Tính giá trị của biểu thức: Q =\((\dfrac{1}{99}+\dfrac{12}{999}+\dfrac{123}{999})\cdot(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6})\)
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
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.
\(Q=\left(\dfrac{1}{99}+\dfrac{12}{999}+\dfrac{123}{9999}\right).\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}\right)\)
\(Q=\left(\dfrac{1}{99}+\dfrac{12}{999}+\dfrac{123}{9999}\right).0\)
\(Q=0\)
Áp dụng các tính chất của phép nhân phân số để tính nhanh :
\(M=\dfrac{8}{3}.\dfrac{2}{5}.\dfrac{3}{8}.10.\dfrac{19}{92}\)
\(N=\dfrac{5}{7}.\dfrac{5}{11}+\dfrac{5}{7}.\dfrac{2}{11}-\dfrac{5}{7}.\dfrac{14}{11}\)
\(Q=\left(\dfrac{1}{99}+\dfrac{12}{999}-\dfrac{123}{9999}\right).\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}\right)\)
\(M=\dfrac{8}{3}\cdot\dfrac{2}{5}\cdot\dfrac{3}{8}\cdot10\cdot\dfrac{19}{92}\\ =\dfrac{8\cdot2\cdot3\cdot10\cdot19}{3\cdot5\cdot8\cdot92}\\ =\dfrac{8\cdot2\cdot3\cdot2\cdot5\cdot19}{3\cdot5\cdot8\cdot2\cdot2\cdot23}\\ =\dfrac{19}{23}\)
\(N=\dfrac{5}{7}\cdot\dfrac{5}{11}+\dfrac{5}{7}\cdot\dfrac{2}{11}-\dfrac{5}{7}\cdot\dfrac{14}{11}\\ =\dfrac{5}{7}\cdot\left(\dfrac{5}{11}+\dfrac{2}{11}-\dfrac{14}{11}\right)\\ =\dfrac{5}{7}\cdot\left(-\dfrac{7}{11}\right)\\ =-\dfrac{5}{11}\)
\(Q=\left(\dfrac{1}{99}+\dfrac{12}{999}-\dfrac{123}{9999}\right)\cdot\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}\right)\\ =\left(\dfrac{1}{99}+\dfrac{12}{999}-\dfrac{123}{9999}\right)\cdot\left(\dfrac{3}{6}-\dfrac{2}{6}-\dfrac{1}{6}\right)\\ =\left(\dfrac{1}{99}+\dfrac{12}{999}-\dfrac{123}{9999}\right)\cdot\left(\dfrac{1}{6}-\dfrac{1}{6}\right)\\ =\left(\dfrac{1}{99}+\dfrac{12}{999}-\dfrac{123}{9999}\right)\cdot0\\ =0\)
Bài 1 : Tìm x , biết :
a, \(\dfrac{-3}{x}=\dfrac{x}{-27}\) b, \(\dfrac{2}{3}\)của x là ( -150 ) c,\(\dfrac{2}{2.4}+\dfrac{2}{4.6}+...+\dfrac{2}{x.\left(x+2\right)}=\dfrac{4}{9}\)
Bài 2 : Tính :
A = \(\left(\dfrac{1}{99}+\dfrac{12}{999}+\dfrac{123}{999}\right)\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}\right)\)
Bài 1: a) Ta có : \(\dfrac{-3}{x}=\dfrac{x}{-27}\Leftrightarrow\left(-3\right).\left(-27\right)=x.x\Leftrightarrow81=x^2\Leftrightarrow9^2=x^2\Leftrightarrow x=9\)
b) Do \(\dfrac{2}{3}\) của x là -150 nên x là: (-150) : \(\dfrac{2}{3}\) = -225
c) \(\dfrac{2}{2.4}+\dfrac{2}{4.6}+...+\dfrac{2}{x\left(x+2\right)}=\dfrac{4}{9}\)
\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{x}-\dfrac{1}{x+2}=\dfrac{4}{9}\)
\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{x+2}=\dfrac{4}{9}\)
\(\Leftrightarrow\dfrac{1}{x+2}=\dfrac{1}{2}-\dfrac{4}{9}\)
\(\Leftrightarrow\dfrac{1}{x+2}=\dfrac{1}{18}\)
\(\Leftrightarrow x+2=18\)
\(\Leftrightarrow x=16\)
Bài 2:
\(A=\left(\dfrac{1}{99}+\dfrac{12}{999}+\dfrac{123}{999}\right)\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}\right)\)
\(A=\left(\dfrac{1}{99}+\dfrac{12}{999}+\dfrac{123}{999}\right)\left(\dfrac{1}{6}-\dfrac{1}{6}\right)\)
\(A=\left(\dfrac{1}{99}+\dfrac{12}{999}+\dfrac{123}{999}\right).0\)
\(A=0\)
Bài 1 - Đặt tính rồi tính
15 423 : 9
306 427 : 23
706 999 : 201
7 406 x 2
36 200 x 206
9 731 x 27
Bài 2 - Tính giá trị của biểu thức
\(a,\) \(\dfrac{2}{5}\) + \(\dfrac{2}{7}\) x \(\dfrac{2}{3}\)
\(b,\) \(\dfrac{2}{6}\) : \(\dfrac{7}{5}\) + \(\dfrac{2}{6}\)
\(c,\) \((\) \(\dfrac{2}{3}\) - \(\dfrac{1}{6}\) \()\) x \(\dfrac{6}{7}\)
Giúp tớ nhé!!!!
em đăng bên box toán để mấy bạn kia trả lời nhé =)))))
Tính giá trị của các biểu thức sau
1) \(A=1+2+2^2+...+2^{2015}\)
2) \(B=\left(\dfrac{1}{4}-1\right)\cdot\left(\dfrac{1}{9}-1\right)\cdot\left(\dfrac{1}{16}-1\right)\cdot\cdot\cdot\cdot\cdot\left(\dfrac{1}{400}-1\right)\)
3) \(C=\left(\dfrac{1}{4\cdot9}+\dfrac{1}{9\cdot14}+\dfrac{1}{14\cdot19}+...+\dfrac{1}{44\cdot49}\right)\cdot\dfrac{1-3-5-7-...-49}{89}\)
4) \(D=\dfrac{2^{12}\cdot3^5-4^6\cdot9^2}{\left(2^2\cdot3\right)^6+8^4\cdot3^5}-\dfrac{5^{10}\cdot7^3-25^5\cdot49^2}{\left(125\cdot7\right)^3+5^9\cdot14^3}\)
5) \(E=\dfrac{\dfrac{1}{2003}+\dfrac{1}{2004}-\dfrac{1}{2005}}{\dfrac{5}{2003}+\dfrac{5}{2004}-\dfrac{5}{2005}}-\dfrac{\dfrac{2}{2002}+\dfrac{2}{2003}-\dfrac{2}{2004}}{\dfrac{3}{2002}+\dfrac{3}{2003}-\dfrac{3}{2004}}\)
6) Cho 13+23+...+103=3025
Tính S= 23+43+63+...+203
Tính giá trị của các biểu thức sau 1) \(A=1+2+2^2+...+2^{2015}\) 2) \(B=\left(\dfrac{1}{4}-1\right)\cdot\left(\dfrac{1}{9}-1\right)\cdot\left(\dfrac{1}{16}-1\right)\cdot\cdot\cdot\cdot\cdot\left(\dfrac{1}{400}-1\right)\) 3) \(C=\left(\dfrac{1}{4\cdot9}+\dfrac{1}{9\cdot14}+\dfrac{1}{14\cdot19}+...+\dfrac{1}{44\cdot49}\right)\cdot\dfrac{1-3-5-7-...-49}{89}\) 4) \(D=\dfrac{2^{12}\cdot3^5-4^6\cdot9^2}{\left(2^2\cdot3\right)^6+8^4\cdot3^5}-\dfrac{5^{10}\cdot7^3-25^5\cdot49^2}{\left(125\cdot7\right)^3+5^9\cdot14^3}\) 5) \(E=\dfrac{\dfrac{1}{2003}+\dfrac{1}{2004}-\dfrac{1}{2005}}{\dfrac{5}{2003}+\dfrac{5}{2004}-\dfrac{5}{2005}}-\dfrac{\dfrac{2}{2002}+\dfrac{2}{2003}-\dfrac{2}{2004}}{\dfrac{3}{2002}+\dfrac{3}{2003}-\dfrac{3}{2004}}\) 6) Cho 13+23+...+103=3025 Tính S= 23+43+63+...+203
Tính giá trị biểu thức A , biết rằng A = M : N
Mà M = \(\dfrac{\dfrac{1}{99}+\dfrac{2}{98}+\dfrac{3}{97}+\dfrac{4}{96}+...+\dfrac{97}{3}+\dfrac{98}{2}+\dfrac{99}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{100}}\)
N = \(\dfrac{92-\dfrac{1}{9}-\dfrac{2}{10}-\dfrac{3}{11}-...-\dfrac{90}{98}-\dfrac{91}{99}-\dfrac{92}{100}}{\dfrac{1}{45}+\dfrac{1}{50}+\dfrac{1}{55}+...+\dfrac{1}{495}+\dfrac{1}{500}}\)
Ta có: \(M=\dfrac{\dfrac{1}{99}+\dfrac{2}{98}+\dfrac{3}{97}+\dfrac{4}{96}+...+\dfrac{97}{3}+\dfrac{98}{2}+\dfrac{99}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{100}}\)
\(=\dfrac{\left(1+\dfrac{1}{99}\right)+\left(1+\dfrac{2}{98}\right)+\left(1+\dfrac{3}{97}\right)+\left(1+\dfrac{4}{96}\right)+...+\left(1+\dfrac{98}{2}\right)+1}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{100}}\)
\(=\dfrac{\dfrac{100}{99}+\dfrac{100}{98}+\dfrac{100}{97}+...+\dfrac{100}{1}+\dfrac{100}{2}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{100}}\)
=100
Ta có: \(N=\dfrac{92-\dfrac{1}{9}-\dfrac{2}{10}-\dfrac{3}{11}-...-\dfrac{90}{98}-\dfrac{91}{99}-\dfrac{92}{100}}{\dfrac{1}{45}+\dfrac{1}{50}+\dfrac{1}{55}+...+\dfrac{1}{495}+\dfrac{1}{500}}\)
\(=\dfrac{\left(1-\dfrac{1}{9}\right)+\left(1-\dfrac{2}{10}\right)+\left(1-\dfrac{3}{11}\right)+...+\left(1-\dfrac{90}{98}\right)+\left(1-\dfrac{91}{99}\right)+\left(1-\dfrac{92}{100}\right)}{\dfrac{1}{5}\left(\dfrac{1}{9}+\dfrac{1}{10}+\dfrac{1}{11}+...+\dfrac{1}{99}+\dfrac{1}{100}\right)}\)
\(=\dfrac{\dfrac{8}{9}+\dfrac{8}{10}+\dfrac{8}{11}+...+\dfrac{8}{99}+\dfrac{8}{100}}{\dfrac{1}{5}\left(\dfrac{1}{9}+\dfrac{1}{10}+\dfrac{1}{11}+...+\dfrac{1}{99}+\dfrac{1}{100}\right)}\)
\(=\dfrac{8}{\dfrac{1}{5}}=40\)
\(\Leftrightarrow\dfrac{M}{N}=\dfrac{100}{40}=\dfrac{5}{2}\)