tìm x:
\(\frac{5^5}{5^x}\)\(=\)\(5^{18}\)
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a, \(x-\frac{2}{5}=\frac{5}{7}\)
\(x=\frac{5}{7}+\frac{2}{5}\)
\(x=\frac{39}{35}\)
b, \(\frac{5}{12}-x=\frac{7}{18}\)
\(x=\frac{5}{12}-\frac{7}{18}\)
\(x=\frac{1}{36}\)
x-2/5=-5/2
x=-5/2+2/5
x=-21/10
vậy x=21/10
5/12-x=7/18
x=5/12-7/18
x=1/36
vậy x=1/35
\(\frac{10^{18}}{2^{18}}=\frac{\left(2.5\right)^{18}}{2^{18}}=\frac{2^{18}.5^{18}}{2^{18}}=5^{18}\)
vậy 5x.5x+1.5x+2=5x+(x+1)+(x+2)<518=>x+(x+1)+(x+2) <18
vậy 3x+3<18 thì3x<15 =>x<5
x <5 vay65 x thuộc 1;2;3;4
vậy x=1;2;3;4
\(\frac{1}{5^{18}}=\frac{5^{18}}{5^x}\)
\(\Rightarrow5^x=5^{18}.5^{18}\)
\(\Rightarrow5^x=5^{36}\)
\(\Rightarrow x=36\)
Chúc bn học tốt
\(5^{18}=\frac{5^{18}}{5^x}\)
\(\Rightarrow5^{18}.5^x=5^{18}\)
\(\Rightarrow5^x=1\)
\(\Rightarrow5^x=5^0\)
\(\Rightarrow x=0\)
Vậy \(x=0\)
Tham khảo nhé~
Ta có: \(\dfrac{-5}{x}=\dfrac{-y}{8}=\dfrac{-18}{72}\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{-5}{x}=-\dfrac{18}{72}=\dfrac{-1}{4}\\\dfrac{-y}{8}=\dfrac{-18}{72}=\dfrac{-1}{4}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=20\\y=2\end{matrix}\right.\)
Vậy: (x,y)=(20;2)
\(\dfrac{x+1}{x-5}+\dfrac{x-18}{x-5}+\dfrac{x+2}{x-5}\)
\(=\dfrac{x+1+x-18+x+2}{x-5}\)
\(=\dfrac{3x-15}{x-5}\)
\(=\dfrac{3\left(x-5\right)}{x-5}\)
\(=\dfrac{3}{1}\)
\(=3\)
\(\dfrac{x+1}{x-5}+\dfrac{x-18}{x-5}+\dfrac{x+2}{x-5}\\ =\dfrac{x+1+x-18+x+2}{x-5}\\ =\dfrac{\left(x+x+x\right)+\left(1-18+2\right)}{x-5}\\ =\dfrac{3x-15}{x-5}=\dfrac{3\left(x-5\right)}{x-5}=3\)
Tính lần lượt 2 vế ta được:
-5 < x< -0,4
=> x = -4, -3, -2, -1, 0
5-13 nhá bạn
55/5x=518
=> 5x=55:518
=> 5x =8,192x10 -10