a: \(\left(2x-3\right)\cdot12=6^2\)

\(\Leftrightarrow2x-3=3\)

hay x=3

a) (2x-3).12=6²

=> (2x-3).12= 36

=> ( 2x-3)= 36 : 12

=> 2x-3 = 3

=> 2x = 3+3

=> 2x = 6

=> x = 6:2

=> x= 3

b)( 5x - 2 ) . 6³= 3. 6⁵
=> ( 5x - 2 ). 216= 23328

=> ( 5x - 2 ) = 23328 : 216

=> ( 5x - 2) = 108

=> 5x = 108 + 2

=> 5x = 110

=> x = 110 : 5

=> x = 22

GIÚP MÌNH VỚI MN ƠI

\(C=\dfrac{6^3+3\cdot6^2+3^3}{13}=\dfrac{3^3\cdot8+3^3\cdot4+3^3}{13}=27\)

MN ƠI GIÚP MÌNH VỚI MÌNH CẦN GẤP

\(A=\dfrac{6^3+3\cdot6^2+3^3}{13}\)

\(=\dfrac{3^3\cdot8+3^3\cdot4+3^3}{13}\)

=27

\(\dfrac{6^3+3\cdot6^2+3^5}{-13}=\dfrac{3^3\cdot2^3+3\cdot2^2\cdot3^2+3^5}{-13}\)

\(=\dfrac{3^3\left(2^3+2^2+3^2\right)}{-13}=\dfrac{3^3\cdot21}{-13}=\dfrac{-567}{13}\)

=2700

=> 2700

a/3125

(0.6)^5/(0.2)^6 
= [(0.2)^5. 3^5]/(0.2)^6 
= 3^5/0.2 
=243/0.2 
= 1215 

6^3+3.6^2+3^3/-13 
= 2^3.3^3+3.2^2.3^2+3^3/-13 
=3^3(2^3+2^2+1)/-13 
=3^3.13/-13 
=-27

\(\frac{6^3+3\cdot6^2+3^3}{-13}=\frac{\left(3\cdot2\right)^3+3\cdot\left(3\cdot2\right)^2+3^3}{-13}=\frac{3^3\cdot2^3+3\cdot3^2\cdot2^2+3^3}{-13}=\frac{3^3\cdot\left(2^3+2^2+1\right)}{-13}=\frac{27\cdot13}{-13}=-27\)

`@` `\text {Ans}`

`\downarrow`

`a.`

\(0,3-\dfrac{4}{9}\div\dfrac{4}{3}\cdot\dfrac{6}{5}+1\)

`=`\(0,3-\dfrac{1}{3}\cdot\dfrac{6}{5}+1\)

`=`\(0,3-0,4+1\)

`= -0,1 + 1`

`= 0,9`

`b.`

\(1+2\div\left(\dfrac{2}{3}-\dfrac{1}{6}\right)\cdot\left(-2,25\right)\)

`=`\(1+2\div\dfrac{1}{2}\cdot\left(-2,25\right)\)

`=`\(1+4\cdot\left(-2,25\right)\)

`= 1+ (-9) = -8`

`c.`

\(\left[\left(\dfrac{1}{4}-0,5\right)\cdot2+\dfrac{8}{3}\right]\div2\)

`=`\(\left(-\dfrac{1}{4}\cdot2+\dfrac{8}{3}\right)\div2\)

`=`\(\left(-\dfrac{1}{2}+\dfrac{8}{3}\right)\div2\)

`=`\(\dfrac{13}{6}\div2\)

`=`\(\dfrac{13}{12}\)

`d.`

\(\left[\left(\dfrac{3}{8}-\dfrac{5}{12}\right)\cdot6+\dfrac{1}{3}\right]\cdot4\)

`=`\(\left(-\dfrac{1}{24}\cdot6+\dfrac{1}{3}\right)\cdot4\)

`=`\(\left(-\dfrac{1}{4}+\dfrac{1}{3}\right)\cdot4\)

`=`\(\dfrac{1}{12}\cdot4=\dfrac{1}{3}\)

`e.`

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

`=`\(-\dfrac{1}{5}\div\dfrac{3}{5}-\dfrac{1}{3}\)

`=`\(-\dfrac{1}{3}-\dfrac{1}{3}=-\dfrac{2}{3}\)

`f.`

\(0,8\div\left\{0,2-7\left[\dfrac{1}{6}+\left(\dfrac{5}{21}-\dfrac{5}{14}\right)\right]\right\}\)

`=`\(0,8\div\left[0,2-7\left(\dfrac{1}{6}-\dfrac{5}{42}\right)\right]\)

`=`\(0,8\div\left(0,2-7\cdot\dfrac{1}{21}\right)\)

`=`\(0,8\div\left(0,2-\dfrac{1}{3}\right)\)

`= 0,8 \div (-2/15)`

`=-6`

`@` `yHGiangg.`