(-10/3)5 . (-6/5)4
giải chi tiết giùm mk nhoa
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\(\frac{5^4.20^4}{25^5.4^5}=\frac{5^4.5^4.4^4}{5^{10}.4^5}=\frac{5^8.4^4}{5^{10}.4^5}=\frac{1}{5^2.4}=\frac{1}{25.4}=\frac{1}{100}\)
\(\frac{5^4.20^4}{25^5.4^5}=\frac{5^4.4^4.5^4}{\left(5^2\right)^5.4^4.4}=\frac{5^4.4^4.5^4}{5^{10}.4^4.4}=\frac{5^4.4^4.5^4}{5^4.5^4.5^2.4^4.4}=\frac{1}{5^2.4}=\frac{1}{100}\)
=>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
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
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 \(\dfrac{6^5.12^6}{4^8.\left(-4\right).3^{10}}\) \(=\dfrac{2^5.3^5.2^{12}.3^6}{2^{16}.\left(-4\right).3^{10}}\) \(=\dfrac{2^{17}.3^{11}}{2^{16}.\left(-4\right).3^{10}}=\dfrac{-6}{4}=\dfrac{-3}{2}\)
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
\(\left(-\frac{10}{3}\right)^5.\left(-\frac{6}{5}\right)^4=\frac{-\left(2.5\right)^5}{3^5}.\frac{\left(3.2\right)^4}{5^4}=\frac{-2^5.5^5.3^4.2^4}{3^5.5^4}=\frac{-2^9.5}{3}\)
Ta có: \(\left(-\frac{10}{3}\right)^5.\left(-\frac{6}{5}\right)^4=\frac{\left(-10\right)^5}{3^5}.\frac{\left(-6\right)^4}{5^4}=\frac{\left(-2.5\right)^5}{3^5}.\frac{\left(-2.3\right)^4}{5^4}=\frac{\left(-2\right)^5.5^5}{3^5}.\frac{\left(-2\right)^5.3^4}{5^4}=\frac{\left(-2\right)^5.5^5.\left(-2\right)^4.3^4}{3^5.5^4}\)\(=\frac{\left[\left(-2\right)^5.\left(-2\right)^4\right].5^5.3^4}{3^5.5^4}=\frac{\left(-2\right)^9.5}{3}=\frac{-512.5}{3}=-\frac{2560}{3}\)
Chuk bn hk tốt!