\(x\) + \(\dfrac{1}{10}\) + \(x\) + \(\dfrac{1}{11}\) = \(x\) + \(\dfrac{1}{21}\)

\(x+x-x\) = \(\dfrac{1}{21}\) - \(\dfrac{1}{10}\) - \(\dfrac{1}{11}\)

\(x\)             = -  \(\dfrac{331}{2310}\)

\(x+\dfrac{1}{10}+x+\dfrac{1}{11}=x+\dfrac{1}{21}\)

\(x+x-x=\dfrac{1}{21}-\dfrac{1}{10}-\dfrac{1}{11}\)

\(x=\dfrac{-11}{210}-\dfrac{1}{11}\)

\(x=\dfrac{-331}{2310}\)

f)

`(2x+1)^3=343`

`(2x+1)^3=7^3`

`=>2x+1=7`

`2x=7-1`

`2x=6`

`x=6:2`

`x=3`

g)

`(x-1)^3 =125`

`(x-1)^3 =5^3`

`=>x-1=5`

`x=6`

h)

`2^(x+2)-2^x=96`

`2^x *2^2 -2^x =96`

`2^x (2^2 -1)=96`

`2^x *3=96`

`2^x =32`

`2^x =2^5`

`=>x=5`

i)

`(x-5)^4 =(x-5)^6` (`x>=5`)

`(x-5)^6 -(x-5)^4 =0`

`(x-5)^4 [(x-5)^2 -1]=0`

`=>x-5=0` hoặc `(x-5)^2 -1=0`

`<=>x=5` hoặc `(x-5)^2 =1`

`<=>x=5` hoặc `x-5=1` hoặc `x-5=-1`

`<=>x=5` hoặc `x=6` hoặc `x=4`

j)

`720:[41-(2x-5)]=2^3 *5`

`720:[41-(2x-5)]=8*5`

`720:[41-(2x-5)]=40`

`41-(2x-5)=720:40`

`41-(2x-5)=18`

`2x-5=41-18`

`2x-5=23`

`2x=28`

`x=14`

\(x\) + \(\dfrac{1}{4}\) + \(x\) + \(\dfrac{3}{2}\) = \(x\) + \(\dfrac{2}{3}\) + \(x\) + 4

\(x\) + \(x\) - \(x\) - \(x\) = \(\dfrac{2}{3}\) + 4 - \(\dfrac{1}{4}\) - \(\dfrac{3}{2}\)

                0 = \(\dfrac{8}{12}\) + \(\dfrac{48}{12}\) - \(\dfrac{3}{12}\) - \(\dfrac{18}{12}\)

               0  = \(\dfrac{35}{12}\) (vô lý)

Vậy   \(x\in\) \(\varnothing\) 

\(...\Rightarrow x+x-x-x=\dfrac{2}{3}+4-\dfrac{1}{4}-\dfrac{3}{2}\)

\(\Rightarrow0x=\dfrac{8}{12}+\dfrac{48}{12}-\dfrac{3}{12}-\dfrac{18}{12}\)

\(\Rightarrow0x=\dfrac{19}{12}\) (vô lý)

\(\Rightarrow x\in\varnothing\)

\(x\) + \(\dfrac{1}{10}\) + \(x\) + \(\dfrac{1}{10}\) = \(x\) + \(\dfrac{1}{21}\)

\(x+x\) - \(x\)  =  \(\dfrac{1}{21}\) -  \(\dfrac{1}{10}\) - \(\dfrac{1}{10}\) 

         \(x\)    = \(\dfrac{1}{21}\)  - \(\dfrac{1}{5}\) 

          \(x\) = - \(\dfrac{16}{105}\) 

\(27^5:9^6\)

\(=\left(3^3\right)^5:\left(3^2\right)^6\)

\(=3^{15}:3^{12}\)

\(=3^{15-12}\)

\(=3^3\)

\(\left(7\dfrac{4}{9}+4\dfrac{7}{11}\right)-3\dfrac{4}{9}\)

\(=\dfrac{67}{9}+\dfrac{51}{11}-\dfrac{31}{9}\)

\(=\dfrac{67}{9}-\dfrac{31}{9}+\dfrac{51}{11}\)

\(=4+\dfrac{51}{11}\)

\(=\dfrac{95}{11}\)

Chúc bạn học tốt

`@` `\text {Ans}`

`\downarrow`

\(\left(x-\dfrac{1}{3}\right)^2=x-\dfrac{1}{3}\)

`\Rightarrow`\(\left(x-\dfrac{1}{3}\right)^2-\left(x-\dfrac{1}{3}\right)^1=0\)

`\Rightarrow`\(\left(x-\dfrac{1}{3}\right)\left(x-\dfrac{1}{3}-1\right)=0\)

`\Rightarrow`\(\left[{}\begin{matrix}x-\dfrac{1}{3}=0\\x-\dfrac{4}{3}=0\end{matrix}\right.\)

`\Rightarrow`\(\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=\dfrac{4}{3}\end{matrix}\right.\)

Vậy, `x \in`\(\left\{\dfrac{1}{3};\dfrac{4}{3}\right\}.\)

`@` `\text {Ans}`

`\downarrow`

`a)`

\(5^x+5^{x+2}=650\)

`\Rightarrow 5^x + 5^x . 5^2 = 650`

`\Rightarrow 5^x . (1 + 5^2) = 650`

`\Rightarrow 5^x . 26 = 650`

`\Rightarrow 5^x = 650 \div 26`

`\Rightarrow 5^x = 25`

`\Rightarrow 5^x = 5^2`

`\Rightarrow x = 2`

Vậy, `x = 2`

`b)`

`(4x + 1)^2 = 25.9`

`\Rightarrow (4x + 1)^2 = 225`

`\Rightarrow (4x + 1)^2 = (+-15^2)`

`\Rightarrow`\(\left[{}\begin{matrix}4x-1=15\\4x-1=-15\end{matrix}\right.\)

`\Rightarrow `\(\left[{}\begin{matrix}4x=16\\4x=-14\end{matrix}\right.\)

`\Rightarrow `\(\left[{}\begin{matrix}x=4\\x=-\dfrac{7}{2}\end{matrix}\right.\)

Vậy, `x \in`\(\left\{-\dfrac{7}{2};4\right\}\)

`c)`

\(2^x+2^{x+3}=144\)

`\Rightarrow 2^x + 2^x . 2^3 = 144`

`\Rightarrow 2^x . (1 + 2^3) = 144`

`\Rightarrow 2^x . 9 = 144`

`\Rightarrow 2^x = 144 \div 9`

`\Rightarrow 2^x = 16`

`\Rightarrow 2^x = 2^4`

`\Rightarrow x = 4`

Vậy, `x = 4`

`d)`

\(3^{2x+2}=9^{x+3}\)

`\Rightarrow `\(3^{2x+2}=\left(3^2\right)^{x+3}\)

`\Rightarrow `\(3^{2x+2}=3^{2x+6}\)

`\Rightarrow 2x + 2 = 2x + 6`

`\Rightarrow 2x - 2x = 6 - 2`

`\Rightarrow 0 = 4 (\text {vô lý})`

Vậy, `x` không có giá trị nào thỏa mãn.

`e)`

\(x^{15}=x^2\)

`\Rightarrow `\(x^{15}-x^2=0\)

`\Rightarrow `\(x^2\cdot\left(x^{13}-1\right)=0\)

`\Rightarrow `\(\left[{}\begin{matrix}x^2=0\\x^{13}-1=0\end{matrix}\right.\)

`\Rightarrow `\(\left[{}\begin{matrix}x=0\\x^{13}=1\end{matrix}\right.\)

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

Vậy, `x \in`\(\left\{0;1\right\}.\)

`@` `\text {Ans}`

`\downarrow`

`a)`

\(\dfrac{7}{5}\cdot\dfrac{8}{19}+\dfrac{7}{5}\cdot\dfrac{12}{19}-\dfrac{7}{5}\cdot\dfrac{1}{19}\)

`=`\(\dfrac{7}{5}\cdot\left(\dfrac{8}{19}+\dfrac{12}{19}-\dfrac{1}{19}\right)\)

`=`\(\dfrac{7}{5}\cdot\dfrac{19}{19}=\dfrac{7}{5}\cdot1=\dfrac{7}{5}\)

`b)`

\(-\dfrac{3}{5}\cdot\dfrac{5}{7}+\left(-\dfrac{3}{5}\right)\cdot\dfrac{3}{7}+\left(-\dfrac{3}{5}\right)\cdot\dfrac{6}{7}\)

`=`\(-\dfrac{3}{5}\cdot\left(\dfrac{5}{7}+\dfrac{3}{7}+\dfrac{6}{7}\right)\)

`=`\(-\dfrac{3}{5}\cdot\dfrac{14}{7}\)

`=`\(-\dfrac{3}{5}\cdot2=-\dfrac{6}{5}\)

`c)`

\(10\dfrac{2}{9}+\left(2\dfrac{2}{5}-7\dfrac{2}{9}\right)\)

`=`\(10\dfrac{2}{9}+2\dfrac{2}{5}-7\dfrac{2}{9}\)

`=`\(\left(10\dfrac{2}{9}-7\dfrac{2}{9}\right)+2\dfrac{2}{5}\)

`=`\(3+2\dfrac{2}{5}=\dfrac{27}{5}\)

`d)`

\(6\dfrac{3}{10}-\left(3\dfrac{4}{7}+2\dfrac{3}{10}\right)\)

`=`\(6\dfrac{3}{10}-3\dfrac{4}{7}-2\dfrac{3}{10}\)

`=`\(\left(6\dfrac{3}{10}-2\dfrac{3}{10}\right)-3\dfrac{4}{7}\)

`=`\(4-3\dfrac{4}{7}=\dfrac{3}{7}\)

a) \(\dfrac{7}{5}.\dfrac{8}{19}+\dfrac{7}{5}.\dfrac{12}{19}-\dfrac{7}{5}.\dfrac{1}{19}\)

\(=\dfrac{7}{5}.\left(\dfrac{8}{19}+\dfrac{12}{19}-\dfrac{1}{19}\right)\)

\(=\dfrac{7}{5}.1\)

\(=\dfrac{7}{5}\)

b) \(\dfrac{-3}{5}.\dfrac{5}{7}+\dfrac{-3}{5}.\dfrac{3}{7}+\dfrac{-3}{5}.\dfrac{6}{7}\)

\(=\dfrac{-3}{5}.\left(\dfrac{5}{7}+\dfrac{3}{7}+\dfrac{6}{7}\right)\)

\(=\dfrac{-3}{5}.2\)

\(=\dfrac{-6}{5}\)

c) \(10\dfrac{2}{9}+\left(2\dfrac{2}{5}-7\dfrac{2}{9}\right)\)

\(=\dfrac{92}{9}+\dfrac{12}{5}-\dfrac{65}{9}\)

\(=\dfrac{92}{9}-\dfrac{65}{9}+\dfrac{12}{5}\)

\(=3+\dfrac{12}{5}\)

\(=\dfrac{15}{5}+\dfrac{12}{5}\)

\(=\dfrac{27}{5}\)

d) \(6\dfrac{3}{10}-\left(3\dfrac{4}{7}+2\dfrac{3}{10}\right)\)

\(=\dfrac{63}{10}-\dfrac{25}{7}-\dfrac{23}{10}\)

\(=\dfrac{63}{10}-\dfrac{23}{10}-\dfrac{25}{7}\)

\(=4-\dfrac{25}{7}\)

\(=\dfrac{28}{7}-\dfrac{25}{7}\)

\(=\dfrac{3}{7}\)

Chúc bạn học tốt

`@` `\text {Ans}`

`\downarrow`

`a)`

\(\left(\dfrac{7}{8}-\dfrac{3}{4}\right)\cdot1\dfrac{1}{3}-\dfrac{2}{3}\cdot0,5\)

`=`\(\dfrac{1}{8}\cdot\dfrac{4}{3}-\dfrac{1}{3}\)

`=`\(\dfrac{1}{6}-\dfrac{1}{3}=-\dfrac{1}{6}\)

`b)`

\(\left(2+\dfrac{5}{6}\right)\div1\dfrac{1}{5}+\left(-\dfrac{7}{12}\right)\)

`=`\(\dfrac{17}{6}\div1\dfrac{1}{5}-\dfrac{7}{12}\)

`=`\(\dfrac{85}{36}-\dfrac{7}{12}=\dfrac{16}{9}\)

`c)`

\(75\%-1\dfrac{1}{2}+0,5\div\dfrac{5}{12}\)

`=`\(-\dfrac{3}{4}+\dfrac{6}{5}=\dfrac{9}{20}\)

a) \(\left(\dfrac{7}{8}-\dfrac{3}{4}\right).1\dfrac{1}{3}-\dfrac{2}{3}.0,5\)

\(=\left(\dfrac{7}{8}-\dfrac{6}{8}\right).\dfrac{4}{3}-\dfrac{2}{3}.\dfrac{1}{2}\)

\(=\dfrac{1}{8}.\dfrac{4}{3}-\dfrac{2}{3}.\dfrac{1}{2}\)

\(=\dfrac{1}{6}-\dfrac{1}{3}\)

\(=\dfrac{-1}{6}\)

b) \(\left(2+\dfrac{5}{6}\right):1\dfrac{1}{5}+\dfrac{-7}{12}\)

\(=\left(\dfrac{12}{6}+\dfrac{5}{6}\right):\dfrac{6}{5}+\dfrac{-7}{12}\)

\(=\dfrac{17}{6}.\dfrac{5}{6}+\dfrac{-7}{12}\)

\(=\dfrac{85}{36}+\dfrac{-7}{12}\)

\(=\dfrac{16}{9}\)

c) \(75\%-1\dfrac{1}{2}+0,5:\dfrac{5}{12}\)

\(=\dfrac{3}{4}-\dfrac{3}{2}+\dfrac{1}{2}.\dfrac{12}{5}\)

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

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

\(=\dfrac{9}{20}\)