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19 tháng 11 2017

\(2011^{2011}.\left(7^{10}:7^8-3.2^4-2^{2011}:2^{2011}\right)\)

\(=2011^{2011}.\left(7^{10-8}-3.2^4-2^{2011-2011}\right)\)

\(=2011^{2011}.\left(7^2-3.2^4-2^0\right)\)

\(=2011^{2011}.\left(49-3.16-1\right)\)

\(=2011^{2011}.\left(49-48-1\right)\)

\(=2011^{2011}.\left(1-1\right)\)

\(=2011^{2011}.0\)

\(=0\)

19 tháng 11 2017

đây là toán mà

19 tháng 11 2017

=2011^2011.(7^2-3.2^4-1)

=2011^2011.(49-3.16-1)

=2011^2011.(49-48-1)

=2011^2011.0

=0

Vote nhá

22 tháng 3 2018

A = 0 

B= 3/11

C= -1 

D= -9/10

22 tháng 11 2021

\(\dfrac{1}{\sqrt{k}+\sqrt{k+1}}=\dfrac{\sqrt{k}-\sqrt{k+1}}{k-k-1}=\sqrt{k+1}-\sqrt{k}\\ \Leftrightarrow\text{Đặt}\text{ }A=\dfrac{1}{3\left(\sqrt{2}+\sqrt{1}\right)}+\dfrac{1}{5\left(\sqrt{3}+\sqrt{2}\right)}+...+\dfrac{1}{4021\left(\sqrt{2011}+\sqrt{2010}\right)}< \dfrac{1}{2\left(\sqrt{2}+\sqrt{1}\right)}+\dfrac{1}{2\left(\sqrt{3}+\sqrt{2}\right)}+...+\dfrac{1}{2\left(\sqrt{2011}+\sqrt{2010}\right)}\\ \Leftrightarrow A< \dfrac{1}{2}\left(\dfrac{1}{\sqrt{2}+\sqrt{1}}+\dfrac{1}{\sqrt{3}+\sqrt{2}}+...+\dfrac{1}{\sqrt{2011}+\sqrt{2010}}\right)\)

\(\Leftrightarrow A< \dfrac{1}{2}\left(\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+...+\sqrt{2011}-\sqrt{2010}\right)\\ \Leftrightarrow A< \dfrac{1}{2}\left(\sqrt{2011}-1\right)< \dfrac{1}{2}\cdot\dfrac{\sqrt{2011}-1}{\sqrt{2011}}=\dfrac{1}{2}\left(1-\dfrac{1}{\sqrt{2011}}\right)\)

28 tháng 1 2016

kho qua

28 tháng 1 2016

quá khó là đằng khác

6 tháng 7 2017

a) \(5\dfrac{4}{23}.27\dfrac{3}{47}+4\dfrac{3}{47}.\left(-5\dfrac{4}{23}\right)\)

\(=5\dfrac{4}{23}.27\dfrac{3}{47}+\left(-4\dfrac{3}{47}\right).5\dfrac{4}{23}\)

\(=5\dfrac{4}{23}.\left[27\dfrac{3}{47}+\left(-4\dfrac{3}{47}\right)\right]\)

\(=5\dfrac{4}{23}.\left(27\dfrac{3}{47}-4\dfrac{3}{27}\right)\)

\(=5\dfrac{4}{23}.23\)

\(=\dfrac{119}{23}.23\)

\(=\dfrac{119}{23}\)

b) \(4.\left(\dfrac{-1}{2}\right)^3+\dfrac{3}{2}\)

\(=4.\dfrac{-1}{6}+\dfrac{3}{2}\)

\(=\dfrac{-4}{6}+\dfrac{3}{2}\)

\(=\dfrac{-2}{3}+\dfrac{3}{2}\)

\(=\dfrac{-4}{6}+\dfrac{9}{6}\)

\(=\dfrac{5}{6}\)

c) \(\left(\dfrac{1999}{2011}-\dfrac{2011}{1999}\right)-\left(\dfrac{-12}{1999}-\dfrac{12}{2011}\right)\)

\(=\dfrac{1999}{2011}-\dfrac{2011}{1999}-\dfrac{-12}{1999}+\dfrac{12}{2011}\)

\(=\left(\dfrac{1999}{2011}+\dfrac{12}{2011}\right)-\left(\dfrac{2011}{1999}+\dfrac{-12}{1999}\right)\)

\(=\dfrac{2011}{2011}-\dfrac{1999}{1999}\)

\(=1-1\)

\(=0\)

d) \(\left(\dfrac{-5}{11}+\dfrac{7}{22}-\dfrac{-4}{33}-\dfrac{5}{44}\right):\left(\dfrac{381}{22}-39\dfrac{7}{22}\right)\)

(đợi đã, mình chưa tìm được hướng làm...)

6 tháng 7 2017

quy đồng lên

9 tháng 8 2020

Ta có : \(\left(\sqrt{a^2+2011}+a\right).\left(\sqrt{a^2+2011}-a\right)\)

\(=\left(\sqrt{a^2+2011}\right)^2-a^2\)

\(=a^2+2011-a^2=2011\)

Nên : \(\left(\sqrt{a^2+2011}+a\right).\left(\sqrt{a^2+2011}-a\right)=2011\)

Mà theo bài ta có : \(\left(\sqrt{a^2+2011}+a\right).\left(\sqrt{a^2+2011}+b\right)=2011\)

Nên : \(\sqrt{a^2+2011}+b=\sqrt{a^2+2011}-a\) ( đpcm )

5 tháng 8 2015

\(\left(1-\frac{1}{7}\right).\left(1-\frac{1}{8}\right).\left(1-\frac{1}{9}\right)......\left(1-\frac{1}{2011}\right)\)

\(=\frac{6}{7}.\frac{7}{8}.\frac{8}{9}.....\frac{2010}{2011}\)

\(=\frac{6.7.8.9.....2010}{7.8.9.10.....2011}\)

\(=\frac{6}{2011}\)