\(\frac{5^4.20^4}{25^5.4^5}=\frac{5^4.5^4.4^4}{5^{10}.4^5}=\frac{5^8.4^4}{5^{10}.4^5}=\frac{1}{5^2.4}=\frac{1}{25.4}=\frac{1}{100}\)

\(\frac{5^4.20^4}{25^5.4^5}=\frac{5^4.4^4.5^4}{\left(5^2\right)^5.4^4.4}=\frac{5^4.4^4.5^4}{5^{10}.4^4.4}=\frac{5^4.4^4.5^4}{5^4.5^4.5^2.4^4.4}=\frac{1}{5^2.4}=\frac{1}{100}\)

\(\frac{5^4.20^4}{25^5.4^5}=\frac{5^4.5^4.4^4}{\left(5^2\right)^5.4^5}=\frac{5^8.4^4}{5^{10}.4^5}=\frac{1}{25.4}=\frac{1}{100}\)

=>1/4:(x-2/3)=2

=>(x-2/3)=1/8

=>x=1/8+2/3=3/24+16/24=19/24

13/4 - 1/4 : ( x - 2/3 )= 5/4

\(\Rightarrow\) 1/4 : ( x- 2/3 ) = 13/4 - 5/4

\(\Rightarrow\) 1/4 : ( x- 2/3)= 2

\(\Rightarrow\) x - 2/3 = 1/4 :2

\(\Rightarrow\) x- 2/3 = 1/8

\(\Rightarrow\) x= 1/8 +2/3 =19/24

Vậy x = 19/24

\(\dfrac{13}{4}-\dfrac{1}{4}\div\left(x-\dfrac{2}{3}\right)=\dfrac{5}{4}\\ \Rightarrow\dfrac{1}{4}\div\left(x-\dfrac{2}{3}\right)=\dfrac{13}{4}-\dfrac{5}{4}\\ \Rightarrow\dfrac{1}{4}\div\left(x-\dfrac{2}{3}\right)=\dfrac{8}{4}=2\\ \Rightarrow x-\dfrac{2}{3}=\dfrac{1}{4}\div2\\ \Rightarrow x-\dfrac{2}{3}=\dfrac{1}{4}\times\dfrac{1}{2}\\ \Rightarrow x-\dfrac{2}{3}=\dfrac{1}{8}\\ \Rightarrow x-\dfrac{16}{24}=\dfrac{3}{24}\\ \Rightarrow x=\dfrac{3}{24}+\dfrac{16}{24}\\ \Rightarrow x=\dfrac{19}{24}\)

Vậy \(x=\dfrac{19}{24}\) 

Ta có : \(\frac{x}{3}=\frac{y}{5}\)

\(\Rightarrow x=3k\) ; \(y=5k\)

Thay \(x=3k\) và \(y=5k\) vào biểu thức \(x+2y=10\) ta có :

\(3k+2\times5k=10\)

\(3k+10k=10\)

\(\left(3+10\right)k=10\)

\(13k=10\)

\(\Rightarrow k=\frac{10}{13}\)

Vậy : 

\(\hept{\begin{cases}x=3k=3\times\frac{10}{13}=\frac{30}{13}\\y=5k=5\times\frac{10}{13}=\frac{50}{13}\end{cases}}\)

Mk ko biết đúng ko, đúng thì k cho mk nha

Đặt x/3=y/5=k
=> x=3k và y=5k
x+2y=10
3k+2.5k=10
3k+10k=10
13k=10
k=10/13
x=3k=3.10/13=30/13
y=5k=5.10/13=50/13

C2 

TCCDTSBN 

\(\frac{x}{3}=\frac{y}{5}=\frac{x+2y}{3+2.5}=\frac{10}{13}\)

=> x \(\frac{10}{13}.3=\frac{30}{13}\)

=> y =\(\frac{10}{13}.5=\frac{50}{13}\)

Ta có: \(\dfrac{6^5\cdot\left(-12\right)^6}{\left(-4\right)^9\cdot\left(-3\right)^{10}}\)

\(=-\dfrac{3^5\cdot2^5\cdot12^6}{4^9\cdot3^{10}}\)

\(=-\dfrac{2^5\cdot3^6\cdot4^6}{4^9\cdot3^5}\)

\(=-\dfrac{2^5\cdot3}{4^3}\)

\(=-\dfrac{2^5}{2^6}\cdot3=-\dfrac{3}{2}\)

 ta được  \(\dfrac{6^5.12^6}{4^8.\left(-4\right).3^{10}}\) \(=\dfrac{2^5.3^5.2^{12}.3^6}{2^{16}.\left(-4\right).3^{10}}\) \(=\dfrac{2^{17}.3^{11}}{2^{16}.\left(-4\right).3^{10}}=\dfrac{-6}{4}=\dfrac{-3}{2}\)

cau 1x=-31/60

cau 2 x=-5/8

Bài 1:

Ta có: \(\frac{497}{-499}=-\frac{497}{499}>-\frac{499}{499}=-1\left(1\right)\)

\(-\frac{2345}{2341}< -\frac{2341}{2341}=-1\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\frac{497}{-499}>-\frac{2345}{2341}\)

Bài 2:

\(\frac{x+5}{2005}+\frac{x+6}{2004}=\frac{x+7}{2003}+3=0\)

\(\Rightarrow\frac{x+5}{2005}+\frac{x+6}{2004}+\frac{x+7}{2003}+3=0\)

\(\Rightarrow\frac{x+5}{2005}+1+\frac{x+6}{2004}+1+\frac{x+7}{2003}+1=0\)

\(\Rightarrow\frac{x+2010}{2005}+\frac{x+2010}{2004}+\frac{x+2010}{2003}=0\)

\(\Rightarrow\left(x+2010\right)\times\left(\frac{1}{2005}+\frac{1}{2004}+\frac{1}{2003}\right)=0\)

Vì \(\left(\frac{1}{2005}+\frac{1}{2004}+\frac{1}{2003}\right)\ne0\Rightarrow x+2010=0\)

\(\Rightarrow x=0-2010=-2010\)

Vậy x = -2010