Ta có 2^{x + 1} . 3^y = 10^x

=> 2^x.3^y.2 = 10^x

=> 3^y.2 = 5^x

=> 3^y.2 = (...5)

=> 3^y = (...5) : 2

Vì 5^y tận cùng là 5

=> 5^y không chia hết cho 2 

=> Không tồn tại x;y \(\inℕ\)thỏa mãn

=> \(x;y\in\varnothing\)

b) 10^x : 5^y = 20^y

=> 10^x = 4^y

=> x = y = 0

c) (2x - 15)⁵ = (2x - 15)³

(2x - 15)⁵ - (2x - 15)³ = 0

=> (2x - 15)³[(2x - 15)² - 1] = 0

=> \(\orbr{\begin{cases}\left(2x-15\right)^3=0\\\left(2x-15\right)^2=1\end{cases}}\Rightarrow\orbr{\begin{cases}2x-15=0\\2x-15=\pm1\end{cases}}\Rightarrow2x-15\in\left\{0;1;-1\right\}\)

=> \(x\in\left\{7,5;8;7\right\}\)

Vì x là số tự nhiên => \(x\in\left\{7;8\right\}\)

b, \(\dfrac{x-3}{4}=\dfrac{15}{20}\)

<=> \(\dfrac{x-3}{4}=\dfrac{3}{4}\)

=> x-3=3

<=> x=6

Vậy x=6

\(a,\dfrac{x}{15}=\dfrac{4}{y}=\dfrac{-2}{5}\)

* \(\dfrac{x}{15}=\dfrac{-2}{5}\)

\(\Rightarrow\dfrac{x}{15}=\dfrac{-6}{15}\)

\(\Rightarrow x=-6\)

*\(\dfrac{4}{y}=\dfrac{-2}{5}\)

\(\Rightarrow\dfrac{4}{y}=\dfrac{4}{-10}\)

\(\Rightarrow y=-10\)

Vậy x = - 6 ; y = - 10

\(b,\dfrac{x-3}{4}=\dfrac{15}{20}\)

=> ( x - 3 ) . 20 = 4. 15

=> 20x - 60 = 60

=> 20x = 60 + 60

=> 20x = 120

=> x = 120 : 20

=> x = 6

Vậy x = 6

\(c,\dfrac{-5}{9}+\dfrac{-8}{15}+\dfrac{22}{-9}+\dfrac{-7}{15}< x\le\dfrac{-1}{3}+\dfrac{-1}{4}+\dfrac{-5}{12}\)

\(\Rightarrow\dfrac{-5}{9}+\dfrac{-8}{15}+\dfrac{-22}{9}+\dfrac{-7}{15}< x\le\dfrac{-4}{12}+\dfrac{-3}{12}+\dfrac{-5}{12}\)

\(\Rightarrow\left(\dfrac{-5}{9}+\dfrac{-22}{9}\right)+\left(\dfrac{-8}{15}+\dfrac{-7}{15}\right)< x\le-1\)

\(\Rightarrow-3+\left(-1\right)< x\le-1\)

\(\Rightarrow-4< x\le-1\)

\(\Rightarrow x=-3;-2;-1\)

a)Ta có: \(\frac{-2}{5}+\frac{6}{5}.\left(y-\frac{2}{3}\right)=\frac{-4}{15}\)

\(\Rightarrow\frac{6}{5}.\left(y-\frac{2}{3}\right)=\frac{-4}{15}-\frac{-2}{15}\)

\(\Rightarrow\frac{6}{5}.\left(y-\frac{2}{3}=\right)\frac{-2}{5}\)

\(\Rightarrow y-\frac{2}{3}=\frac{-2}{5}:\frac{6}{5}=\frac{-1}{3}\)

\(\Rightarrow y=\frac{-1}{3}+\frac{2}{3}=\frac{1}{3}\)

Vậy x = \(\frac{1}{3}\)

b) Ta có: \(\frac{-2}{5}+\frac{2}{3}x+\frac{1}{6}x=\frac{-4}{15}\)

        \(\Rightarrow\frac{-2}{5}+x.\left(\frac{2}{3}+\frac{1}{6}\right)=\frac{-4}{15}\)

        \(\Rightarrow x.\frac{5}{6}=\frac{-4}{15}-\frac{-2}{15}\)

         \(x.\frac{5}{6}=\frac{-2}{15}\)

\(\Rightarrow x=\frac{-2}{15}:\frac{5}{6}=\frac{-4}{25}\)

Vậy x = \(\frac{-4}{25}\)

c) Ta có: \(\frac{3}{2}x+\frac{-2}{5}-\frac{2}{3}.x=\frac{-4}{15}\)

\(\Rightarrow\frac{3}{2}x-\frac{2}{3}x+\frac{-2}{5}=\frac{-4}{15}\)

\(\Rightarrow x.\left(\frac{3}{2}-\frac{2}{4}\right)=\frac{-4}{15}-\frac{-2}{15}\)

\(\Rightarrow x.\frac{5}{6}=\frac{-2}{15}\)

\(\Rightarrow x=\frac{-2}{15}:\frac{5}{6}=\frac{-4}{25}\)

Vậy x = \(\frac{-4}{25}\)

Ủng hộ tớ nha m.n

a)Ta có: \(\frac{- 2}{5} + \frac{6}{5} . \left(\right. y - \frac{2}{3} \left.\right) = \frac{- 4}{15}\)

\(\Rightarrow \frac{6}{5} . \left(\right. y - \frac{2}{3} \left.\right) = \frac{- 4}{15} - \frac{- 2}{15}\)

\(\Rightarrow \frac{6}{5} . \left(\right. y - \frac{2}{3} = \left.\right) \frac{- 2}{5}\)

\(\Rightarrow y - \frac{2}{3} = \frac{- 2}{5} : \frac{6}{5} = \frac{- 1}{3}\)

\(\Rightarrow y = \frac{- 1}{3} + \frac{2}{3} = \frac{1}{3}\)

Vậy x = \(\frac{1}{3}\)

b) Ta có: \(\frac{- 2}{5} + \frac{2}{3} x + \frac{1}{6} x = \frac{- 4}{15}\)

        \(\Rightarrow \frac{- 2}{5} + x . \left(\right. \frac{2}{3} + \frac{1}{6} \left.\right) = \frac{- 4}{15}\)

        \(\Rightarrow x . \frac{5}{6} = \frac{- 4}{15} - \frac{- 2}{15}\)

         \(x . \frac{5}{6} = \frac{- 2}{15}\)

\(\Rightarrow x = \frac{- 2}{15} : \frac{5}{6} = \frac{- 4}{25}\)

Vậy x = \(\frac{- 4}{25}\)

c) Ta có: \(\frac{3}{2} x + \frac{- 2}{5} - \frac{2}{3} . x = \frac{- 4}{15}\)

\(\Rightarrow \frac{3}{2} x - \frac{2}{3} x + \frac{- 2}{5} = \frac{- 4}{15}\)

\(\Rightarrow x . \left(\right. \frac{3}{2} - \frac{2}{4} \left.\right) = \frac{- 4}{15} - \frac{- 2}{15}\)

\(\Rightarrow x . \frac{5}{6} = \frac{- 2}{15}\)

\(\Rightarrow x = \frac{- 2}{15} : \frac{5}{6} = \frac{- 4}{25}\)

Vậy x = \(\frac{- 4}{25}\)

các ý a,b,c c chỉ cần nhan chéo cho nhau

d) \(x.\left(y+2\right)-y=15\)

\(\Rightarrow x.\left(y+2\right)=15+y\)

\(\Rightarrow x=\frac{y+15}{y+2}=\frac{y+2+13}{y+2}=1+\frac{13}{y+2}\)

y + 2 là ước nguyên của 13

\(y+2=1\Rightarrow y=-1\Rightarrow x=14\)

\(y+2=-1\Rightarrow y=-3\Rightarrow x=-12\)

\(y+2=13\Rightarrow y=11\Rightarrow x=2\)

\(y+2=-13\Rightarrow y=-15\Rightarrow x=0\)

Ai thấy đúng thì ủng hộ, mink chỉ làm được vậy thuu

|x|-5=13

|x|=13-5

|x|=8

Ta có hai trường hợp:

x=-8 hoặc 8

a) \(\dfrac{-5}{6}.\dfrac{120}{25}< x< \dfrac{-7}{15}.\dfrac{9}{14}\)

\(\Rightarrow-4< x< \dfrac{-3}{10}\)

\(\Rightarrow\dfrac{-40}{10}< x< \dfrac{-3}{10}\)

\(\Rightarrow x\in\left\{\dfrac{-39}{10};\dfrac{-38}{10};\dfrac{-37}{10};...;\dfrac{-5}{10};\dfrac{-4}{10}\right\}\)

b) \(\left(\dfrac{-5}{3}\right)^2< x< \dfrac{-24}{35}.\dfrac{-5}{6}\)

\(\Rightarrow\dfrac{25}{9}< x< \dfrac{4}{7}\)

\(\Rightarrow\dfrac{175}{63}< x< \dfrac{36}{63}\)

\(\Rightarrow x=\varnothing\)

c) \(\dfrac{1}{18}< \dfrac{x}{12}< \dfrac{y}{9}< \dfrac{1}{4}\)

\(\Leftrightarrow\dfrac{2}{36}< \dfrac{3x}{36}< \dfrac{4y}{36}< \dfrac{9}{36}\)

\(\Rightarrow x\in\left\{1;2\right\}\)

+) Với \(x=1\)

\(\Rightarrow y\in\left\{1;2\right\}\)

+) Với \(x=2\)

\(\Rightarrow y=2\)

Vậy \(x=1\) thì \(y\in\left\{1;2\right\}\); \(x=2\) thì \(y=8\).

a) \(\frac{-2}{5}+\frac{5}{6}.x=\frac{-4}{15}\)

\(\frac{5}{6}.x=\frac{-4}{15}-\frac{-2}{5}\)

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

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

\(x=\frac{4}{25}\)

b) \(\left(x-\frac{1}{5}\right)\left(y+\frac{1}{2}\right)\left(z-3\right)=0\)

\(x-\frac{1}{5}=0\)

\(x=0+\frac{1}{5}\)

\(x=\frac{1}{5}\)