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2 tháng 7 2015

B=1/1x4+1/4x3+1/3x8+...+1/7x16+1/16x9+1/9x20

2B=2x(1/4+1/12+1/24+...+1/112+1/144+1/180

2B=2/8+2/24+2/48+...+2/224+2/288+2/360

2B=2/2x4+2/4x6+2/6x8+...+2/14x16+2/16x18+2/18x20

2B=1/2-1/4+1/4-1/6+1/6-1/8+...+1/14-1/16+1/16-1/18+1/18-1/20

2B=1/2-1/20

2B=9/20

B=9/20:2

B=9/40

 

17 tháng 6 2016

\(\text{a) Ta co }\) \(4^{x+3}-3.4^{x+1}=13.4^{11}\)

\(\Rightarrow\)       \(4^{x+1}\left(16-3\right)=13.4^{11}\)

\(\Rightarrow4^{x+1}.13=13.4^{11}\)

\(\Rightarrow4^{x+1}=4^{11}\)

\(\Rightarrow x+1=11\)

\(\Rightarrow\text{x=10}\)

17 tháng 6 2016

a) 

\(4^{x+3}-3.4^{x+1}=13.4^{11}\)

<=> \(4^{x+1}\left(16-3\right)=13.4^{11}\)

<=> \(4^{x+1}.13=13.4^{11}\)

<=> \(4^{x+1}=4^{11}\)

<=> \(x+1=11\)

<=> x=10

17 tháng 6 2016

=> \(11.6^{x-1}+2.6^{x-1+2}=11.6^{11}+2.6^{11+2}\)

=>\(11.6^{x-1}+2.6^{x-1}.6^2=11.6^{11}+2.6^{11}.6^2\)

=>\(6^{x-1}\left(11+2.6^2\right)=6^{11}\left(11+2.6^2\right)\)

=>\(6^{x-1}=6^{11}\)

=>\(x-1=11\)

=>\(x=12\)

Tíck giùm mik nha !

13 tháng 9 2015

a/ (1/3)^5 . 3^5 = 1/243 . 243 = 1

b/ (1,5)^3 .8 = 3,375 . 8= 27

4 tháng 7 2018

Đặt \(A=\frac{1}{1\cdot4}+\frac{1}{4\cdot7}+\frac{1}{7\cdot10}+...+\frac{1}{x\left(x+3\right)}=\frac{49}{148}\)

\(3\left(\frac{1}{1\cdot4}+\frac{1}{4\cdot7}+\frac{1}{7\cdot10}+\frac{1}{\left(x+3\right)}\right)=3\cdot\frac{49}{148}\)

\(\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+...+\frac{3}{x\left(x+3\right)}=\frac{147}{148}\)

\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{147}{148}\)

\(1-\frac{1}{x-1}=\frac{147}{148}\)

\(\frac{1}{x-1}=1-\frac{147}{148}\)

\(\frac{1}{x-1}=\frac{1}{148}\)

\(\Rightarrow x-1=148\)

\(\Leftrightarrow x=148+1\)

\(\Leftrightarrow x=149\)

Vậy x=149

4 tháng 7 2018

\(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{x.\left(x+3\right)}=\frac{49}{148}\)

\(\Rightarrow\frac{1}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{x.\left(x+3\right)}\right)=\frac{49}{148}\)

\(\Rightarrow\frac{1}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{49}{148}\)

\(\Rightarrow\frac{1}{3}.\left(1-\frac{1}{x+3}\right)=\frac{49}{148}\)

\(\Rightarrow1-\frac{1}{x+3}=\frac{49}{148}:\frac{1}{3}\)

\(\Rightarrow1-\frac{1}{x+3}=\frac{147}{148}\)

\(\Rightarrow\frac{1}{x+3}=1-\frac{147}{148}\)

\(\Rightarrow\frac{1}{x+3}=\frac{1}{148}\)

\(\Rightarrow x+3=148\)

\(\Rightarrow x=148-3\)

\(\Rightarrow x=145\)

Vậy x = 145

_Chúc bạn học tốt_