câu 1 thiếu đề
câu 2:
Ta có: 2150=(26)25=6425
3100=(34)25=8125
Vì 6425<8125 nên 2150<3100
x o dau vay???
2^150 =(2^3)^50=8^ 50
3^100= (3^2)^50 =9^50
ma 8^50< 9^50=> 2^150<3^100
câu 1 thiếu đề
câu 2:
Ta có: 2150=(26)25=6425
3100=(34)25=8125
Vì 6425<8125 nên 2150<3100
x o dau vay???
2^150 =(2^3)^50=8^ 50
3^100= (3^2)^50 =9^50
ma 8^50< 9^50=> 2^150<3^100
Tim x \(\in\) Z thoa
\(3\frac{1}{3}:2\frac{1}{2}\) <x<\(7\frac{2}{3}.\frac{3}{7}+\frac{5}{2}\)
Tim x,y,z:\(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}\) và 2x+3y-z=95
Tim x,y\(\in\) z biet :
a)\(\frac{x}{3}-\frac{2}{y}=\frac{1}{5}\) b)\(\frac{2}{x}-\frac{y}{3}=\frac{5}{6}\)
c)\(\frac{5}{x}-\frac{y}{3}=\frac{1}{6}\) d) \(\frac{x}{6}-\frac{2}{y}=\frac{1}{4}\)
Tim x biet :
a, \(4\frac{1}{3}\left(\frac{1}{6}-\frac{1}{2}\right)\le x\le\frac{2}{3}\left(\frac{1}{2}-\frac{1}{3}-\frac{3}{4}\right)\) x thuoc Z
b , \(|x-3|+1=x\)
tim x \(\in\)Z, de
a, A= \(\frac{x+2}{3}\)\(\in\)Z
b, B = \(\frac{7}{x-1}\)\(\in\) Z
c, C = \(\frac{x+1}{x-1}\)\(\in\)Z
tim cac so nguyen duong x y thoa man \(\frac{x}{2}+\frac{x}{y}-\frac{3}{2}=\frac{10}{y}\)
Tim x thuoc Z, biet: \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\left(x+1\right):2}=\frac{2009}{2011}\)
Tìm \(x\in Z\) \(3\frac{2}{3}.\left(\frac{1}{5}-\frac{1}{2}\right)\le x\le\frac{3}{11}.\left(\frac{1}{5}+\frac{2}{3}-\frac{1}{2}\right)\)
Giúp với
bài 1 tim x y thuộc z
\(\frac{3}{x-5}=\frac{x-5}{27}\)
\(\frac{3}{x}=\frac{y}{35}=\frac{-36}{84}\)
\(\frac{3x+1}{-2}=\frac{-8}{3x+1}\)
\(\frac{3}{x-5}=\frac{-4}{x+2}\)