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Bài 1 : Thực hiện phép tính(1) D = \(1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+...+\frac{1}{16}\left(1+2+...+16\right)\)(2) M =\(\frac{\frac{1}{99}+\frac{2}{98}+\frac{3}{97}+...+\frac{99}{1}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}\)Bài 2 : Tìm x biết(1) \(x-\left\{x-\left[x-\left(-x+1\right)\right]\right\}=1\)(2) \(\left[\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}\right]\cdot...
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Bài 1 : Thực hiện phép tính

(1) D = \(1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+...+\frac{1}{16}\left(1+2+...+16\right)\)

(2) M =\(\frac{\frac{1}{99}+\frac{2}{98}+\frac{3}{97}+...+\frac{99}{1}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}\)

Bài 2 : Tìm x biết

(1) \(x-\left\{x-\left[x-\left(-x+1\right)\right]\right\}=1\)

(2) \(\left[\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}\right]\cdot x=\frac{2015}{1}+\frac{2014}{2}+...+\frac{1}{2015}\)

(3) \(\frac{x}{\left(a+5\right)\left(4-a\right)}=\frac{1}{a+5}+\frac{1}{4-a}\)

(4) \(\frac{x+2}{11}+\frac{x+2}{12}+\frac{x+2}{13}=\frac{x+2}{14}+\frac{x+2}{15}\)

(5) \(\frac{x+1}{2015}+\frac{x+2}{2014}+\frac{x+3}{2013}+\frac{x+4}{2012}+4=0\)

Bài 3 : 

(1) Cho : A =\(\frac{9}{1}+\frac{8}{2}+\frac{7}{3}+...+\frac{1}{9}\); B =\(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{10}\)

CMR : \(\frac{A}{B}\)Là 1 số nguyên

(2) Cho : D =\(\frac{1}{1001}+\frac{1}{1002}+\frac{1}{1003}+...+\frac{1}{2000}\)CMR : \(D< \frac{3}{4}\)

Bài 4 : Ký hiệu [x] là số nguyên lớn nhất không vượt quá x , gọi là phần nguyên của x.

VD : [1.5] =1 ; [3] =3 ; [-3.5] = -4

(1) Tính :\(\left[\frac{100}{3}\right]+\left[\frac{100}{3^2}\right]+\left[\frac{100}{3^3}\right]+\left[\frac{100}{3^4}\right]\)

(2) So sánh : A =\(\left[X\right]+\left[X+\frac{1}{5}\right]+\left[X+\frac{2}{5}\right]+\left[X+\frac{3}{5}\right]+\left[X+\frac{4}{5}\right]\)và B = [5x]. Biết x=3.7

0
9 tháng 9 2019

Gửi tạm trước 2 câu !

\(a,\text{ }3^2\cdot\frac{1}{243}\cdot81^2\cdot3^{-3}=3^2\cdot\frac{1}{3^5}\cdot\left(3^4\right)^2\cdot\frac{1}{3^3}=3^2\cdot\frac{1}{3^5}\cdot3^8\cdot\frac{1}{3^3}=3^2=9\)\(b,\text{ }\frac{\left(-3\right)^{10}\cdot15^5}{25^3\cdot\left(-9\right)^7}=\frac{3^{10}\cdot\left(3\cdot5\right)^5}{\left(5^2\right)^3\cdot\left(-3\cdot3\right)^7}=\frac{3^{10}\cdot3^5\cdot5^5}{5^6\cdot3^7\cdot\left(-3\right)^7}=\frac{3^{15}\cdot5^5}{5^6\cdot3^7\cdot\left(-3\right)^7}=\frac{3}{-5}\)

9 tháng 9 2019

Trả lời :

\(a,\text{ }3^2\cdot\frac{1}{243}\cdot81^2\cdot3^{-3}=3^2\cdot\frac{1}{3^5}\cdot\left(3^4\right)^2\cdot\frac{1}{3^3}=3^2\cdot\frac{1}{3^5}\cdot3^8\cdot\frac{1}{3^3}=3^2=9\)\(b,\text{ }\frac{\left(-3\right)^{10}\cdot15^5}{25^3\cdot\left(-9\right)^7}=\frac{3^{10}\cdot\left(3\cdot5\right)^5}{\left(5^2\right)^3\cdot\left(-3\cdot3\right)^7}=\frac{3^{10}\cdot3^5\cdot5^5}{5^6\cdot3^7\cdot\left(-3\right)^7}=\frac{3^{15}\cdot5^5}{5^6\cdot3^7\cdot\left(-3\right)^7}=\frac{3}{-5}\)

A=\(\left(\frac{1}{4}-1\right).\left(\frac{1}{9}-1\right).\left(\frac{1}{16}-1\right).............\left(\frac{1}{9801}-1\right).\left(\frac{1}{10000}-1\right)\)

A=\(\left(\frac{1-4}{4}\right).\left(\frac{1-9}{9}\right).\left(\frac{1-16}{16}\right).............\left(\frac{1-9801}{9801}\right).\left(\frac{1-10000}{10000}\right)\)

A=\(\frac{-3}{4}.\frac{-8}{9}.\frac{-15}{16}.....................\frac{-9800}{9801}.\frac{-9999}{10000}\)

A=\(\frac{-1.3}{2^2}.\frac{-2.4}{3^2}.\frac{-3.5}{4^2}.....................\frac{-98.100}{99^2}.\frac{-99.101}{100^2}\)

A=\(\frac{\left[\left(-1\right).\left(-2\right).\left(-3\right)....................\left(-98\right).\left(-99\right)\right].\left(3.4.5............100.101\right)}{\left(2.3.4.........99.100\right).\left(2.3.4...............99.100\right)}\)

A=\(\frac{1.101}{100.2}\)=\(\frac{101}{200}\)

2

\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+.................+\frac{2}{x.\left(x+1\right)}=\frac{2015}{2017}\)

\(\frac{1}{3.2}+\frac{1}{6.2}+\frac{1}{10.2}+.................+\frac{2}{2.x.\left(x+1\right)}=\frac{1}{2}.\frac{2015}{2017}\)

\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+.................+\frac{1}{x.\left(x+1\right)}=\frac{2015}{2017}.\frac{1}{2}\)

\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+..................+\frac{1}{x.\left(x+1\right)}=\frac{2015}{2017}.\frac{1}{2}\)

\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+..............+\frac{1}{x}-\frac{1}{x+1}=\frac{2015}{2017}.\frac{1}{2}\)

\(\frac{1}{2}-\frac{1}{x+1}=\frac{2015}{2017}.\frac{1}{2}\)

\(\frac{x+1}{2.\left(x+1\right)}-\frac{2}{2.\left(x+1\right)}=\frac{2015}{2017}.\frac{1}{2}\)

\(\frac{\left(x+1\right)-2}{2.\left(x+1\right)}=\frac{2015}{2017}.\frac{1}{2}\)

\(\frac{x-1}{2.\left(x+1\right)}=\frac{2015}{2017}.\frac{1}{2}\)

=>\(\frac{x-1}{x+1}=\frac{2015}{2017}.\frac{1}{2}:\frac{1}{2}\)

\(\frac{x-1}{x+1}=\frac{2015}{2017}\)

=>x+1=2017

=>x=2018-1

=>x=2016

Vậy x=2016

Còn bài 3 em ko biết làm em ms lớp 6

Chúc anh học tốt

12 tháng 11 2015

3n+2 - 2n+2 +3n - 2n = 3n . 32 - 2n. 22 +3n -2n

                             = 3n(32+1) - (2n.22 +2n)

                             =3n . 10 - 2n .5

                             =3n.10 - 2n-1 .2 .5

                             = 3n.10 - 2n-1 .10

                             = 10(3n - 2n-1)

vì 10 chia hết cho 10 nên 10(3n-2n-1) chia hết cho 10

                         =>  3n+2 - 2n+2 +3n -2n chia hết cho 10

                           

12 tháng 11 2015

Ai làm nhanh nhất mình sẽ **** xin cảm ơn các bạn mình đang cần gấp

 

17 tháng 1 2016

\(S=1+\frac{1}{2}+\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+...+\frac{101}{2}\)

\(S=1+\frac{1+2+3+4+...+101}{2}\)

\(S=1+\frac{10201}{2}=...\)

tick cho mink nha!

7 tháng 12 2019

a, \(\left(x-\frac{1}{2}\right)^3=\frac{1}{27}\)\(\Rightarrow\left(x-\frac{1}{2}\right)^3=\left(\frac{1}{3}\right)^3\)\(\Rightarrow x-\frac{1}{2}=\frac{1}{3}\)\(\Rightarrow x=\frac{5}{6}\)

b, \(\left(x-1\right)^{x+2}=\left(x-1\right)^{x+6}\)

\(\Rightarrow\left(x-1\right)^{x+2}-\left(x-1\right)^{x+6}=0\)

\(\Rightarrow\left(x-1\right)^{x+2}\left[1-\left(x-1\right)^4\right]=0\)

\(\Rightarrow\orbr{\begin{cases}\left(x-1\right)^{x+2}=0\\1-\left(x-1\right)^4=0\end{cases}}\Rightarrow\orbr{\begin{cases}x-1=0\\\left(x-1\right)^4=1\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\\left(x-1\right)^4=1\end{cases}}\)

Giải: \(\left(x-1\right)^4=1\)\(\Rightarrow\orbr{\begin{cases}x-1=1\\x-1=-1\end{cases}}\Rightarrow\orbr{\begin{cases}x=2\\x=0\end{cases}}\)

c, Vì \(\left(x+20\right)^{100}\ge0\)\(\forall x\inℝ\)\(\left|y+4\right|\ge0\)\(\forall y\inℝ\)

\(\Rightarrow\left(x+20\right)^{100}+\left|y+4\right|\ge0\)\(\forall x,y\inℝ\)

Dấu " = " xảy ra <=> \(\hept{\begin{cases}x+20=0\\y+4=0\end{cases}}\Rightarrow\hept{\begin{cases}x=-20\\y=-4\end{cases}}\)

d, \(2^{x-1}=16\)\(\Rightarrow2^{x-1}=2^4\)=> x - 1 = 4 => x = 5 

Bài 1

\(a,\left(\frac{3}{5}\right)^2-\left[\frac{1}{3}:3-\sqrt{16}.\left(\frac{1}{2}\right)^2\right]-\left(10.12-2014\right)^0\)

\(=\frac{9}{25}-\left[\frac{1}{9}-4.\frac{1}{4}\right]-1\)

\(=\frac{9}{25}-\left(-\frac{8}{9}\right)-1\)

\(=\frac{9}{25}+\frac{8}{9}-1\)

\(=\frac{56}{225}\)

\(b,|-\frac{100}{123}|:\left(\frac{3}{4}+\frac{7}{12}\right)+\frac{23}{123}:\left(\frac{9}{5}-\frac{7}{15}\right)\)

\(=\frac{100}{123}:\left(\frac{4}{3}\right)+\frac{23}{123}:\frac{4}{3}\)

\(=\left(\frac{100}{123}+\frac{23}{123}\right):\frac{4}{3}\)

\(=1:\frac{4}{3}=\frac{3}{4}\)

Phần c đăng riêng vì mk chưa tìm đc cách giải bt mỗi đáp án :v 

\(c,\frac{\left(-5\right)^{32}.20^{43}}{\left(-8\right)^{29}.125^{25}}\)

\(=\frac{\left(-5\right)^{32}.\left(4.5\right)^{43}}{\left[4.\left(-2\right)\right]^{29}.\left(-5^3\right)^{25}}\)

\(=\frac{-5^{32}.4^{43}.5^{43}}{4^{29}.\left(-2\right)^{29}.\left(5\right)^{75}}\)

\(=\frac{\left(-5^4\right)^8.4^{43}.5^{43}}{4^{29}.\left(-2\right)^{29}.\left(5^3\right)^{25}}\)

\(=-\frac{1}{2}\)

5 tháng 3 2020

\(B=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)\)

\(+...+\frac{1}{20}\left(1+2+3+...+20\right)\)

\(=1+\frac{1}{2}.\frac{2.3}{2}+\frac{1}{3}.\frac{3.4}{2}\)\(+...+\frac{1}{20}.\frac{20.21}{2}\)

\(=\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+...+\frac{21}{2}\)

\(=\frac{\frac{\left(21+2\right)\left[\left(21-2\right)+1\right]}{2}}{2}=115\)