LT
1
K
Khách
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PU
3
20 tháng 9 2019
\(\frac{x+1}{2009}+\frac{x+2}{2008}+\frac{x+3}{2007}=\frac{x+10}{2000}+\frac{x+11}{1999}+\frac{x+12}{1998}\)
\(\Rightarrow\frac{x+1}{2009}+1+\frac{x+2}{2008}+1+\frac{x+3}{2007}+1=\frac{x+10}{2000}+1+\frac{x+11}{1999}+1+\frac{x+12}{1998}+1\)
\(\Rightarrow\frac{x+2010}{2009}+\frac{x+2010}{2008}+\frac{x+2010}{2007}=\frac{x+1010}{2000}+\frac{x+2010}{1999}+\frac{x+2010}{1998}\)
\(\Rightarrow\left(x+2010\right)\left(\frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}\right)=\left(x+2010\right)\left(\frac{1}{2000}+\frac{1}{1999}+\frac{1}{1998}\right)\)
\(\Rightarrow x+2010=0\) vì \(0< \frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}< \frac{1}{2000}+\frac{1}{1999}+\frac{1}{1998}\)
\(\Rightarrow x=-2010\)
F
20 tháng 9 2019
Bài giải
\(\frac{x+1}{2009}+\frac{x+2}{2008}+\frac{x+3}{2007}=\frac{x+10}{2000}+\frac{x+11}{1999}+\frac{x+12}{1998}\)
\(\Rightarrow\left(\frac{x+1}{2009}+1\right)+\left(\frac{x+2}{2008}+1\right)+\left(\frac{x+3}{2007}+1\right)=\left(\frac{x+10}{2000}+1\right)+\left(\frac{x+11}{1999}+1\right)+\left(\frac{x+12}{1998}+1\right)\)
\(\Rightarrow\frac{x+2010}{2009}+\frac{x+2010}{2008}+\frac{x+2010}{2007}=\frac{x+2010}{2000}+\frac{x+2010}{1999}+\frac{x+2010}{1998}\)
\(\Rightarrow\frac{x+2010}{2009}+\frac{x+2010}{2008}+\frac{x+2010}{2007}-(\frac{x+2010}{2000}+\frac{x+2010}{1999}+\frac{x+2010}{1998})=0\)
\(\Rightarrow\frac{x+2010}{2009}+\frac{x+2010}{2008}+\frac{x+2010}{2007}-\frac{x+2010}{2000}-\frac{x+2010}{1999}-\frac{x+2010}{1998}=0\)
\(\left(x+2010\right)\left(\frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}-\frac{1}{2000}-\frac{1}{1999}-\frac{1}{1998}\right)=0\)
Vì \(\left(\frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}-\frac{1}{2000}-\frac{1}{1999}-\frac{1}{1998}\right)\ne0\) nên \(x+2010=0\)
\(x=0-2010=-2010\)
KB
1
8 tháng 7 2017
\(\dfrac{x+1}{2009}+\dfrac{x+2}{2008}+\dfrac{x+3}{2007}=\dfrac{x+10}{2000}+\dfrac{x+11}{1999}+\dfrac{x+12}{1998}\)
\(\Rightarrow\left(\dfrac{x+1}{2009}+1\right)+\left(\dfrac{x+2}{2008}+1\right)+\left(\dfrac{x+3}{2007}+1\right)=\left(\dfrac{x+10}{2000}+1\right)+\left(\dfrac{x+11}{1999}+1\right)+\left(\dfrac{x+12}{1998}+1\right)\)
\(\Rightarrow\dfrac{x+2010}{2009}+\dfrac{x+2010}{2008}+\dfrac{x+2010}{2007}=\dfrac{x+2010}{2000}+\dfrac{x+2010}{1999}+\dfrac{x+2010}{1998}\)\(\Rightarrow\dfrac{x+2010}{2009}+\dfrac{x+2010}{2008}+\dfrac{x+2010}{2007}-\dfrac{x+2010}{2000}-\dfrac{x+2010}{1999}-\dfrac{x+2010}{1998}=0\)\(\Rightarrow\left(x+2010\right)\left(\dfrac{1}{2009}+\dfrac{1}{2010}+\dfrac{1}{2007}-\dfrac{1}{2000}-\dfrac{1}{1999}-\dfrac{1}{1998}\right)=0\)\(\Rightarrow x+2010=0\Rightarrow x=-2010\)
25 tháng 9 2020
a.
\(\left(1\frac{1}{4}+\frac{3}{5}\right):\left(-\frac{11}{12}\right)+\left(\frac{3}{8}-1\frac{2}{5}\right):\left(-\frac{11}{12}\right)\)
\(=\left(\frac{5}{4}+\frac{3}{5}+\frac{3}{8}-\frac{7}{5}\right):\left(-\frac{11}{12}\right)\)
\(=\left(\frac{13}{8}-\frac{4}{5}\right):\left(-\frac{11}{12}\right)\)
\(=\frac{33}{40}:\left(-\frac{11}{12}\right)\)
\(=\frac{33}{40}\cdot\left(-\frac{12}{11}\right)\)
\(=\frac{-9}{10}\)
b.
\(\left(\frac{3}{8}-1\frac{2}{5}\right):\left(-\frac{11}{15}\right)+\left(1\frac{1}{4}+\frac{3}{5}\right):\left(-\frac{11}{15}\right)\)
\(=\left(\frac{3}{8}-\frac{7}{5}+\frac{5}{4}+\frac{3}{5}\right):\left(-\frac{11}{15}\right)\)
\(=\left(\frac{13}{8}-\frac{4}{5}\right):\left(-\frac{11}{15}\right)\)
\(=\frac{33}{40}:\left(-\frac{11}{15}\right)\)
\(=\frac{33}{40}\cdot\left(-\frac{15}{11}\right)\)
\(=\frac{-9}{8}\)
LN
0
9 tháng 7 2023
a: N=(7-8)+(9-10)+...+(2009-2010)
=(-1)+(-1)+....+(-1)
=-1*1002=-1002
b: Đặt A=2+3+4+...+2023
Số số hạng là 2023-2+1=2022(số)
Tổng là (2023+2)*2022/2=2047275
=>P=1-2047275=-2047274
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
9 tháng 7 2023
A = 7 - 8 + 9 -10 + 11 - 12 +...+ 2009 - 2010
A = (7-8) + (9 - 10) + ( 11 - 12) +...+ ( 2009 - 2010)
Xét dãy số: 7; 9; 11;...; 2009
Dãy số trên là dãy số cách đều với khoảng cách là: 9 - 7 = 2
Dãy số trên có số số hạng là: (2009 - 7) : 2 + 1 = 1002
Vậy tổng A có 1002 nhóm mỗi nhóm có giá trị là: 7 - 8 = -1
A = -1 \(\times\) 1002 = - 1002
B = 1 - 2 - 3 - 4 -...- 2022 - 2023
B = 1 - ( 2 + 3 + 4 +...+ 2022 + 2023)
B = 1 - (2 + 2023).{ ( 2023 - 2): 1 + 1}: 2 = -2047274
(1+2+3+...+2009)(12.6-36.2):\(\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\right)\)
=(1+2+3+...+2009)(12.3.2-12.3.2):\(\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\right)\)
=(1+2+3+...+2009).0:\(\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\right)\)
=0