Áp dụng t/c của dãy tỉ số bằng nhau:

   \(\dfrac{x}{12}=\dfrac{y}{3}=\dfrac{x-y}{12-3}=\dfrac{36}{9}=4\)

\(\bullet \dfrac{x}{12}=4=>x=12.4=48\)

\(\bullet \dfrac{y}{3}=4=>y=3.4=12\)

x/12 = y/3 và x-y= 36

= x/12=y/3 

 = x-y/12-3

= 36/9= 4

=> x = 12.4= 48

=> y = 3.4 = 12

`|x + 1/2 | - 2/3 =` \(\sqrt{\dfrac{16}{9}}\)

`|x + 1/2 | - 2/3 =4/3`

`|x+1/2| = 4/3+2/3`

`|x+1/2| = 2`

`=>` \(\left[{}\begin{matrix}x+\dfrac{1}{2}=2\\x+\dfrac{1}{2}=-2\end{matrix}\right.\)

`=>` \(\left[{}\begin{matrix}x=2-\dfrac{1}{2}\\x=-2-\dfrac{1}{2}\end{matrix}\right.\)

`=>` \(\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{5}{2}\end{matrix}\right.\)

Vậy..

`#H.J`

\(\left|x+\dfrac{1}{2}\right|=\dfrac{4}{3}+\dfrac{2}{3}=2\Leftrightarrow\left[{}\begin{matrix}x=2-\dfrac{1}{2}=\dfrac{3}{2}\\x=-2-\dfrac{1}{2}=-\dfrac{5}{2}\end{matrix}\right.\)

1) \(\sqrt{144}-\sqrt{81}+\sqrt{\dfrac{169}{25}}=12-9+\dfrac{13}{5}=\dfrac{28}{5}\)

2) \(\dfrac{20^3\cdot\left(-49\right)^2}{14^3\cdot5^4}=\dfrac{8000\cdot2401}{2744\cdot625}=\dfrac{19208000}{1715000}=\dfrac{19208}{1715}=\dfrac{56}{5}\)

3) \(-1\dfrac{1}{2}x-\left(\dfrac{3}{4}\right)^2=1\Leftrightarrow\dfrac{-3}{2}x-\dfrac{9}{16}=1\Leftrightarrow\dfrac{-3}{2}x=\dfrac{25}{16}\Leftrightarrow x=\dfrac{-25}{24}\)

4) \(\dfrac{x-2}{2023}=\dfrac{2023}{x-2}\Leftrightarrow\left(x-2\right)^2=2023^2\Leftrightarrow x-2=2023\Leftrightarrow x=2025\)

1, \(=12-9+\dfrac{13}{5}=3+\dfrac{13}{5}=\dfrac{15+13}{5}=\dfrac{28}{5}\)

2, \(=\dfrac{5^3.4^3.7^2.7^2}{7^2.7.5^3.5}=\dfrac{7.4^3}{5}=\dfrac{448}{5}\)

3, \(=-\dfrac{3}{2}x-\dfrac{9}{16}+1=-\dfrac{3}{2}x+\dfrac{7}{16}\)

4, \(\left(x-2\right)^2=2023^2\Leftrightarrow\left[{}\begin{matrix}x=2025\\x=-2021\end{matrix}\right.\)

16^x = 64¹⁰

16^x = 1152921504606846976

=> 16^x = x = 15.

16^x = 64¹⁰

(4²)^x = (4³)¹⁰

4^2x = 4³⁰

2x = 30 

x = 30 : 2

x = 15 

\(\dfrac{4}{5}x=-3-\dfrac{1}{5}=\dfrac{-16}{5}\Leftrightarrow x=-\dfrac{16}{5}:\dfrac{4}{5}=-4\)

`1/5 + 4/5x = -3`

`4/5x = -3 - 1/5`

`4/5x = -15/5 - 1/5`

`4/5x=-16/5`

`x=-16/5:4/5`

`x=-16/5 xx 5/4`

`x=-4`

\(3^{x+1}-3^x=54\Leftrightarrow3^x\left(3-1\right)=54\Leftrightarrow3^x=27\Leftrightarrow3^x=3^3\Rightarrow x=3\)

3^x+1 = 54 + 3^x

<=> 3^x+1 - 3^x = 54

<=> 3^x . 2 = 54

<=> 3^x = 27

=> 3^x = 3³ => x = 3 

\(a+b=c+d\Leftrightarrow\left(a+b\right)^3=\left(c+d\right)^3\)

\(\Leftrightarrow a^3+b^3+3ab\left(a+b\right)=c^3+d^3+3cd\left(c+d\right)\)

\(\Leftrightarrow ab\left(a+b\right)=cd\left(c+d\right)\)

\(\Leftrightarrow ab=cd\)

\(\Leftrightarrow\dfrac{a}{c}=\dfrac{d}{b}=t\Leftrightarrow a=ct,d=bt\)

\(\left(a+b\right)^2\left(a^3+b^3\right)=a^5+b^5+2ab\left(a^3+b^3\right)+a^2b^2\left(a+b\right)\)

\(\left(c+d\right)^2\left(c^3+d^3\right)=c^5+d^5+2cd\left(c^3+d^3\right)+c^2d^2\left(c+d\right)\)

mà \(a+b=c+d,a^3+b^3=c^3+d^3,ab=cd\)

suy ra \(a^5+b^5=c^5+d^5\).

   9 ≤ 3ⁿ < 1000

  3² ≤ 3ⁿ < 10³

3² ≤ 3ⁿ  ≤  9³ < 10³

3² ≤ 3ⁿ ≤  3⁶ < 10³

2 ≤ n ≤ 6