Tính :
1 + \(\frac{1}{2}\) x ( 1 + 2 ) + \(\frac{1}{3}\) x ( 1 + 2 + 3 ) + ........ + \(\frac{1}{20}\) x ( 1 + 2 + 3 + ....... + 20 )
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
14 tháng 6 2018
Ta có
1+1/2(1+2)+....+1/20(1+2+3+...+20)
= 2/2+3/2+.....+19/2
=(2+3+....+19)/2
= 189/2
KT
19 tháng 2 2018
\(\frac{x+2}{327}+\frac{x+3}{326}+\frac{x+4}{325}+\frac{x+5}{324}+\frac{x+349}{5}=0\)
\(\Leftrightarrow\)\(\frac{x+2}{327}+1+\frac{x+3}{326}+1+\frac{x+4}{325}+1+\frac{x+5}{324}+1 +\frac{x+349}{5}-4=0\)
\(\Leftrightarrow\)\(\frac{x+329}{327}+\frac{x+329}{326}+\frac{x+329}{325}+\frac{x+329}{324}+\frac{x+329}{5}=0\)
\(\Leftrightarrow\)\(\left(x+329\right)\left(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\right)=0\)
\(\Leftrightarrow\)\(x+329=0\) (vì 1/327 + 1/326 + 1/325 + 1/324 + 1/5 khác 0 )
\(\Leftrightarrow\)\(x=-329\)
19 tháng 2 2018
Bài 1 :
\(\frac{x+2}{327}+\frac{x+3}{326}+\frac{x+4}{325}+\frac{x+5}{324}+\frac{x+349}{5}=0\)
\(\Leftrightarrow\)\(\left(\frac{x+2}{327}+1\right)+\left(\frac{x+3}{326}+1\right)+\left(\frac{x+4}{325}+1\right)+\left(\frac{x+5}{324}+1\right)+\left(\frac{x+349}{5}-4\right)=0\)
\(\Leftrightarrow\)\(\frac{x+329}{327}+\frac{x+329}{326}+\frac{x+329}{325}+\frac{x+329}{324}+\frac{x+329}{5}=0\)
\(\Leftrightarrow\)\(\left(x+329\right)\left(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\right)=0\)
Vì \(\left(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\right)\ne0\)
\(\Rightarrow\)\(x+329=0\)
\(\Rightarrow\)\(x=-329\)
Vậy \(x=-329\)
23 tháng 2 2017
bài 1+2: phân tích mẫu thành nhân tử r` áp dụng
1/ab=1/a-1/b
bài 3+4: quy đồng rút gọn blah...
21 tháng 7 2017
[ 19/20 x 3/4 + 1/20 x 3/4 ] x [ 1/2 x 3/4 -2/4 x 3/4 ]
=[ 3/4 x [19/20 + 1/20 ] x [3/4 x [2/4 - 1/2]
=[ 3/4 x 1 ] x [3/4 x 0 ]
=0
21 tháng 7 2017
\(\left(\frac{19}{20}.\frac{3}{4}+\frac{1}{20}.\frac{3}{4}\right).\left(\frac{1}{2}.\frac{3}{4}-\frac{2}{4}.\frac{3}{4}\right)\)
\(=\left(\frac{19}{20}.\frac{3}{4}+\frac{1}{20}.\frac{3}{4}\right).\left(\frac{1}{2}.\frac{3}{4}-\frac{1}{2}.\frac{3}{4}\right)\)
\(=\left(\frac{19}{20}.\frac{3}{4}+\frac{1}{20}.\frac{3}{4}\right).0\)
\(=0\)
A
14 tháng 12 2018
Thặc vler .V
A/\(\frac{1}{x^2+3x+2}+\frac{1}{x^2+5x+6}+\frac{1}{x^2+7x+12}+\frac{1}{x^2+9x+20}\)
\(=\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+4\right)}+\frac{1}{\left(x+4\right)\left(x+5\right)}\)
\(=\left[\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}\right]+\left[\frac{1}{\left(x+3\right)\left(x+4\right)}+\frac{1}{\left(x+4\right)\left(x+5\right)}\right]\)
\(=\left[\frac{x+3}{\left(x+1\right)\left(x+2\right)\left(x+3\right)}+\frac{x+1}{\left(x+1\right)\left(x+2\right)\left(x+3\right)}\right]+\left[\frac{x+5}{\left(x+3\right)\left(x+4\right)\left(x+5\right)}+\frac{x+3}{\left(x+3\right)\left(x+4\right)\left(x+5\right)}\right]\)
\(=\frac{2x+4}{\left(x+1\right)\left(x+2\right)\left(x+3\right)}+\frac{2x+8}{\left(x+3\right)\left(x+4\right)\left(x+5\right)}\)
\(=\frac{2\left(x+2\right)}{\left(x+1\right)\left(x+2\right)\left(x+3\right)}+\frac{2\left(x+4\right)}{\left(x+3\right)\left(x+4\right)\left(x+5\right)}\)
\(=\frac{2}{\left(x+1\right)\left(x+3\right)}+\frac{2}{\left(x+3\right)\left(x+5\right)}\)
\(=\frac{2x+10}{\left(x+1\right)\left(x+3\right)\left(x+5\right)}+\frac{2x+2}{\left(x+1\right)\left(x+3\right)\left(x+5\right)}\)
\(=\frac{4x+12}{\left(x+1\right)\left(x+3\right)\left(x+5\right)}\)
\(=\frac{4\left(x+3\right)}{\left(x+1\right)\left(x+3\right)\left(x+5\right)}\)
\(=\frac{4}{\left(x+1\right)\left(x+5\right)}\)
B/\(\frac{x-1}{x-2}+\frac{1}{2-x}\)
\(=\frac{x-1}{x-2}-\frac{1}{x-2}\)
\(=\frac{x-1-1}{x-2}\)
\(=\frac{x-2}{x-2}\)
\(=1\)
28 tháng 6 2018
Đáp án: thiếu đề
@#@
mời bn xem xét lại đề bài.
~hok tốt~
Đặt biểu thức là A, ta có:
\(A=1+\frac{1}{2}.\left(1+2\right)+\frac{1}{3}.\left(1+2+3\right)+...+\frac{1}{20}.\left(1+2+3+...+20\right)\)
\(A=1+\frac{1}{2}.2.3:2+\frac{1}{3}.3.4:2+...+\frac{1}{20}.20.21:2=\frac{2}{2}+\frac{3}{2}+...+\frac{21}{2}\)
\(A=\frac{2+3+4+...+21}{2}=\frac{230}{2}=115\)