chứng tỏ rằng :
a) |x.y| = |x| . | y| b)\(\left|\dfrac{x}{y}\right|\) = \(\left|\dfrac{x}{y}\right|\) với y ≠ 0
tìm x biết :
a) \(\left|x+\dfrac{1}{2}\right|\)=\(\dfrac{5}{2}\) b) \(\left|2x-\dfrac{2}{3}\right|\)+\(\dfrac{1}{3}\)=0 c) |x-2| = 2x + 1
1.Tính nhanh:
A= \(\dfrac{\dfrac{2}{3}-\dfrac{1}{4}+\dfrac{5}{11}}{\dfrac{5}{12}+1-\dfrac{7}{11}}\)
2. Cho: B =\(\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{19}\) .Hãy chứng tỏ rằng B > 1.
3. Rút gọn:
a) C= \(\left(1-\dfrac{1}{2}\right).\left(1-\dfrac{1}{3}\right).\left(1-\dfrac{1}{4}\right)....\left(1-\dfrac{1}{20}\right)\)
b) D= \(1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2012}}\)
4. So sánh: E=\(\dfrac{20^{10}+1}{20^{10}-1}\) và F =\(\dfrac{20^{10}-1}{20^{10}-3}\)
5. Tính giá trị của biểu thức:
M= \(\dfrac{\dfrac{3}{5}+\dfrac{3}{7}-\dfrac{3}{11}}{\dfrac{4}{5}+\dfrac{4}{7}-\dfrac{4}{11}}\)
Tìm x, biết:
a).\(\left(\dfrac{1}{1.101}+\dfrac{1}{2.102}+...+\dfrac{1}{10.110}\right).x=\dfrac{1}{1.11}+\dfrac{1}{2.12}+...+\dfrac{1}{100.110}\)
b).\(x-\dfrac{20}{11.13}-\dfrac{20}{13.15}-\dfrac{20}{15.17}-...-\dfrac{20}{53.55}=\dfrac{3}{11}\)
c).\(\dfrac{x-1}{99}+\dfrac{x-2}{98}+\dfrac{x-5}{95}=3+\dfrac{1}{99}+\dfrac{1}{98}+\dfrac{1}{95}\)
Mấy bạn tính nhanh, hợp lí, giải ra từng bước dùm mik nha
Thanks m.n
Tìm số nguyên \(x\), biết rằng :
\(4\dfrac{1}{3}.\left(\dfrac{1}{6}-\dfrac{1}{2}\right)\le x\le\dfrac{2}{3}.\left(\dfrac{1}{3}-\dfrac{1}{2}-\dfrac{3}{4}\right)\)
Cho ba phân số bằng nhau \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}\) . Chứng minh rằng : \(^{\left(\dfrac{a}{b}\right)^3}\)= \(\dfrac{a}{d}\)
Bài 1 : Tính nhanh
A=\(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{10}}\)
Bài 2:Tìm x biết
\(\dfrac{1}{5.8}+\dfrac{1}{8.11}+\dfrac{1}{11.14}+...+\dfrac{1}{x\left(x+3\right)}=\dfrac{101}{1540}\)
\(1+\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{1}{x\left(x+1\right):2}=1\dfrac{2007}{2009}\)
Tìm các số nguyên x, y biết:
\(^{\left(x^2+1\right)\cdot\left(x+1\right)=3^y}\)
Tìm x :
\(\dfrac{1}{1.2}.\left|2x-1\right|+\dfrac{1}{2.3}.\left|2x-1\right|+\dfrac{1}{3.4}+..+\dfrac{1}{1996.1997}.\left|2x-1\right|=1996\)