a) Ta có: \(\frac{2}{3}x-\frac{1}{2}=\frac{1}{10}\)

\(\Leftrightarrow x\cdot\frac{2}{3}=\frac{1}{10}+\frac{1}{2}=\frac{6}{10}\)

hay \(x=\frac{6}{10}:\frac{2}{3}=\frac{6}{10}\cdot\frac{3}{2}=\frac{18}{20}=\frac{9}{10}\)

Vậy: \(x=\frac{9}{10}\)

b) Ta có: \(5\frac{4}{7}:x=13\)

\(\Leftrightarrow\frac{39}{7}:x=13\)

\(\Leftrightarrow x=\frac{39}{7}:13=\frac{39}{7}\cdot\frac{1}{13}=\frac{3}{7}\)

Vậy: \(x=\frac{3}{7}\)

c) Ta có: \(\left(2\frac{4}{5}x-50\right):\frac{2}{3}=51\)

\(\Leftrightarrow\frac{14}{5}x-50=51\cdot\frac{2}{3}=34\)

\(\Leftrightarrow x\cdot\frac{14}{5}=84\)

\(\Leftrightarrow x=84:\frac{14}{5}=84\cdot\frac{5}{14}=\frac{420}{14}=30\)

Vậy: x=30

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

\(\Leftrightarrow\frac{1}{3}:x=\frac{3}{5}-\frac{2}{3}=\frac{-1}{15}\)

hay \(x=\frac{1}{3}:\frac{-1}{15}=\frac{1}{3}\cdot\left(-15\right)=\frac{-15}{3}=-5\)

Vậy: x=-5

e) Ta có: \(8\frac{2}{3}:x-10=-8\)

\(\Leftrightarrow\frac{26}{3}:x=2\)

hay \(x=\frac{26}{3}:2=\frac{26}{3}\cdot\frac{1}{2}=\frac{26}{6}=\frac{13}{3}\)

Vậy: \(x=\frac{13}{3}\)

g) Ta có: \(x+30\%=-1.3\)

\(\Leftrightarrow x+\frac{3}{10}=\frac{-13}{10}\)

hay \(x=\frac{-13}{10}-\frac{3}{10}=\frac{-16}{10}=\frac{-8}{5}\)

Vậy: \(x=\frac{-8}{5}\)

i) Ta có: \(3\frac{1}{3}x+16\frac{3}{4}=-13.25\)

\(\Leftrightarrow x\cdot\frac{10}{3}+\frac{67}{4}=-\frac{53}{4}\)

\(\Leftrightarrow x\cdot\frac{10}{3}=\frac{-53}{4}-\frac{67}{4}=-30\)

\(\Leftrightarrow x=-30:\frac{10}{3}=-30\cdot\frac{3}{10}=\frac{-90}{10}=-9\)

Vậy: x=-9

k) Ta có: \(\left(2\frac{4}{5}x-50\right):\frac{2}{3}=51\)

\(\Leftrightarrow x\cdot\frac{14}{5}-50=51\cdot\frac{2}{3}=34\)

\(\Leftrightarrow x\cdot\frac{14}{5}=34+50=84\)

hay \(x=84:\frac{14}{5}=84\cdot\frac{5}{14}=30\)

Vậy: x=30

m) Ta có: \(\left|2x-1\right|=\left(-4\right)^2\)

\(\Leftrightarrow\left|2x-1\right|=16\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=16\\2x-1=-16\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=17\\2x=-15\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{17}{2}\\x=\frac{-15}{2}\end{matrix}\right.\)

Vậy: \(x\in\left\{\frac{17}{2};\frac{-15}{2}\right\}\)

thank you nha!

1, x^2 . x^3 = 3^7 : 3^2

       x^5     = 3^5

=> x = 3

2, x - 18 : 3 = 12

    x - 6       = 12

    x            = 12 + 6

    x            = 18

3, (x - 18) : 2 = 12

     x - 18       = 12 . 2

     x - 18       = 24

     x             = 24 + 18

     x             = 42

4, x^2 = 16

    x^2 = 4^2

=> x = 4

x=4 nha bn

k k nha

mk k lại cho!

b) \(\frac{3}{x-5}=\frac{-4}{x+2}\)

=> 3.(x + 2) = (x - 5). (-4)

= 3x + 6 = -4x + 20

= 3x - 4x = 26

=> -x = 26

=> x = -26

c) \(\frac{x}{-2}=\frac{-8}{x}\)

=> x.x = -2 . (-8)

 => x² = 16

=> x = 4

t i c k nha!!! 645646677778879078452352543546456457567564546567

còn câu a

cái câu a) kiểu j đấy

Bài 1

a, 2³ + ( x - 3² ) = 1

x - 3² = 1 - 2³ = -7

x = -7 + 3²

x = 2

b, 5 . (x+7) -10 = 40

5 . (x+7) = 50

x+7 = 50 :5 =10

x = 10 - 7

x = 3

9) \(\dfrac{x}{4}=\dfrac{9}{x}\)

Theo định nghĩa về hai phân số bằng nhau, ta có:

\(4\cdot9=x^2\\ 36=x^2\Rightarrow\left[{}\begin{matrix}x=6\\x=-6\end{matrix}\right.\)

8)

\(x:\dfrac{5}{3}+\dfrac{1}{3}=-\dfrac{2}{5}\\ x:\dfrac{5}{3}=-\dfrac{2}{5}+\dfrac{1}{3}\\ x:\dfrac{5}{3}=-\dfrac{1}{15}\\ x=\dfrac{1}{15}\cdot\dfrac{5}{3}\\ x=\dfrac{1}{9}\)

7)

\(2x-16=40+x\\ 2x-x=40+16\\ x\left(2-1\right)=56\\ x=56\)

6)

\(1\dfrac{1}{2}+x=\dfrac{3}{2}-7\\ \dfrac{3}{2}+x=\dfrac{3}{2}-7\\ \dfrac{3}{2}-\dfrac{3}{2}=-7-x\\ -7-x=0\\ x=-7-0\\ x=-7\)

5)

\(3\dfrac{1}{2}-\dfrac{1}{2}x=\dfrac{2}{3}\\ \dfrac{7}{2}-\dfrac{1}{2}x=\dfrac{2}{3}\\ \dfrac{1}{2}x=\dfrac{7}{2}-\dfrac{2}{3}\\ \dfrac{1}{2}x=\dfrac{17}{6}\\ x=\dfrac{17}{6}:\dfrac{1}{2}\\ x=\dfrac{17}{3}\)

4)

\(x\cdot\left(x+1\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)

3)

\(\left(\dfrac{2x}{5}+2\right):\left(-4\right)=-1\dfrac{1}{2}\\ \left(\dfrac{2x}{5}+2\right):\left(-4\right)=-\dfrac{3}{2}\\ \dfrac{2x}{5}+2=-\dfrac{3}{2}\cdot\left(-4\right)\\ \dfrac{2x}{5}+2=6\\ \dfrac{2x}{5}=6-2\\ \dfrac{2x}{5}=4\\ 2x=4\cdot5\\ 2x=20\\ x=20:2\\ x=10\)

2)

\(\dfrac{1}{3}+\dfrac{1}{2}:x=-0,25\\ \dfrac{1}{3}+\dfrac{1}{2}:x=-\dfrac{1}{4}\\ \dfrac{1}{2}:x=-\dfrac{1}{4}-\dfrac{1}{3}\\ \dfrac{1}{2}:x=-\dfrac{7}{12}\\ x=\dfrac{1}{2}:-\dfrac{7}{12}\\ x=-\dfrac{6}{7}\)

1)

\(\dfrac{4}{3}+x=\dfrac{2}{15}\\ x=\dfrac{2}{15}-\dfrac{4}{3}x=-\dfrac{6}{5}\)

Bài 2:

a) \(\left(x-3\right)^3+27=0\)

\(\Leftrightarrow\left(x-3\right)^3=0-27\)

\(\Leftrightarrow\left(x-3\right)^3=-27\)

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

\(\Leftrightarrow x-3=-3\)

\(\Leftrightarrow x=\left(-3\right)+3\)

\(\Leftrightarrow x=0\)

b) \(-125-\left(x+1\right)^3=0\)

\(\Leftrightarrow\left(x+1\right)^3=-125-0\)

\(\Leftrightarrow\left(x+1\right)^3=-125\)

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

\(\Leftrightarrow x+1=-5\)

\(\Leftrightarrow x=\left(-5\right)-1\)

\(\Leftrightarrow x=-6\)

c) \(\left(2x-\dfrac{1}{4}\right)^2-\dfrac{1}{16}=0\)

\(\Leftrightarrow\left(2x-\dfrac{1}{4}\right)^2=0+\dfrac{1}{16}\)

\(\Leftrightarrow\left(2x-\dfrac{1}{4}\right)^2=\dfrac{1}{16}\)

\(\Leftrightarrow\left(2x-\dfrac{1}{4}\right)^2=\left(\dfrac{1}{4}\right)^2\)

\(\Leftrightarrow2x-\dfrac{1}{4}=\dfrac{1}{4}\)

\(\Leftrightarrow2x=\dfrac{1}{4}+\dfrac{1}{4}\)

\(\Leftrightarrow2x=\dfrac{1}{2}\)

\(\Leftrightarrow x=\dfrac{1}{2}:2\)

\(\Leftrightarrow x=\dfrac{1}{4}\)

d) \(2^x+2^{x+1}=24\)

\(\Leftrightarrow2^x+2^x.2=24\)

\(\Leftrightarrow2^x\left(1+2\right)=24\)

\(\Leftrightarrow2^x.3=24\)

\(\Leftrightarrow2^x=24:3\)

\(\Leftrightarrow2^x=8\)

\(\Leftrightarrow2^x=2^3\)

\(\Rightarrow x=3\)

e) \(\left|x+\dfrac{1}{5}\right|-\dfrac{1}{2}=1\)

\(\Leftrightarrow\left|x+\dfrac{1}{5}\right|=1+\dfrac{1}{2}\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{5}=-\dfrac{3}{2}\\x+\dfrac{1}{5}=\dfrac{3}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{17}{10}\\x=\dfrac{13}{10}\end{matrix}\right.\)

g) \(\left|x-3\right|+2x=10\)

\(\Leftrightarrow\left|x-3\right|=10-2x\)

\(\Leftrightarrow\left|x-3\right|=2.5-2x\)

\(\Leftrightarrow\left|x-3\right|=2\left(5-x\right)\)

(không chắc có nên làm tiếp câu g không, thấy đề cứ là lạ, có j sai sai...)

Bài 1:

a) \(2^7+2^9⋮10\)

Ta có: \(2^7+2^9=2^{4.1}.2^3+2^{4.2}.2\)

\(\Leftrightarrow\overline{A6}.2^3+\overline{B6}.2\)

\(\Leftrightarrow\overline{A6}.8+\overline{B6}.2\)

\(\Leftrightarrow\overline{C8}+\overline{D2}\)

\(\Leftrightarrow\overline{E0}\)

Mà \(\overline{E0}⋮10\) \(\Rightarrow2^7+2^9⋮10\)

b) \(8^{24}.25^{10}⋮2^{36}.5^{20}\)

Ta có: \(8^{24}.25^{10}=\left(2^3\right)^{24}.\left(5^2\right)^{10}\)

\(\Leftrightarrow2^{72}.5^{20}\)

Do \(2^{72}⋮2^{36}\) và \(5^{20}⋮5^{20}\) \(\Rightarrow8^{24}.25^{10}⋮2^{36}.5^{20}\)

c) \(3^{10}+3^{12}⋮30\)

Ta có: \(3^{10}+3^{12}=3^{4.2}.3^2+3^{4.3}\)

\(\Leftrightarrow\overline{A1}.3^2+\overline{B1}\)

\(\Leftrightarrow\overline{A1}.9+\overline{B1}\)

\(\Leftrightarrow\overline{C9}+\overline{B1}\)

\(\Leftrightarrow\overline{D0}⋮10\)

(Chứng minh chia hết cho 10 rồi chứng minh chia hết cho 3, mình chưa tìm được cách làm, chờ chút)

Bài 1:

a, \(\frac{1}{-16}-\frac{3}{45}=\frac{-1}{16}-\frac{1}{15}\)

\(=\frac{-15}{240}-\frac{16}{240}\)

\(=\frac{-31}{240}\)

b, \(=\frac{-10}{12}-\frac{-12}{12}\)

\(=\frac{2}{12}=\frac{1}{6}\)

c, \(=\frac{-30}{6}-\frac{1}{6}\)

\(=\frac{-31}{6}\)

Bài 2:

a, \(x=-\frac{1}{2}-\frac{3}{4}\)

\(x=-\frac{1}{4}\)

b,   \(\frac{1}{2}+x=-\frac{11}{2}\)

\(x=-\frac{11}{2}-\frac{1}{2}\)

\(x=-6\)

Bạn nhớ k đúng và chọn câu trả lời này nhé!!!! Mình giải đúng và chính xác hết ^_^