`a, 2/3 +3/4 = (8+9)/12=17/12.`

`1 1/3+4/5 = 4/3 + 4/5 = (20+12)/15=32/15`.

`=> x=2.`

`b, 5/6-1/4=(20-6)/24=7/12`.

`2 1/3-2/5= 7/3-2/5 = (35-6)/15=29/15`.

`=> x=1`.

a) \(\dfrac{2}{3}+\dfrac{3}{4}=\dfrac{8+9}{12}=\dfrac{17}{12}\)

-> 1 1/3 + 4/5 = 4/3 + 4/5 =  20+12/15 = 32/15

vậy x có thể = 14/14 = 1 (x thuộc N)

a: \(x+\dfrac{3}{9}=\dfrac{7}{6}\cdot\dfrac{2}{3}\)

=>\(x+\dfrac{1}{3}=\dfrac{14}{18}=\dfrac{7}{9}\)

=>\(x=\dfrac{7}{9}-\dfrac{1}{3}=\dfrac{7}{9}-\dfrac{3}{9}=\dfrac{4}{9}\)

b: \(x-\dfrac{2}{3}=\dfrac{1}{8}:\dfrac{5}{4}\)

=>\(x-\dfrac{2}{3}=\dfrac{1}{8}\cdot\dfrac{4}{5}=\dfrac{1}{10}\)

=>\(x=\dfrac{1}{10}+\dfrac{2}{3}=\dfrac{3+20}{30}=\dfrac{23}{30}\)

bạn đã kiểm tra kĩ chưa vậy?mình đọc đề câu B mà loạn não luôn á;-;

mik kiểm tra rùi

Bài 4: 

b: Ta có: \(2x\left(x-\dfrac{1}{4}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{4}\end{matrix}\right.\)

\(a,\Leftrightarrow-\dfrac{1}{2}x=\dfrac{1}{4}\Leftrightarrow x=-\dfrac{1}{2}\\ b,\Leftrightarrow\dfrac{1}{6}:x=\dfrac{1}{2}+\dfrac{1}{3}=\dfrac{5}{6}\Leftrightarrow x=\dfrac{1}{6}:\dfrac{5}{6}=\dfrac{1}{5}\\ c,\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{5}=3\\x+\dfrac{1}{5}=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{14}{5}\\x=-\dfrac{16}{5}\end{matrix}\right.\)

\(d,\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2=\dfrac{22}{9}-\dfrac{7}{3}=\dfrac{1}{9}\\ \Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{1}{3}\\x+\dfrac{1}{2}=-\dfrac{1}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{6}\\x=-\dfrac{5}{6}\end{matrix}\right.\\ e,\Leftrightarrow2\left|x\right|=2-\dfrac{1}{2}=\dfrac{3}{2}\\ \Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{3}{2}\\2x=-\dfrac{3}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=-\dfrac{3}{4}\end{matrix}\right.\)

\(f,\Leftrightarrow\left|x+\dfrac{1}{2}\right|=1+\dfrac{1}{6}=\dfrac{7}{6}\\ \Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{7}{6}\\x+\dfrac{1}{2}=-\dfrac{7}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-\dfrac{5}{3}\end{matrix}\right.\)

e: ta có: \(2\left|x\right|+\dfrac{1}{2}=2\)

\(\Leftrightarrow2\left|x\right|=\dfrac{3}{2}\)

\(\Leftrightarrow\left|x\right|=\dfrac{3}{4}\)

hay \(x\in\left\{\dfrac{3}{4};-\dfrac{3}{4}\right\}\)

\(\left(x+\dfrac{1}{3}\right)\times\dfrac{9}{14}\times\dfrac{7}{3}-\dfrac{1}{3}=1:\dfrac{9}{5}\\ \Rightarrow\left(x+\dfrac{1}{3}\right)\times\dfrac{3}{2}-\dfrac{1}{3}=\dfrac{5}{9}\\ \Rightarrow\left(x+\dfrac{1}{3}\right)\times\dfrac{3}{2}=\dfrac{5}{9}+\dfrac{1}{3}\\ \Rightarrow\left(x+\dfrac{1}{3}\right)\times\dfrac{3}{2}=\dfrac{8}{9}\\ \Rightarrow x+\dfrac{1}{3}=\dfrac{8}{9}:\dfrac{3}{2}\\ \Rightarrow x+\dfrac{1}{3}=\dfrac{16}{27}\\ \Rightarrow x=\dfrac{16}{27}-\dfrac{1}{3}\\ \Rightarrow x=\dfrac{7}{27}\)

5(x-1)=125-25

5(x-1)=100

x-1=100:5

x-1=20

x=20+1

x=21

12(x-1):3=64+8

12(x-1):3=72

12(x-1)=72.3

12(x-1)=216

x-1=216:12

x-1=18

x=18+1

x=19

(x-1)^3=5^3

=>x-1=5

x=5+1

x=6

\(a,\Leftrightarrow x^3=\dfrac{20}{3}\Leftrightarrow x=\sqrt[3]{\dfrac{20}{3}}\\ b,\Leftrightarrow x-1=9\Leftrightarrow x=10\\ c,\Leftrightarrow\left[{}\begin{matrix}x-1=5\\x-1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\\ d,\Leftrightarrow2x+1=5\Leftrightarrow x=2\\ e,\Leftrightarrow2x-4=4\Leftrightarrow x=4\)

Câu a) xem lại đề giùm nhé em

b) \(\left(x-1\right)^3=9^3\)

\(x-1=9\)

\(x=10\)

Vậy \(x=10\)

c) \(\left(x-1\right)^2=25\)

\(x-1=5\) hoặc \(x-1=-5\)

* \(x-1=5\)

\(x=6\)

* \(x-1=-5\)

\(x=-4\)

Vậy \(x=-4\); \(x=6\)

d) \(\left(2x+1\right)^3=125\)

\(\left(2x+1\right)^3=5^3\)

\(2x+1=5\)

\(2x=4\)

\(x=2\)

Vậy \(x=2\)

e) Sửa đề: \(\left(2x+4\right)^3=64\)

\(\left(2x+4\right)^3=4^3\)

\(2x+4=4\)

\(2x=0\)

\(x=0\)

Vậy \(x=0\)

\(a,\Leftrightarrow\left(5x+1\right)\left(x-4\right)-\left(x-4\right)=0\\ \Leftrightarrow\left(x-4\right)\left(5x+1-x\right)=0\\ \Leftrightarrow5x\left(x-4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\\ b,\Leftrightarrow2x^2-10x-2x^2-3x=26\\ \Leftrightarrow-13x=26\\ \Leftrightarrow x=-2\\ c,\Leftrightarrow x^3+1-x^3+3x=15\\ \Leftrightarrow3x=14\\ \Leftrightarrow x=\dfrac{14}{3}\)

\(d,\Leftrightarrow x^3-5x+2x^2-10+5x-2x^2-17=0\\ \Leftrightarrow x^3-27=0\\ \Leftrightarrow x^3=27\\ \Leftrightarrow x=3\)