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Bài 3 :
b) Ta có 1+ 2 + 3 +4 + ...+ x =15
Nên \(\frac{x\left(x+1\right)}{2}=15\)
\(x\left(x+1\right)=30\)
=> \(x\left(x+1\right)=5.6\)
=> x = 5
a ) 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128 + 1/256
Đạt A = 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128 + 1/256
A x 2 = 2 x ( 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128 + 1/256)
A x 2 = 1 + 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128
Lấy A x 2 - A ta có :
A x 2 - A = 1 + 1/2 + ..... + 1/128 - 1/2 + 1/4 + ........ + 1/256
A x ( 2 - 1 ) = 1 - 1/ 256
A = 255/256
b) 1/3 + 1/9 + 1/27 + 1/81 + 1/243 + 1/729
Đặt A = 1/3 + 1/9 + 1/27 + 1/81 + 1/243 + 1/729
A x 3 = 3 x ( 1/3 + 1/9 + 1/27 + 1/81 + 1/243 + 1/729)
= 1 + 1/ 3 + 1/9 + 1/27 + 1/81 + 1/243
Lấy A x 3 - A ta có :
A x 3 - A = 1 + 1/3 + 1/9 +..... + 1/243 - 1/3 + 1/9 +........+ 1/243 + 1/29
A x ( 3 - 1 ) = 1 - 1/29
A x2 = 28/29
A = 28/29 : 2 ( tự tính
Đặt A = \(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+.....+\frac{1}{256}\)
\(\Rightarrow2A=1+\frac{1}{2}+\frac{1}{4}+.....+\frac{1}{128}\)
\(\Rightarrow2A-A=1-\frac{1}{256}\)
\(\Rightarrow A=\frac{255}{256}\)
a) 2006/2005 = 1/2005 ; 2005/2006 =1/2006
Vì: 1/2005 > 1/2006 ; nên 2006/2005 < 2005/2006
b) 99/1000 = 990/10000
Vì: 990/10000 <999/10000 ; nên 99/1000 < 999/10000
c) 13/15 =1- 2/15 ;1333/1555 = 1- 222/555 => 1- 2/5
Vì: 2/15 < 2/5 ; nên 13/15 >1333/1555
d) 650650/480480 = 650/480 =65/48 ; 222222/144144 = 222/144
Vì: 65/48 < 222/144 ; nên 650650/480480 < 222222/144144
(câu a với câu c là tính theo kiểu phần bù các bạn nhớ nhé!)
(1/2003+1/2004+1/2005) / (2/2003+2/2004+2/2005)
= (1/2003+1/2004+1/2005) / 2(1/2003+1/2004+1/2005)
= 1/2
bài2 \(x\times\dfrac{15}{16}-x\times\dfrac{4}{16}=2\)
\(x\times\dfrac{11}{16}=2\)
\(x=2:\dfrac{11}{16}\)
\(x=\dfrac{32}{11}\)
Bài 1 :
\(\dfrac{x}{16}\times\left(2017-1\right)=2\)
\(\dfrac{x}{16}\times2016=2\)
\(\dfrac{x}{16}=\dfrac{2}{2016}\)
\(x=\dfrac{2}{2016}\times16\)
\(x=\dfrac{1}{63}\)
1a,\(\frac{8}{5}\)=1\(\frac{3}{5}\)b,\(\frac{9}{2}\)=4\(\frac{1}{2}\)c,\(\frac{7}{4}\)=1\(\frac{3}{4}\)d,\(\frac{10}{3}\)
2a,1\(\frac{3}{4}\)+1\(\frac{5}{8}\)=\(\frac{7}{4}\)+\(\frac{13}{8}\)=\(\frac{14}{8}\)+\(\frac{13}{8}\)=\(\frac{27}{8}\)b,3\(\frac{4}{5}\)x2\(\frac{6}{8}\)=\(\frac{19}{5}\)x\(\frac{22}{8}\)=\(\frac{418}{40}\)=\(\frac{209}{20}\)
3,3\(\frac{2}{5}\)=\(\frac{17}{5}\),3\(\frac{4}{5}\)=\(\frac{19}{5}\) , 8\(\frac{2}{4}\)=\(\frac{34}{4}\),10\(\frac{6}{7}\)=\(\frac{76}{7}\) , 5\(\frac{1}{2}\)=\(\frac{11}{2}\),5\(\frac{3}{4}\)=\(\frac{23}{4}\)
=>\(\frac{17}{5}\)<\(\frac{19}{5}\)=>3\(\frac{2}{5}\)<3\(\frac{4}{5}\) =>\(\frac{34}{4}\)<\(\frac{76}{7}\)=>8\(\frac{2}{4}\)<10\(\frac{6}{7}\) =>\(\frac{11}{2}\)<\(\frac{23}{4}\)=>5\(\frac{1}{2}\)<5\(\frac{3}{4}\)
k mik nha ......
Đặt \(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{128}\)
\(2A=1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{64}\)
\(2A-A=\left(1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{64}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{128}\right)\)
\(A=1-\frac{1}{128}=\frac{127}{128}\)
Ủng hộ mk nha ^_-
\(2\cdot A=1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{64}\)
\(A=\left(1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{64}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{128}\right)\)
\(A=1-\frac{1}{128}=\frac{128}{128}-\frac{1}{128}=\frac{127}{128}\)