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26 tháng 5 2018
Làm từng bài nha bạn
\(a)\) \(\left|x\right|=3,5\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=3,5\\x=-3,5\end{cases}}\)
Vậy \(x=3,5\) hoặc \(x=-3,5\)
Chúc bạn học tốt ~
15 tháng 5 2017
\(\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+2}\right)=\frac{5}{11}.\)
\(\frac{1}{2}\left(1-\frac{1}{x+2}\right)=\frac{5}{11}\)
\(1-\frac{1}{x+2}=\frac{5}{11}:\frac{1}{2}\)
\(1-\frac{1}{x+2}=\frac{5}{11}\cdot2=\frac{10}{11}\)
\(\frac{1}{x+2}=1-\frac{10}{11}\)
\(\frac{1}{x+2}=\frac{1}{11}\)
\(\Rightarrow x+2=11\)
\(\Rightarrow x=11-2=9\)
vậy x=9
ko viết đề bài nha
26 tháng 10 2019
Bài 1: 25 + 3(x - 8) = 106
3(x - 8) = 106 - 25
3(x - 8) = 81
(x - 8) = 81 : 3
(x - 8) = 27
x = 27 + 8
x = 25
Bài 2: 720 : [41 - (2x - 5)] = 23 . 5
720 : [41 - (2x - 5)] = 8 . 5
720 : [41 - (2x - 5)] = 40
[41 - (2x - 5)] = 720 : 40
[41 - (2x - 5)] = 18
(2x - 5) = 41 - 18
(2x - 5) = 23
2x = 23 + 5
2x = 28
x = 28 : 2
x = 14
15 tháng 8 2018
\(\left(1+\frac{1}{1\cdot3}\right)\left(1+\frac{1}{2\cdot4}\right)\left(1+\frac{1}{3\cdot5}\right)+...+\left(1+\frac{1}{2014\cdot2016}\right)=\frac{x}{1008}\)
\(\Rightarrow\frac{4}{3}\cdot\frac{9}{8}\cdot\frac{16}{15}\cdot...\cdot\frac{4060225}{4060224}=\frac{x}{1008}\)
\(\Rightarrow\frac{\left(2\cdot2\right)\left(3\cdot3\right)\left(4\cdot4\right)...\left(2015\cdot2015\right)}{\left(1\cdot3\right)\left(2\cdot4\right)\left(3\cdot5\right)...\left(2014\cdot2016\right)}=\frac{x}{1008}\)
\(\Rightarrow\frac{\left(2\cdot3\cdot4\cdot...\cdot2015\right)\left(2\cdot3\cdot4\cdot...\cdot2015\right)}{\left(1\cdot2\cdot3\cdot...\cdot2014\right)\left(3\cdot4\cdot5\cdot...\cdot2016\right)}=\frac{x}{1008}\)
\(\Rightarrow\frac{2015\cdot2}{1\cdot2016}=\frac{x}{1008}\)
\(\Rightarrow\frac{2015}{1008}=\frac{x}{1008}\)
\(\Rightarrow x=2015\)
16 tháng 3 2019
a) \(x+\)\(\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{41.45}=\frac{-37}{45}\)
\(\Rightarrow x+\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{41}-\frac{1}{45}\right)=\frac{-37}{45}\)
\(\Rightarrow x+\frac{1}{5}-\frac{1}{45}=\frac{-37}{45}\)
\(\Rightarrow x+\frac{1}{5}=-\frac{4}{5}\)
\(\Rightarrow x=\frac{-3}{5}\)
b) Đặt \(A=\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{2003.2005}\)
\(\Rightarrow2A=\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{2003.2005}\)
\(\Rightarrow2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2003}-\frac{1}{2005}\)
\(\Rightarrow2A=1-\frac{1}{2005}\)
\(\Rightarrow2A=\frac{2004}{2005}\)
\(\Rightarrow A=\frac{1002}{2005}\)
T
16 tháng 3 2019
Tính tổng:
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2003.2005}\)
= \(\frac{1}{2}\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{2003+2005}\right)\)
= \(\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+..+\frac{1}{2003}-\frac{1}{2005}\right)\)
= \(\frac{1}{2}\left(1-\frac{1}{2005}\right)\)
= \(\frac{1}{2}\cdot\frac{2004}{2005}\)
= \(\frac{1002}{2005}\)
k nha
22 tháng 4 2018
\(\frac{2}{3}x-\frac{1}{2}=\frac{1}{10}\)
\(\frac{2}{3}x\) = \(\frac{1}{10}+\frac{1}{2}\)
\(\frac{2}{3}x\) = \(\frac{3}{5}\)
\(x\) = \(\frac{3}{5}:\frac{2}{3}\)
\(x\) = \(\frac{9}{10}\)
12 tháng 8 2020
TA CÓ THỂ THẤY, VẾ TRÁI CÓ: 12 CẶP
=> \(12x+\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{23.25}\right)=11x+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^5}\)
<=> \(x+\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{23.25}\right)=\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^5}\) (****)
Ta xét: \(A=\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{23.25}\)
=> \(2A=\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{23.25}\)
=> \(2A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{23}-\frac{1}{25}\)
=> \(2A=1-\frac{1}{25}=\frac{24}{25}\)
=> \(A=\frac{12}{25}\)
Ta tiếp tục xét: \(B=\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^5}\)
=> \(3B=1+\frac{1}{3}+...+\frac{1}{3^4}\)
=> \(3B-B=\left(1+\frac{1}{3}+...+\frac{1}{3^4}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^5}\right)\)
=> \(2B=1-\frac{1}{3^5}=\frac{242}{243}\)
=> \(B=\frac{121}{243}\)
THAY CÁC GIÁ TRỊ A; B VÀO PT (****) TA ĐƯỢC:
=> \(x+\frac{12}{25}=\frac{121}{243}\)
<=> \(x=\frac{121}{243}-\frac{12}{25}=\frac{109}{6075}\)
29 tháng 4 2018
\(\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{\left\{\left(2x+1\right).\left(2x+3\right)\right\}}\right)=\frac{49}{99}\)
\(\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2x+1}-\frac{1}{2x+3}\right)=\frac{49}{99}\)
\(\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{2x+3}\right)=\frac{49}{99}\)
\(\frac{1}{2}.\cdot\left(\frac{2x+3}{2x+3}-\frac{1}{2x+3}\right)=\frac{49}{99}\)
\(\frac{1}{2}.\frac{2x+2}{2x+3}=\frac{49}{99}\)
\(\frac{2x+2}{2x+3}=\frac{49}{99}:\frac{1}{2}\)
\(\frac{2x+2}{2x+3}=\frac{98}{99}\)
=) \(2x+2=98\)và \(2x+3=99\)
TH1 : \(2x+2=98\)
\(2x=98-2\)
\(2x=96\)
\(x=96:2\)
\(x=48\)( THỎa mãn )
TH2 :
\(2x+3=99\)
\(2x=99-3\)
\(2x=96\)
\(x=96:2\)
\(x=48\)( THỎa mãn )
Vậy x = 48
3 . 5^(x+1) + 6250 = 25^3
→ 3 . 5^(x+1) = (5^2)^3 - 10.5^4
→ 3 . 5^(x+1) = 5^6 - 10 . 5^4
→ 3 . 5^(x+1) = 5^4 . ( 5^2 - 10 )
→ 3 . 5^x . 5^1 = 5^4 . ( 25 - 10 )
→ 15 . 5^x = 5^4 . 15
→ x = 4
3.5ˣ⁺¹ + 6250 = 25³
3.5ˣ⁺¹ + 6250 = 15625
3.5ˣ⁺¹ = 15625 - 6250
3.5ˣ⁺¹ = 9375
5ˣ⁺¹ = 9375 : 3
5ˣ⁺¹ = 3125
5ˣ⁺¹ = 5⁵
x + 1 = 5
x = 5 - 1
x = 4