Ta có : 1, 1/ 2.(x+1) = 1/6 .
⇒ 6 = 2.(x+1) .
⇒ x + 1 = 6 : 2 .
⇒ x + 1 = 3 .
⇒ x = 3 - 1 .
⇒ x = 2 .
2, 1 < x/5 ≤ 2
⇒ 5/5 < x/5 ≤ 10/5 .
⇒ 5 < x ≤ 10 .
⇒ x = 6 ; 7 ; 8 ; 9 ; 10 .
Ta có : 1, 1/ 2.(x+1) = 1/6 .
⇒ 6 = 2.(x+1) .
⇒ x + 1 = 6 : 2 .
⇒ x + 1 = 3 .
⇒ x = 3 - 1 .
⇒ x = 2 .
2, 1 < x/5 ≤ 2
⇒ 5/5 < x/5 ≤ 10/5 .
⇒ 5 < x ≤ 10 .
⇒ x = 6 ; 7 ; 8 ; 9 ; 10 .
Tính:
a) \(\dfrac{-5}{6}+\dfrac{1}{6}\)
b) \(\dfrac{1}{3}-\dfrac{5}{4}.\dfrac{4}{15}\)
c) \(4.\left(-5\right)^2+\left(-2\right)^3.25\)
d) \(\dfrac{-2}{7}+\dfrac{1}{5}:\dfrac{7}{15}\)
Tính
a, \(1\dfrac{7}{20}\) : 2,7+2,7:1,35+(0,4:\(2\dfrac{1}{2}\) ) . (4,2-\(1\dfrac{3}{40}\) )
b, \(\left(6\dfrac{3}{5}-3\dfrac{3}{14}\right).5\dfrac{5}{6}:\left(21-1.25\right):2,5\)
c,\(\dfrac{0,125-\dfrac{1}{5}+\dfrac{1}{7}}{0,375-\dfrac{3}{5}+\dfrac{3}{7}}+\dfrac{\dfrac{1}{2}+\dfrac{1}{3}-0,2}{\dfrac{3}{4}+0,5-\dfrac{3}{10}}\)
1) Chứng minh rằng : \(\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+...+\dfrac{1}{2^n}< 1\)
2) Cho : \(A=\dfrac{8n+193}{4n+3}\)
Tìm n ϵ N để : a) A là số tự nhiên.
b) A là phân số tối giản.
3) Tìm các số nguyên tố x, y biết : \(\left(x-2\right)^2.\left(y-3\right)^2=-4\)
4) Tìm x ∈ N biết : \(\left(x-5\right).\dfrac{30}{100}=\dfrac{20.x}{100}+5\)
Tìm số nguyên x,y biết
a \(\dfrac{1}{x}=\dfrac{1}{6}+\dfrac{y}{3}\)
b\(\dfrac{x}{8}-\dfrac{2}{y}=\dfrac{3}{4}\)
c\(\dfrac{2x+5}{5}=\dfrac{x}{5}\)
a,1+5+9+13+16+....+x=501501
b,\(\left(x-\dfrac{1}{3}\right)^2-\dfrac{1}{4}=0\)
c,\(2^x+642=5^y\)
d,x+(x+1)+(x+2)+....+(x+30)=1240
1)Cho B=\(\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{19}\).Chứng minh B>1
2)Tính nhanh:M=\(\dfrac{\dfrac{3}{5}+\dfrac{3}{7}-\dfrac{3}{11}}{\dfrac{4}{5}+\dfrac{4}{7}-\dfrac{4}{11}}\)
tính bằng cách hợp lý
A=\(\dfrac{-2}{3}+\dfrac{5}{3}:10-\dfrac{1}{27}\times\left(-3\right)^2\)
Tìm x:
a) \(x\) + \(\dfrac{-3}{7}=\dfrac{4}{7}\).
b) \(\dfrac{1}{2}x-75\%=\dfrac{1}{4}.\)
c) \(\left|x-\dfrac{2}{3}\right|+2,25=\dfrac{3}{4}\).
Chứng tỏ rằng B = \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+\dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+\dfrac{1}{8^2}< 1\)