\(\dfrac{4}{5}=\dfrac{12}{15}\\ \dfrac{15}{21}=\dfrac{5}{7}\\ \dfrac{3}{8}=\dfrac{12}{32}\\ \dfrac{36}{24}=\dfrac{6}{4}\)

4/5 = 12/15

15/21 = 5/7

3/8 =12/32

36/24 =6/4

4/5 = 12/15

15/21 = 5/7

3/8 =12/32

36/24 =6/4

=> (x - 3) - 8 = 12 : 4
=> (x - 3) - 8 = 3
=> x - 3 = 8 + 3 
=> x - 3 = 11
=> x = 11 + 3 
=> x = 14

12 : (x - 3) - 8 = 4

12 : ( x - 3 ) = 4 + 8

12 : ( x - 3 ) = 12

x - 3 = 12 : 12

x - 3 = 1

x = 1 + 3

x = 4

\(5\sqrt{4x-16}-\dfrac{7}{3}\sqrt{9x-36}=36-3\sqrt{x-4}\)

\(\Leftrightarrow10\sqrt{x-4}-7\sqrt{x-4}+3\sqrt{x-4}=36\)

\(\Leftrightarrow\sqrt{x-4}=6\)

\(\Leftrightarrow x-4=36\)

hay x=40

giúp mình đi mọi người

2/3=8/12

5/6=10/12

3/4=9/12

giúp mình đi mọi người ơi

\(\dfrac{3^8\cdot20^5-3^9\cdot5^5\cdot2^9}{6^8\cdot10^4-3^8\cdot2^9\cdot5^4}=\dfrac{3^8\cdot2^{10}\cdot5^5-3^9\cdot5^5\cdot2^9}{2^8\cdot3^8\cdot2^4\cdot5^4-3^8\cdot2^9\cdot5^4}\\ =\dfrac{3^8\cdot2^9\cdot5^5\left(2-3\right)}{2^9\cdot3^8\cdot5^4\left(2^3-1\right)}=\dfrac{-5}{2^3-1}=\dfrac{-5}{7}\)

tham khảo 

Vì a chia cho 8 dư 6⇒(a+2)⋮8

a chia cho 12 dư 10 ⇒(a+2)⋮12

Do đó (a+2)∈BC(12;8) mà BCNN(12,8)=24.

Do đó (a+2)⋮24⇒a chia cho 24 dư 22

nó tương tự á

hey , đề bài sai ròi .

A = \(\dfrac{1}{1+3}\) + \(\dfrac{1}{1+3+5}\) + \(\dfrac{1}{1+3+5+7}\) + ... + \(\dfrac{1}{1+3+5+7+...+2021}\)

\(\Leftrightarrow\) A = \(\dfrac{1}{\dfrac{\left(1+3\right).2}{2}}\) + \(\dfrac{1}{\dfrac{\left(1+5\right).3}{2}}\) + \(\dfrac{1}{\dfrac{\left(1+7\right).4}{2}}\) + ... + \(\dfrac{1}{\dfrac{\left(1+2021\right).1011}{2}}\)

= \(\dfrac{2}{2.4}\) + \(\dfrac{2}{3.6}\) + \(\dfrac{2}{4.8}\) + ... + \(\dfrac{2}{1011.2021}\)

= \(\dfrac{1}{2.2}\) + \(\dfrac{1}{3.3}\) + \(\dfrac{1}{4.4}\) + ... + \(\dfrac{1}{2021.2021}\)

A < \(\dfrac{1}{4}\) + ( \(\dfrac{1}{2.3}\) + \(\dfrac{1}{3.4}\) + ... + \(\dfrac{1}{2020.2021}\) )

< \(\dfrac{1}{4}\) + ( \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{4}\) + ... + \(\dfrac{1}{2020}\) - \(\dfrac{1}{2021}\) )

< \(\dfrac{1}{4}\) + ( \(\dfrac{1}{2}\) - \(\dfrac{1}{2021}\) ) < \(\dfrac{1}{4}\) + \(\dfrac{1}{2}\) = \(\dfrac{3}{4}\)

Kiểu như vậy hả ?

a = |2x-1/3|-7/4

   Do |2x-1/3| \(\ge\) 0

         |2x-1/3|-7/4 \(\ge\)  7/4 

Dấu = xảy ra <=> 2x-1/3=0. =>. x= 1/6

b    1/3|x-2|+2|3-1/2 y|+4

 Do |x-2| \(\ge\) 0

      |3-1/2y| \(\ge\) 0

   => 1/3|x-2|+2|3-1/2 y|+4 \(\ge\) 4

Dấu = xảy ra <=>\(\left\{{}\begin{matrix}x-2=0\\3-\dfrac{1}{2}y=0\end{matrix}\right.\)

<=>\(\left\{{}\begin{matrix}x=2\\y=6\end{matrix}\right.\)

a: Ta có: \(\left|2x-\dfrac{1}{3}\right|\ge0\forall x\)

\(\Leftrightarrow\left|2x-\dfrac{1}{3}\right|-\dfrac{7}{4}\ge-\dfrac{7}{4}\forall x\)

Dấu '=' xảy ra khi \(x=\dfrac{1}{6}\)

b: Ta có: \(\dfrac{1}{3}\left|x-2\right|\ge0\forall x\)

\(2\left|3-\dfrac{1}{2}y\right|\ge0\forall y\)

Do đó: \(\dfrac{1}{3}\left|x-2\right|+2\left|3-\dfrac{1}{2}y\right|\ge0\forall x,y\)

\(\Leftrightarrow\left|x-2\right|\cdot\dfrac{1}{3}+\left|3-\dfrac{1}{2}y\right|\cdot2+4\ge4\forall x,y\)

Dấu '=' xảy ra khi x=2 và y=6

b: 84/96=42/48=21/24=7/8

d: 126/72=21/12=7/4

43/30

1/16

10/21

5/3

43/30

1/16

10/21

5/3

Bài 1: Tính

a, 5/6 + 3/5=40/30

b, 11/16 - 5/8=1/16

c, 5/9 x 6/7=30/63

d, 5/4 : 3/4=5/3

giúp me