\(B=\)\(\dfrac{9}{9}+\dfrac{99}{99}+\dfrac{999}{999}+\dfrac{9999}{9999}+....+\dfrac{99999999999999999999}{99999999999999999999}+\dfrac{999999999999999999999}{999999999999999999991}=?\)
Tính B
\(\dfrac{9^9+99^{99}+999^{999}+9999^{9999}}{9^9+99^{99}+999^{999}+9999^{9999}}=?\)
1
Cộng xong tử và mẫu cùng bằng nhau nên bằng 1
Đấy là ý thôi bạn, cần cách trình bày. Bạn tự nghĩ nhé
Tính nhanh :
Q = \(\left(\dfrac{1}{99}+\dfrac{12}{999}+\dfrac{123}{9999}\right).\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}\right)\)
Ta có \(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}=\dfrac{1}{6}-\dfrac{1}{6}=0\) nên Q = 0.
\(Q=\left(\dfrac{1}{99}+\dfrac{12}{999}+\dfrac{123}{9999}\right).\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}\right)\)
\(Q=\left(\dfrac{1}{99}+\dfrac{12}{999}+\dfrac{123}{9999}\right).0\)
\(Q=0\)
Tính giá trị của biểu thức: \(Q=\left(\dfrac{1}{99}+\dfrac{12}{999}+\dfrac{123}{9999}\right)\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}\right)\)
\(Q=\left(\dfrac{1}{99}+\dfrac{12}{999}+\dfrac{123}{9999}\right)\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}\right)\)
\(Q=\left(\dfrac{1}{99}+\dfrac{12}{999}+\dfrac{123}{9999}\right)\left(\dfrac{3}{6}-\dfrac{2}{6}-\dfrac{1}{6}\right)\)
\(Q=\left(\dfrac{1}{99}+\dfrac{12}{999}+\dfrac{123}{9999}\right)\cdot\dfrac{0}{6}\)
\(Q=\left(\dfrac{1}{99}+\dfrac{12}{999}+\dfrac{123}{9999}\right)\cdot0\)
\(Q=0\)
1.Tính nhanh
A =( \(\dfrac{1}{99}+\dfrac{12}{999}-\dfrac{123}{9999}\)) x \(\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}\right)\)
2.Tìm n ∈ Z để \(\dfrac{n+3}{n-2}\)nhận giá trị nguyên
1. \(\left(\dfrac{1}{99}+\dfrac{12}{999}-\dfrac{123}{9999}\right).\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}\right)\)
\(=\left(\dfrac{1}{99}+\dfrac{12}{999}-\dfrac{123}{9999}\right).0\)
\(=0\)
Bài 1:
Ta có: \(A=\left(\dfrac{1}{99}+\dfrac{12}{999}-\dfrac{123}{9999}\right)\cdot\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}\right)\)
\(=\left(\dfrac{1}{99}+\dfrac{12}{999}-\dfrac{123}{9999}\right)\cdot\left(\dfrac{3}{6}-\dfrac{2}{6}-\dfrac{1}{6}\right)\)
=0
1. Rút gọn phân số ( Các bạn ghi đầy đủ giùm mình nha )
a,\(\dfrac{3.5.7.11.13.37-10101.55}{1212120+40404}\)
b,\(\dfrac{5+55+555+5555}{9+99+999+9999}\)
a, \(\dfrac{3.5.7.11.13.37-10101.55}{1212120+40404}\)
\(=\dfrac{55\left(3.7.13.37-10101\right)}{1212120+40404}\)
\(=\dfrac{55.0}{1212120+40404}=0\)
b, \(\dfrac{5+55+555+5555}{9+99+999+9999}\)
\(=\dfrac{5.\left(1+11+111+1111\right)}{9.\left(1+11+111+1111\right)}=\dfrac{5}{9}\)
Chúc bạn học tốt!!!
1. Rút gọn phân số ( Các bạn ghi đầy đủ giùm mình nha )
a,\(\dfrac{3.5.7.11.13.37-10101.55}{1212120+40404}\)
b,\(\dfrac{5+55=555+5555}{9+99+999+9999}\)
a,\(\dfrac{3.5.7.11.13.37-10101.55}{1212120+40404}\)
\(=\dfrac{55\left(3.7.13.37-10101\right)}{1212120+40404}\)
\(=\dfrac{55.0}{1212120+40404}=0\)
cho biểu thúc A=\(\dfrac{3}{4}\)+\(\dfrac{8}{9}\)+\(\dfrac{15}{16}\)+....+\(\dfrac{9999}{10000}\) chứng minh A<99
Áp dụng các tính chất của phép nhân phân số để tính nhanh :
\(M=\dfrac{8}{3}.\dfrac{2}{5}.\dfrac{3}{8}.10.\dfrac{19}{92}\)
\(N=\dfrac{5}{7}.\dfrac{5}{11}+\dfrac{5}{7}.\dfrac{2}{11}-\dfrac{5}{7}.\dfrac{14}{11}\)
\(Q=\left(\dfrac{1}{99}+\dfrac{12}{999}-\dfrac{123}{9999}\right).\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}\right)\)
\(M=\dfrac{8}{3}\cdot\dfrac{2}{5}\cdot\dfrac{3}{8}\cdot10\cdot\dfrac{19}{92}\\ =\dfrac{8\cdot2\cdot3\cdot10\cdot19}{3\cdot5\cdot8\cdot92}\\ =\dfrac{8\cdot2\cdot3\cdot2\cdot5\cdot19}{3\cdot5\cdot8\cdot2\cdot2\cdot23}\\ =\dfrac{19}{23}\)
\(N=\dfrac{5}{7}\cdot\dfrac{5}{11}+\dfrac{5}{7}\cdot\dfrac{2}{11}-\dfrac{5}{7}\cdot\dfrac{14}{11}\\ =\dfrac{5}{7}\cdot\left(\dfrac{5}{11}+\dfrac{2}{11}-\dfrac{14}{11}\right)\\ =\dfrac{5}{7}\cdot\left(-\dfrac{7}{11}\right)\\ =-\dfrac{5}{11}\)
\(Q=\left(\dfrac{1}{99}+\dfrac{12}{999}-\dfrac{123}{9999}\right)\cdot\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}\right)\\ =\left(\dfrac{1}{99}+\dfrac{12}{999}-\dfrac{123}{9999}\right)\cdot\left(\dfrac{3}{6}-\dfrac{2}{6}-\dfrac{1}{6}\right)\\ =\left(\dfrac{1}{99}+\dfrac{12}{999}-\dfrac{123}{9999}\right)\cdot\left(\dfrac{1}{6}-\dfrac{1}{6}\right)\\ =\left(\dfrac{1}{99}+\dfrac{12}{999}-\dfrac{123}{9999}\right)\cdot0\\ =0\)
Chứng minh 98<\(\dfrac{3}{4}+\dfrac{8}{9}+\dfrac{15}{16}+...+\dfrac{9999}{10000}< 99\)
Đặt \(A=\dfrac{3}{4}+\dfrac{8}{9}+...+\dfrac{9999}{10000}=1-\dfrac{1}{4}+1-\dfrac{1}{9}+...+1-\dfrac{1}{10000}\)
\(=99-\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}\right)=99-B\)
Do \(B=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{100^2}>0\Rightarrow99-B< 99\Rightarrow A< 99\)
Do \(B=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{100^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\)
\(\Rightarrow B< \dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}=1-\dfrac{1}{100}\)
\(\Rightarrow A=99-B>99-\left(1-\dfrac{1}{100}\right)=98+\dfrac{1}{100}>98\)
Vậy \(98< \dfrac{3}{4}+\dfrac{8}{9}+...+\dfrac{9999}{10000}< 99\)