Tính A=1+3+5+...+29/21+23+25+...+29
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HN
0
18 tháng 1 2016
a, <=> (x-5/100) -1 +(x-4/101) -1 +(x-3/102) -1= (x-100/5) -1+(x-101/4) -1 +(x-102/3) -1
<=> (x-105)(1/100 +1/101 +1/102)= (x-105)(1/5+1/4+1/3)
<=> (x-105)(1/100+1/101+1/102-1/5-1/4-1/3)=0
vì 1/100+1/101+1/102-1/5-1/4-1/3 khác 0 <=> x-105=0
<=> x=105
18 tháng 1 2016
b, 29-x/21 +1+27-x/23 +1+25-x/25 +1+23-x/27 +1+21-x/29 +1=0
<=> 50-x/21 +50-x/23 +50-x/25 +50-x/27 +50-x/29=0
<=> (50-x)(1/21 +1/23 +1/25 +1/27 +1/29)=0
vì 1/21+1/23+1/25+1/27+1/29 lớn hơn 0
nên 50-x=0
<=> x=50
21 tháng 3 2019
\(\frac{29-x}{21}+\frac{27-x}{23}+\frac{25-x}{25}+\frac{23-x}{27}+\frac{21-x}{29}=-5.\)
\(\left(\frac{29-x}{21}+1\right)+\left(\frac{27-x}{23}+1\right)+\left(\frac{25-x}{25}+1\right)+\left(\frac{23-x}{27}+1\right)+\left(\frac{21-x}{29}+1\right)\)\(=0\)
\(\frac{50-x}{21}+\frac{50-x}{23}+\frac{50-x}{25}+\frac{50-x}{27}+\frac{50-x}{29}=0\)
\(\left(50-x\right).\left(\frac{1}{21}+\frac{1}{23}+\frac{1}{25}+\frac{1}{27}+\frac{1}{29}\right)=0\)
=> 50 - x = 0 \(\left(\frac{1}{21}+\frac{1}{23}+\frac{1}{25}+\frac{1}{27}+\frac{1}{29}\ne0\right)\)
=> x = 50
PT
1
CM
30 tháng 8 2019
= 1 - 1 3 + 1 3 - 1 5 + 1 5 - 1 7 + 1 7 - 1 11 + 1 11 = 1 - 1 11 = 10 11
23 tháng 7 2018
29-x/21 + 27-x/23 + 25-x/25 + 23-x/27 + 21-x/29 = -5
1 + 29-x/21 + 1 + 27-x/23 + 1 + 25-x/25 + 1 + 23-x/27 + 1 + 21-x/29 = 0
50-x/21 + 50-x/23 + 50-x/25 + 50-x/27 + 50-x/29 = 0
(50-x) (1/21 + 1/23 + 1/25 + 1/27 + 1/29) = 0
Vì: 1/21 + 1/23 + 1/25 + 1/27 + 1/2 > 0
=> 50 - x = 0
x = 50
Vậy x = 50
TN
21 tháng 7 2018
\(\frac{-1}{3}+\frac{0,2-0,3+\frac{5}{11}}{-0,3+\frac{9}{16}-\frac{15}{12}}\)
\(=\frac{-1}{3}+\frac{\frac{2}{10}-\frac{3}{10}+\frac{5}{11}}{\frac{-3}{10}+\frac{9}{16}-\frac{15}{12}}\)
\(=\frac{-1}{3}+\frac{\frac{39}{110}}{\frac{-79}{80}}\)
\(=\frac{-1}{3}-\frac{312}{869}\)
\(=\frac{-1805}{2607}\)