(2013x10-8052x2-4026):(13x12+6)+17x11=
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A6
0
H9
HT.Phong (9A5)
CTVHS
2 tháng 7 2023
\(26\times7-17\times9+13\times26-17\times11\)
\(=26\times\left(7+13\right)-17\left(9+11\right)\)
\(=26\times20-17\times20\)
\(=20\times\left(26-17\right)\)
\(=20\times11\)
\(=220\)
4 tháng 9 2016
a)(125x37x32):4 = 37000
b)374:(17x11)=2 ai k cho minh minh se k lai cho
NT
0
KX
1
AH
Akai Haruma
Giáo viên
19 tháng 7
Lời giải:
$A=-1+2+(-3)+4+(-5)+6+....+(-4025)+4026$
$=[(-1)+2]+[(-3)+4]+[(-5)+6]+...+[(-4025)+4026]$
$=\underbrace{1+1+1+.....+1}_{2013}=1.2013=2013$
21 tháng 1 2020
Đặt \(A=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{2011}{4026}\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{2011}{4026}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2011}{4026}\)
\(=1-\frac{1}{x+1}=\frac{2011}{4026}\)
\(\Rightarrow\)x+1=4026
x=4026-1
x=4025
Vậy x=4025.
XO
21 tháng 1 2020
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{2011}{4026}\)
=> \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{2011}{4026}\)
=> \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2011}{4026}\)
=> \(1-\frac{1}{x+1}=\frac{2011}{4026}\)
=> \(\frac{1}{x+1}=\frac{2015}{4026}\Rightarrow x+1=\frac{4026}{2015}\Rightarrow x=\frac{2011}{2015}\)
NT
7
26 tháng 7 2020
\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{x\left(x+1\right)}=\frac{2011}{4026}\)
=> \(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+...+\frac{1}{x\left(x+1\right)}=\frac{2011}{4026}\)
=> \(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2011}{4026}\)
=> \(\frac{1}{2}-\frac{1}{x+1}=\frac{2011}{4026}\)
=> \(\frac{1}{x+1}=\frac{1}{2}-\frac{2011}{4026}=\frac{1}{2013}\)
=> x + 1 = 2013 => x = 2012
26 tháng 7 2020
Trả lời:
\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{x.\left(x+1\right)}=\frac{2011}{4026}\)
\(\Leftrightarrow\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{x.\left(x+1\right)}=\frac{2011}{4026}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2011}{4026}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2011}{4026}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2013}\)
\(\Leftrightarrow x+1=2013\)
\(\Leftrightarrow x=2012\)
Vậy \(x=2012\)
Lời giải:
$(2013\times 10-8052\times 2-4026):(13\times 12+6)+17\times 11$
$=(2013\times 10-2013\times 4\times 2-2013\times 2):(13\times 12+6)+17\times 11$
$=2013\times (10-4\times 2-2):(13\times 12+6)+17\times 11$
$=2013\times 0:(13\times 12+6)+17\times 11$
$=0+17\times 11=17\times 11=187$