\left(x-5\right)^2+4=68\)

=> \(\left(x-5\right)^2=68-4=64\)

=> \(\left[{}\begin{matrix}x-5=8\\x-5=-8\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=8+5=13\\x=-8+5=-3\end{matrix}\right.\)

d/ \(\left(\left|x\right|-\frac{1}{2}\right)\left(2x+\frac{3}{2}\right)=0\)

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

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

=> \(\left[{}\begin{matrix}\left[{}\begin{matrix}x=\frac{1}{2}\\x=-\frac{1}{2}\end{matrix}\right.\\x=-\frac{3}{2}:2=-\frac{3}{2}.\frac{1}{2}=-\frac{3}{4}\end{matrix}\right.\)

e) \(5x+2=3x+8\)

=> \(5x-3x=8-2=6\)

=> \(2x=6\)

=> \(x=6:2=3\)

f/ \(26-\left(5-2x\right)=27\)

=> \(5-2x=26-27=-1\)

=> \(2x=5-\left(-1\right)=5+1=6\)

=> \(x=6:2=3\)

g/ \(\left(4x-8\right)-\left(2x-6\right)=4\)

=> \(4x-8-2x+6=4\)

=> \(\left(4x-2x\right)+\left(-8+6\right)=4\)

=> \(2x+-2=4\)

=> \(2x=4+2=6\)

=> \(x=6:2=3\)

h/ \(\left(x+3\right)^3:3-1=-10\)

=> \(\left(x+3\right)^3:3=-10+1=-9\)

=> \(\left(x+3\right)^3=-9.3=-27\)

=> \(x+3=-3\)

=> \(x=-3-3=-6\)

Thank

giúp mk đi mà

a) 2x + 34 = 72

2x=38

x=19

b) (4+x) . 2 phần 3 =9

4+x=\(\frac{27}{2}\) 

x=\(\frac{19}{2}\)

c) (2x + 1 ) 3 =36 

2x+1=12

2x=11

x=\(\frac{11}{2}\)

d ) (x + 5 ) : 4 =12

x+5=48

x=43

e ) 1 phần 2 x -- 4 = 15

\(\frac{1}{2}x\)=19

x=38

\(\frac{5}{6}=\frac{x-1}{x}\left(đk:x\ne0\right)\)

\(< =>5x=6\left(x-1\right)< =>5x=6x-6\)

\(< =>6x-5x=6< =>x=6\left(tmđk\right)\)

\(\frac{1}{2}=\frac{x+1}{3x}\left(đk:x\ne0\right)\)

\(< =>3x=2\left(x+1\right)< =>3x=2x+2\)

\(< =>3x-2x=2< =>x=2\left(tmđk\right)\)

\(\frac{3}{x+2}=\frac{5}{2x+1}\left(đk:x\ne-2;-\frac{1}{2}\right)\)

\(< =>3\left(2x+1\right)=5\left(x+2\right)< =>6x+3=5x+10\)

\(< =>6x-5x=10-3< =>x=7\left(tmđk\right)\)

\(\frac{5}{8x-2}=-\frac{4}{7-x}\left(đk:x\ne\frac{1}{4};7\right)\)

\(< =>\frac{5}{8x-2}=\frac{4}{x-7}< =>5\left(x-7\right)=4\left(8x-2\right)\)

\(< =>5x-35=32x-8< =>32x-5x=-35+8\)

\(< =>27x=-27< =>x=-1\)

\(\frac{4}{3}=\frac{2x-1}{3}< =>4.3=\left(2x-1\right).3\)

\(< =>12=6x-3< =>6x=12+3\)

\(< =>6x=15< =>x=\frac{15}{6}=\frac{5}{2}\)

\(\frac{2x-1}{3}=\frac{3x+1}{4}< =>4\left(2x-1\right)=3\left(3x+1\right)\)

\(< =>8x-4=9x+3< =>9x-8x=-4-3\)

\(< =>9x-8x=-7< =>x=-7\)

\(\frac{4}{x+2}=\frac{7}{3x+1}\left(đk:x\ne-2;-\frac{1}{3}\right)\)

\(< =>4\left(3x+1\right)=7\left(x+2\right)< =>12x+4=7x+14\)

\(< =>12x-7x=14-4< =>5x=10\)

\(< =>x=\frac{10}{5}=2\left(tmđk\right)\)

\(-\frac{3}{x+1}=\frac{4}{2-2x}\left(đk:x\ne-1;1\right)\)

\(< =>-3\left(2-2x\right)=4\left(x+1\right)< =>-6+6x=4x+4\)

\(< =>6x-4x=4+6< =>2x=10\)

\(< =>x=\frac{10}{2}=5\left(tmđk\right)\)

\(\frac{x+1}{3}=\frac{3}{x+1}\left(đk:x\ne-1\right)\)

\(< =>\left(x+1\right)\left(x+1\right)=3.3\)

\(< =>x^2+2x+1=9< =>x^2+2x+1-9=0\)

\(< =>x^2+2x-8=0< =>x^2-2x+4x-8=0\)

\(< =>x\left(x-2\right)+4\left(x-2\right)=0< =>\left(x+4\right)\left(x-2\right)=0\)

\(< =>\orbr{\begin{cases}x+4=0\\x-2=0\end{cases}< =>\orbr{\begin{cases}x=-4\\x=2\end{cases}}}\left(tmđk\right)\)

a: =>-2x=90/91

hay x=-45/91

b: =>2x=-7

hay x=-7/2

c: ->-3x=-12

hay x=4

1. \(6x^3-8=40\\ 6x^3=48\\ x^3=8\\ \Rightarrow x=2\)Vậy x = 2

2. \(4x^5+15=47\\ 4x^5=32\\ x^5=8\\ \Rightarrow x\in\varnothing\left(\text{vì }x\in N\right)\)Vậy x ∈ ∅

3. \(2x^3-4=12\\ 2x^3=16\\ x^3=8\\ \Rightarrow x=2\)Vậy x = 2

4. \(5x^3-5=0\\ 5x^3=5\\ x^3=1\\ \Rightarrow x=1\)Vậy x = 1

5. \(\left(x-5\right)^{2016}=\left(x-5\right)^{2018}\\ \Rightarrow\left[{}\begin{matrix}x-5=0\\x-5=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\x=6\end{matrix}\right.\)Vậy \(x\in\left\{5;6\right\}\)

6. \(\left(3x-2\right)^{20}=\left(3x-1\right)^{20}\\ \Rightarrow3x-2=3x-1\\ 3x-3x=2-1\\ 0=1\left(\text{vô lí}\right)\)Vậy x ∈ ∅

7. \(\left(3x-1\right)^{10}=\left(3x-1\right)^{20}\\ \left(3x-1\right)^{10}=\left[\left(3x-1\right)^2\right]^{10}\\ \Rightarrow\left(3x-1\right)^2=3x-1\\ \left(3x-1\right)^2-\left(3x-1\right)=0\\ \left(3x-1\right)\left[\left(3x-1\right)-1\right]=0\\ \left(3x-1\right)\left(3x-2\right)=0\\ \Rightarrow\left[{}\begin{matrix}3x-1=0\\3x-2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}3x=1\\3x=2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{1}{3}\left(\text{loại vì }x\in N\right)\\x=\frac{2}{3}\left(\text{loại vì }x\in N\right)\end{matrix}\right.\)Vậy x ∈ ∅

8. \(\left(2x-1\right)^{50}=2x-1\\ \left(2x-1\right)^{50}-\left(2x-1\right)=0\\ \left(2x-1\right)\left[\left(2x-1\right)^{49}-1\right]=0\\ \Rightarrow\left[{}\begin{matrix}2x-1=0\\\left(2x-1\right)^{49}=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=1\\2x-1=1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\frac{1}{2}\left(\text{loại vì }x\in N\right)\\x=1\left(t/m\right)\end{matrix}\right.\)Vậy x = 1

9. \(\left(\frac{x}{3}-5\right)^{2000}=\left(\frac{x}{3}-5\right)^{2008}\\ \left(\frac{x}{3}-5\right)^{2008}-\left(\frac{x}{3}-5\right)^{2000}=0\\ \left(\frac{x}{3}-5\right)^{2000}\left[\left(\frac{x}{3}-5\right)^8-1\right]=0\\ \Rightarrow\left[{}\begin{matrix}\left(\frac{x}{3}-5\right)^{2000}=0\\\left(\frac{x}{3}-5\right)^8=1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}\frac{x}{3}-5=0\\\frac{x}{3}-5=1\\\frac{x}{3}-5=-1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}\frac{x}{3}=5\\\frac{x}{3}=6\\\frac{x}{3}=4\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=5\cdot3=15\\x=6\cdot3=18\\x=4\cdot3=12\end{matrix}\right.\)Vậy \(x\in\left\{15;18;12\right\}\)

\(1.6x^3-8=40\\ \Leftrightarrow6x^3=48\\ \Leftrightarrow x^3=8\Leftrightarrow x^3=2^3=\left(-2\right)^3\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)

Vậy \(x\in\left\{2;-2\right\}\)

\(2.4x^3+15=47\) (T nghĩ đề là mũ 3)

\(\Leftrightarrow4x^3=32\Leftrightarrow x^3=8=2^3=\left(-2\right)^3\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)

Vậy \(x\in\left\{2;-2\right\}\)

Câu 3, 4 tương tự nhé.

giải nhanh giup mình nhé