a)Sai
b)Đúng

a, sai 

b, đúng

Ct: a(b-c)=a.b-a.c

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`

_____

`4^5 \div 4^3`

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

`= 4^2`

______

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

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

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

______

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

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

`=`\(7^2\)

_______

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

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

`= 17^3`

_______

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

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

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

`= 5`

_______

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

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

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

`= 15`

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

\(x^3=125\)

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

\(\Rightarrow x=3\)

___________________

\(4^5:4^3\)

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

\(=4^2=16\)

___________________

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

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

\(=1\)

___________________

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

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

\(=7^2=49\)

___________________

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

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

\(=17^3=4913\)

___________________

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

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

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

\(=5^1=5\)

___________________

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

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

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

\(=15^1=15\)

__________________

* 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

giải hộ mình nhé

`(5+8/17):x+(-4/17):x+(3+1/7):(17+1/3)=4/11`

`(5+8/17)xx1/x-4/17xx1/x+22/7:52/17=4/11`

`(5+4/17)xx1/x+187/182=4/11`

`89/17xx1/x=-1329/2002`

`x=-118281/34034`

Số to quá ;-;

\(\left(5+\dfrac{8}{17}\right):x+\left(\dfrac{-4}{17}\right):x+\left(3+\dfrac{1}{17}\right):\left(17+\dfrac{1}{3}\right)=\dfrac{4}{11}\) 

                                        \(\left(\dfrac{93}{17}+\dfrac{-4}{17}\right):x+\dfrac{52}{17}:\dfrac{52}{3}=\dfrac{4}{11}\) 

                                                                 \(\dfrac{89}{17}:x+\dfrac{3}{17}=\dfrac{4}{11}\) 

                                                                          \(\dfrac{89}{17}:x=\dfrac{4}{11}-\dfrac{3}{17}\) 

                                                                          \(\dfrac{89}{17}:x=\dfrac{35}{187}\) 

                                                                                  \(x=\dfrac{89}{17}:\dfrac{35}{187}\) 

                                                                                   \(x=\dfrac{979}{35}\) 

Mk thấy đề bài thế này mới đúng nếu sai thì nhắc mk nha!

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

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

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

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

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

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

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

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

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

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

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

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

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

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

\(\Rightarrow x=-3\)

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 \(x\in\left\{-2;-1;0;1;2\right\}\)