giải hộ mình nhé

\(a,0,5x-\frac{2}{3}x=\frac{7}{12}\Rightarrow\frac{1}{2}x-\frac{2}{3}x=\frac{7}{12}\)

\(\Rightarrow x\left(\frac{1}{2}-\frac{2}{3}\right)=\frac{7}{12}\Rightarrow x\cdot\left(\frac{3}{6}-\frac{4}{6}\right)=\frac{7}{12}\)

\(\Rightarrow x\cdot\left(-1\right)=\frac{7}{12}\Rightarrow x=\frac{7}{12}:\left(-1\right)=\frac{7}{-12}\)

\(c,\frac{\left(x-5\right)}{12}\cdot\frac{9}{29}=\frac{-6}{29}\Rightarrow\frac{\left(x-5\right)}{12}=\frac{-6}{29}:\frac{9}{26}\)

\(\frac{\Rightarrow\left(x-5\right)}{12}=\frac{-6}{9}=\frac{-2}{3}\Rightarrow x-5=-\frac{2}{3}\cdot12\)

\(\Rightarrow x-5=\frac{-24}{3}=-8\Rightarrow x=-8+5=-3\)

\(a,0,5x-\frac{2}{3}x=\frac{7}{12}\)

\(\Rightarrow\frac{1}{2}x-\frac{2}{3}x=\frac{7}{12}\)

\(\Rightarrow-\frac{1}{6}x=\frac{7}{12}\)

\(\Rightarrow x=-\frac{7}{2}\)

\(c,\frac{x-5}{12}\cdot\frac{9}{29}=-\frac{6}{29}\)

\(\Rightarrow\frac{x-5}{12}=-\frac{2}{3}\)

\(\Rightarrow x-5=12.\left(-\frac{2}{3}\right)\)

\(\Rightarrow x-5=-8\)

\(\Rightarrow x=-3\)

b)\(\frac{-8}{17}+\frac{5}{17}< \frac{x}{17}< \frac{-6}{17}+\frac{9}{17}\)\(\left(x\in Z\right)\)

\(\Rightarrow\frac{-3}{17}< \frac{x}{17}< \frac{3}{17}\)

\(\Rightarrow-3< x< 3\)

\(\Rightarrow x\in\left\{-2;-1;0;1;2\right\}\)

Vậy \(x\in\left\{-2;-1;0;1;2\right\}\)

14

14

\(\dfrac{13}{17}< \dfrac{x}{17}< \dfrac{15}{17}\Rightarrow x=14\)

`@` `\text {Ans}`

`\downarrow`

`x^3 = 125`

`=> x^3 = 5^3`

`=> x = 5`

Vậy, `x = 5`

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`4^5 \div 4^3`

`=`\(4^{5-3}\)

`= 4^2`

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\(8^{12}\cdot8^3\div8^{15}\)

`=`\(8^{12+3-15}\)

`=`\(8^0=1\)

______

\(7^{15}\div7^{13}\)

`=`\(7^{15-13}\)

`=`\(7^2\)

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\(17^{15}\cdot17^{10}\div17^{22}\)

`=`\(17^{15+10-22}\)

`= 17^3`

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\(5^{10}\cdot25\div5^{11}\)

`=`\(5^{10}\cdot5^2\div5^{11}\)

`=`\(5^{10+2-11}\)

`= 5`

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\(15^{10}\cdot225\div15^{11}\)

`=`\(15^{10}\cdot15^2\div15^{11}\)

`=`\(15^{12}\div15^{11}\)

`= 15`

`@` `\text {Kaizuu lv uuu}`

\(x^3=125\)

\(\Rightarrow x^3=5^3\)

\(\Rightarrow x=3\)

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\(4^5:4^3\)

\(=4^{5-3}\)

\(=4^2=16\)

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\(8^{12}.8^3:8^{15}\)

\(=8^{15}:8^{15}\)

\(=1\)

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\(7^{15}:7^{13}\)

\(=7^{15-13}\)

\(=7^2=49\)

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\(17^{15}.17^{10}:17^{22}\)

\(=17^{25}:17^{22}\)

\(=17^3=4913\)

___________________

\(5^{10}.25:5^{11}\)

\(=5^{10}.5^2:5^{11}\)

\(=5^{12}:5^{11}\)

\(=5^1=5\)

___________________

\(15^{10}.225:15^{11}\)

\(=15^{10}.15^2:15^{11}\)

\(=15^{12}:15^{11}\)

\(=15^1=15\)

__________________

* aⁿ . a^m = a^n+m

 aⁿ : a^m = a^n-m

Chúc bạn học tốt

\(\Leftrightarrow\frac{-2}{17}\le\frac{x}{17}\le\frac{2}{17}\Rightarrow x\in\left(-2;-1;0;1;2\right)\)

\(\Leftrightarrow\frac{-1}{24}\le\frac{x}{24}\le\frac{5}{24}\Rightarrow x\in\left(-1;0;1;2;3;4;5\right)\)

2 câu sau tự làm nha

\(-\frac{5}{17}+\frac{3}{17}\le\frac{x}{17}\le\frac{13}{17}+-\frac{11}{17}\)

\(\frac{-2}{17}\le\frac{x}{17}\le\frac{2}{17}\)

=> \(x\in\left\{-2;-1;0;1;2\right\}\)

thank nha

a) Ta có: \(\left|x-9\right|=8\)

         \(\Leftrightarrow x-9=\pm8\)

         \(\Leftrightarrow\orbr{\begin{cases}x-9=8\\x-9=-8\end{cases}}\)

         \(\Leftrightarrow\orbr{\begin{cases}x=17\left(TM\right)\\x=1\left(TM\right)\end{cases}}\)

Vậy \(x=17\)hoặc \(x=1\)

b);c) Ta có: \(17-\left|x+5\right|=14\)

            \(\Leftrightarrow\left|x+5\right|=3\)

            \(\Leftrightarrow x+5=\pm3\)

            \(\Leftrightarrow\orbr{\begin{cases}x+5=3\\x+5=-3\end{cases}}\)

            \(\Leftrightarrow\orbr{\begin{cases}x=-2\left(TM\right)\\x=-8\left(TM\right)\end{cases}}\)

Vậy \(x=-2\)hoặc \(x=-8\)