tìm x:
\(\frac{5^5}{5^x}\)\(=\)\(5^{18}\)
Tìm x, biết:
\(5^{18}=\frac{5^5}{5^x}\)
5^18 = 5^5-x
=> 18 = 5-x
=> x = 5 - 18 = -13
Vậy,......
\(5^{18}=\frac{5^5}{5^x}\Rightarrow5^{18+x}=5^5\Rightarrow18+x=5\Rightarrow x=-13\)
1) \(\frac{24}{-12}=\frac{x}{5}=\frac{-y}{3}\)Tìm x và y
2) \(\frac{1}{3}+\frac{-2}{5}+\frac{1}{6}+\frac{-5}{25}\le\frac{x}{10}< \frac{-3}{4}+\frac{4}{14}+\frac{-2}{8}+\frac{-3}{5}+\frac{5}{7}\)Tìm x
3) \(\frac{8.x+18}{2.x+6}\)Tìm x
Câu 1: Tìm x
a) \(x-\frac{2}{5}=\frac{5}{7}\)
b) \(\frac{5}{12}-x=\frac{7}{18}\)
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
câu b mình sai nên mình làm lại
5/12-x=7/18
x=5/12-7/18
x=1/36
vậy x=1/36
Tìm số nguyên x biết 5x.5x+1.5x+2<\(\frac{10^{18}}{2^{18}}\)
\(\frac{10^{18}}{2^{18}}=\frac{\left(2.5\right)^{18}}{2^{18}}=\frac{2^{18}.5^{18}}{2^{18}}=5^{18}\)
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
\(\frac{10^{18}}{2^{18}}=\frac{\left(2.5\right)^{18}}{2^{18}}=\frac{2^{18}.5^{18}}{2^{18}}=5^{18}\)
Vậy 5x.5x+1.5x+2=5x+(x+1)-(x+2)<518
=> x+(x+1)-(x+2)<18
Vậy 3x+3<18 thì 3x<15
=> x<5
Vậy x={ 1;2;3;4}
Tìm x:
1/ \(5^{18}=\frac{5^{18}}{5^x}\)
\(\frac{1}{5^{18}}=\frac{5^{18}}{5^x}\)
\(\Rightarrow5^x=5^{18}.5^{18}\)
\(\Rightarrow5^x=5^{36}\)
\(\Rightarrow x=36\)
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~
#Nguyễn Phạm Thi Vân bạn viết 1/ kiểu đó mk tưởng bạn viết 1/518 chứ
\(\frac{2}{\left(x+1\right)\left(x+3\right)}+\frac{5}{\left(x+3\right)+\left(x+8\right)}+\frac{10}{\left(x+8\right)\left(x+18\right)}+\frac{1}{x+18}=\frac{5}{6}\)
Tìm x;y;z
\(\frac{-5}{x}\)=\(\frac{-y}{8}\)=\(\frac{-18}{72}\)
Yêu cầu: tìm x, y
DOGE LIKE
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)
\(\dfrac{-5}{x}=\dfrac{-y}{8}=\dfrac{-18}{72}\)
+)\(\dfrac{-5}{x}=\dfrac{-18}{72}\)
\(\Rightarrow x=\dfrac{-5.72}{-18}=20\)
+)\(\dfrac{-y}{8}=\dfrac{-18}{72}\)
\(\Rightarrow-y=\dfrac{-18.8}{72}=-2\)
\(\Rightarrow y=2\)
Thực hiện phép tính sau:
\(\frac{x+1}{x-5}+\frac{x-18}{x-5}+\frac{x+2}{x-5}\)
\(\dfrac{x+1}{x-5}+\dfrac{x-18}{x-5}+\dfrac{x+2}{x-5}\)
\(=\dfrac{x+1+x-18+x+2}{x-5}\)
\(=\dfrac{3x-15}{x-5}\)
\(=\dfrac{3\left(x-5\right)}{x-5}\)
\(=\dfrac{3}{1}\)
\(=3\)
\(\dfrac{x+1}{x-5}+\dfrac{x-18}{x-5}+\dfrac{x+2}{x-5}\\ =\dfrac{x+1+x-18+x+2}{x-5}\\ =\dfrac{\left(x+x+x\right)+\left(1-18+2\right)}{x-5}\\ =\dfrac{3x-15}{x-5}=\dfrac{3\left(x-5\right)}{x-5}=3\)
tìm x
\(4\frac{5}{9}:2\frac{5}{18}-7< x< \left(3\frac{1}{5}:3,2+4,5\times1\frac{31}{45}\right):\left(-21\frac{1}{2}\right)\)
Tính lần lượt 2 vế ta được:
-5 < x< -0,4
=> x = -4, -3, -2, -1, 0