So sánh
1/3+1/3^2+1/3^3+...+1/3^100 và 1/2
Bài 1 : So sánh
1) - 3 \(\sqrt{13}\) và -9
2) \(\sqrt{15}-1\) và \(\sqrt{10}\)
3)5 và \(\sqrt{8}+1\)
1) `-3\sqrt13=-3\sqrt13`
`-9=-3\sqrt9`
`\sqrt13>\sqrt9`
`=> -3\sqrt13 < -3\sqrt9`
`=> -3\sqrt13 < 9`.
2) `\sqrt15 < \sqrt16`
`<=> \sqrt15-1 < \sqrt16-1`
`<=> \sqrt15-1 < 3 < \sqrt10`
`=> \sqrt15-1 <\sqrt10`
3) `5=4+1=\sqrt16+1`
`\sqrt8+1=\sqrt8+1`
`=> 5>\sqrt8+1`
1) \(-3\sqrt{13}=-\sqrt{117}< -\sqrt{81}=-9\)
3) Ta có: \(5^2=25=9+16\)
\(\left(2\sqrt{2}+1\right)^2=9+4\sqrt{2}\)
mà \(16>4\sqrt{2}\)
nên \(5>2\sqrt{2}+1\)
Bài 1 : so sánh
1) A= 1/143+1/99+1/63+...+1/3 và B=199/206
A=1/1*3+1/3*5+...+1/9*11+1/11*13
=1/2(1-1/3+1/3-1/5+...+1/11-1/13)
=1/2*12/13=6/13<B
1.so sánh
1/2:2/3......2/3:1/2 4/7:2/5......4/7:3/5
4/15:4/7......2/5x10/3 5/6......15/18-11/18
2.viết số thích hợp vào ô trống
2/3=..../18 7/9=49/.... ..../5=18/15
..../3=10/15 5/9=..../45 49/56=7/....
6/8=42/.... 2/9=..../63 49/56=..../8
`1)1/2:2/3 .... 2/3 : 1/2`
`=>1/2xx3/2 .... 2/3xx2`
`=>3/4 .... 4/3`
Vì `3/4 < 1` và `4/3>1`
`=>3/4<4/3`
__
`4/7:2/5 ... 4/7 : 3/5`
`=>4/7xx5/2....4/7xx5/3`
`=>20/14...20/21`
`=>10/7...20/21`
Vì `10/7>1` và `20/21<1`
`=>10/7>20/21`
__
`4/15:4/7....2/5xx10/3`
`=>4/15xx7/4...20/15`
`=>7/15...20/15`
Vì `7<20` nên `7/15<20/15`
__
`5/6...15/18-11/18`
`=>5/6...4/18`
Ta có : MSC : `18`
`5/6 = 15/18`
Vì `15>4` nên `5/6 > 4/18`
`2)2/3=(2xx6)/(3xx6)=12/18`
`7/9=(7xx7)/(9xx7)=49/63`
`6/5=(6xx3)/(5xx3)=18/15`
`2/3=(2xx5)/(3xx5)=10/15`
`5/9=(5xx5)/(9xx5)=25/45`
`49/56=(49:7)/(56:7)=7/8`
`6/8=(6xx7)/(8xx7)=42/56`
`2/9=(2xx7)/(9xx7)=14/63`
`49/56=(49:7)/(56:7)=7/8`
so sánh: 1/3 + 1/3^2 + 1/3^3 + ... + 1/3^99 +1/3^100 và 1/2
so sánh S = 1/3 - 2/3^2 + 3/3^3 -4/3^4 + ... + 99/3^99 -100/3^100 và 1/5
So sánh A và B :
\(A=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{100}}\)
\(B=\dfrac{1}{2}\)
Ta có `3A=1+1/3+....+1/3^99`
`=>3A-A=1-1/3^100`
`=>2A=1-1/3^100`
`=>A=1/2-1/(2.3^100)<1/2`
Hay `A<B`
So sánh
a, 1/3 + 1/3^2 + 1/3^3 +....+ 1/3^99 + 1/3^100 và 1/2
b, 3/1^2*2^2 + 5/2^2 *3^2 +7/3^2*4^2 +......+ 19/9^2*10^2 và 1
a)
\(3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{99}}\)
\(3A-A=\left(1+\frac{1}{3}+...+\frac{1}{3^{99}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{100}}\right)\)
\(2A=1-\frac{1}{3^{100}}\)
\(\Rightarrow2A< 1\)
\(\Rightarrow A< \frac{1}{2}\)
\(A=\dfrac{1}{^{ }3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{100}}\) và \(B=\dfrac{1}{2}\). Hãy so sánh chúng
Ta có: \(3\cdot A=1+\dfrac{1}{3^1}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}\)
\(A=\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{100}}\)
Do đó:
\(3\cdot A-A=1+\dfrac{1}{3^1}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{100}}-\dfrac{1}{3}-\dfrac{1}{3^2}-...-\dfrac{1}{3^{100}}\)
hay \(2\cdot A=1-\dfrac{1}{3^{100}}\)
\(\Leftrightarrow A=\left(1-\dfrac{1}{3^{100}}\right):2\)
\(\Leftrightarrow A=\left(1-\dfrac{1}{3^{100}}\right)\cdot\dfrac{1}{2}\)
\(\Leftrightarrow A=\dfrac{1}{2}-\dfrac{1}{2\cdot3^{100}}< \dfrac{1}{2}\)
hay A<B
Cho S = \(\dfrac{1}{3}-\dfrac{2}{3^2}+\dfrac{3}{3^3}-\dfrac{4}{3^4}+....+\dfrac{99}{3^{99}}-\dfrac{100}{3^{100}}\) so sánh S và \(\dfrac{1}{5}\)