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NT
0
1 tháng 6 2016
\(\text{Ta có: }\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+.....+\frac{3}{\left(x+2\right)\left(x+5\right)}=\frac{3}{20}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+.....+\frac{1}{\left(x+2\right)}-\frac{1}{\left(x+5\right)}=\frac{3}{20}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{\left(x+5\right)}=\frac{3}{20}\)
\(\Rightarrow\frac{1}{\left(x+5\right)}=\frac{1}{2}-\frac{3}{20}\)
MC
0
LA
2
24 tháng 4 2015
a)\(=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{5\cdot6}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\)
\(=\left(1-\frac{1}{6}\right)+\left(\frac{1}{2}-\frac{1}{2}\right)+...+\left(\frac{1}{5}-\frac{1}{5}\right)\)
\(=\left(1-\frac{1}{6}\right)+0+...+0=1-\frac{1}{6}=\frac{6}{6}-\frac{1}{6}=\frac{5}{6}\)
b)\(\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+\frac{3}{11\cdot14}=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}\)
\(=\left(\frac{1}{2}-\frac{1}{14}\right)+\left(\frac{1}{5}-\frac{1}{5}\right)+\left(\frac{1}{8}-\frac{1}{8}\right)+\left(\frac{1}{11}-\frac{1}{11}\right)\)
\(=\left(\frac{1}{2}-\frac{1}{14}\right)+0+...+0=\frac{1}{2}-\frac{1}{14}=\frac{7}{14}-\frac{1}{14}=\frac{6}{14}\)
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22 tháng 10 2019
1. \(6x^3-8=40\\ 6x^3=48\\ x^3=8\\ \Rightarrow x=2\)Vậy x = 2
2. \(4x^5+15=47\\ 4x^5=32\\ x^5=8\\ \Rightarrow x\in\varnothing\left(\text{vì }x\in N\right)\)Vậy x ∈ ∅
3. \(2x^3-4=12\\ 2x^3=16\\ x^3=8\\ \Rightarrow x=2\)Vậy x = 2
4. \(5x^3-5=0\\ 5x^3=5\\ x^3=1\\ \Rightarrow x=1\)Vậy x = 1
5. \(\left(x-5\right)^{2016}=\left(x-5\right)^{2018}\\ \Rightarrow\left[{}\begin{matrix}x-5=0\\x-5=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\x=6\end{matrix}\right.\)Vậy \(x\in\left\{5;6\right\}\)
6. \(\left(3x-2\right)^{20}=\left(3x-1\right)^{20}\\ \Rightarrow3x-2=3x-1\\ 3x-3x=2-1\\ 0=1\left(\text{vô lí}\right)\)Vậy x ∈ ∅
7. \(\left(3x-1\right)^{10}=\left(3x-1\right)^{20}\\ \left(3x-1\right)^{10}=\left[\left(3x-1\right)^2\right]^{10}\\ \Rightarrow\left(3x-1\right)^2=3x-1\\ \left(3x-1\right)^2-\left(3x-1\right)=0\\ \left(3x-1\right)\left[\left(3x-1\right)-1\right]=0\\ \left(3x-1\right)\left(3x-2\right)=0\\ \Rightarrow\left[{}\begin{matrix}3x-1=0\\3x-2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}3x=1\\3x=2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{1}{3}\left(\text{loại vì }x\in N\right)\\x=\frac{2}{3}\left(\text{loại vì }x\in N\right)\end{matrix}\right.\)Vậy x ∈ ∅
8. \(\left(2x-1\right)^{50}=2x-1\\ \left(2x-1\right)^{50}-\left(2x-1\right)=0\\ \left(2x-1\right)\left[\left(2x-1\right)^{49}-1\right]=0\\ \Rightarrow\left[{}\begin{matrix}2x-1=0\\\left(2x-1\right)^{49}=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=1\\2x-1=1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\frac{1}{2}\left(\text{loại vì }x\in N\right)\\x=1\left(t/m\right)\end{matrix}\right.\)Vậy x = 1
9. \(\left(\frac{x}{3}-5\right)^{2000}=\left(\frac{x}{3}-5\right)^{2008}\\ \left(\frac{x}{3}-5\right)^{2008}-\left(\frac{x}{3}-5\right)^{2000}=0\\ \left(\frac{x}{3}-5\right)^{2000}\left[\left(\frac{x}{3}-5\right)^8-1\right]=0\\ \Rightarrow\left[{}\begin{matrix}\left(\frac{x}{3}-5\right)^{2000}=0\\\left(\frac{x}{3}-5\right)^8=1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}\frac{x}{3}-5=0\\\frac{x}{3}-5=1\\\frac{x}{3}-5=-1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}\frac{x}{3}=5\\\frac{x}{3}=6\\\frac{x}{3}=4\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=5\cdot3=15\\x=6\cdot3=18\\x=4\cdot3=12\end{matrix}\right.\)Vậy \(x\in\left\{15;18;12\right\}\)
22 tháng 10 2019
\(1.6x^3-8=40\\ \Leftrightarrow6x^3=48\\ \Leftrightarrow x^3=8\Leftrightarrow x^3=2^3=\left(-2\right)^3\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
Vậy \(x\in\left\{2;-2\right\}\)
\(2.4x^3+15=47\) (T nghĩ đề là mũ 3)
\(\Leftrightarrow4x^3=32\Leftrightarrow x^3=8=2^3=\left(-2\right)^3\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
Vậy \(x\in\left\{2;-2\right\}\)
Câu 3, 4 tương tự nhé.
NB
23 tháng 3 2017
ta có : 1/10 = 1/2.5
1/40 = 1/5.8
suy ra 1/10 + 1/40 + 1/88 + ... + 1/(3x+2).(3x+5) = 3/20 = 1/2.5 + 1/5.8 + 1/8.11 +....+ 1/(3x+2).(3x+5)
= 1/3(1/2 - 1/5) + 1/3(1/5 - 1/8) + ...+ 1/3(1/8 - 1/11) + ....+ 1/3 [(3x+2) (3x+5)] = 3/20
= 1/3(1/2 - 1/5 + 1/5 - 1/8 + 1/8 - 1/11 +.... + 1/3x+2 + 1/3x+5) = 3/20
1/3 ( 1/2 - 1/3x+5) = 3/20
suy ra 1/2 - 1/3x+5 = 3/20 : 1/3 = 9/20
suy ra 1/3x+5 = 1/2 - 9/20 = 1/20
suy ra 3x+5 = 20
suy ra 3x = 20 - 5= 15
suy ra x = 15 : 3 = 5
vay x = 5
nhớ k nha!!!!
làm lại
ta có : \(\frac{1}{10}+\frac{1}{40}+\frac{1}{88}+...+\frac{1}{\left(x+2\right)\left(x+5\right)}=\frac{3}{20}\)
=>\(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{\left(x+2\right)\left(x+5\right)}=\frac{3}{20}\)
=>\(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+....+\frac{3}{\left(x+2\right)\left(x+5\right)}=\frac{9}{20}\)
=>\(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{x+2}-\frac{1}{x+5}=\frac{9}{20}\)
=>\(\frac{1}{2}-\frac{1}{x+5}=\frac{9}{20}\)
=>\(\frac{1}{x+5}=\frac{1}{2}-\frac{9}{20}\)
=>\(\frac{1}{x+5}=\frac{1}{20}\)
=>\(x+5=20\)
=>\(x=20-5\)
=>\(x=15\)
ta có : \(\frac{1}{10}+\frac{1}{40}+\frac{1}{88}+....+\frac{1}{\left(x+2\right)\left(x+5\right)}=\frac{3}{20}\)
=>\(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{\left(x+2\right)\left(x+3\right)}=\frac{3}{20}\)
=>\(3.\left(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+....+\frac{1}{\left(x+2\right)\left(x+3\right)}\right)=3.\frac{3}{20}\)
=>\(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+....+\frac{3}{\left(x+2\right)\left(x+3\right)}=\frac{9}{20}\)
=>\(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{x+2}-\frac{1}{x+3}=\frac{9}{20}\)
=>\(\frac{1}{2}-\frac{1}{x+3}=\frac{9}{20}\)
=>\(\frac{1}{x+3}=\frac{1}{2}-\frac{9}{20}\)
=>\(\frac{1}{x+3}=\frac{1}{20}\)
=>\(x+3=20\)
=>\(x=20-3\)
=>\(x=17\)