6x6=
7x7=
8x7=
5x5=
ai nhanh nhất mình tick cho?
cho mình hỏi là bài tính nhanh 1/2x2/3x3/4x4/5x5/6x6/7x7/8 làm như thế nào ạ
Tính bằng cách thuận tiện nhất:
1/2x2/3x3/4x4/5x5/6x6/7x7/8x8/9
5x5+1=
6x6+1=
ai tcik mình mình tick lại
5x5 +1 = 25+1=26
6 x6 +1 = 36+1 = 37
5 x 5 + 1 = 25 + 1 = 26
6 x 6 + 1 = 12 + 1 = 13
Tổng: 1 + 2x2 + 3x3 + 4x4 + 5x5 + 6x6 + 7x7 + 8x8 + 9x9 + 10x10 là bao nhiêu?
1 + 2 x 2 + 3 x 3 + 4 x 4 + 5 x 5 + 6 x 6 + 7 x 7 + 8 x 8 + 9 x 9 + 10 x 10
= 1 + 4 + 9 + 16 + 25 + 36 + 49 + 64 + 81 + 100
= 385
1 + 2x2 + 3x3 + 4x4 + 5x5 + 6x6 + 7x7 + 8x8 + 9x9 + 10x10
= 1+4+9+16+25+36+49+64+81+100
=(81+9)+(64+16)+(49+1)+)36+4)+25+100
=90+80+50+40+25 +100
=385
các bạn ơi giúp mk voi
so sánh A va B biet
A=1/5x5+1/6x6+1/7x7+.....+1/99x99+1/100x100
b=1/6
ghi ra ca cach làm de hiểu nhất nhé!!!!!!!!!!!!!giúp vs mà mink tick cho may cai cung dc 10 lun
đặt A=1/5x5 +1/6x6 + 1/7x7 + .....+ 1/100x100
=>A>1/5x6 + 1/6x7 +1/7x8 + .... + 1/100x101
=>A>1/5 - 1/6 + 1/6 - 1/7 + +1/7 - 1/8 + ..... + 1/100 - 1/101
=>A> 1/5 - 1/101
=>A>96/505 > 96/576 = 1/6
=>A>1/6
=>A>B
a>1/5x6+1/6x7+...+1/100x101
=1/5-1/6+1/6-1/7+...+1/100-1/101
=1/5-1/101
=101/505-5/101
=96/101
vì 96/101>1/6 nên a>1/6
1x1=
2x2=
3x3=
4x4=
5x5=
6x6=
7x7=
8x8=
9x9=
10x10=
1 x 1 = 1
2 x 2 = 4
3 x 3 = 9
4 x 4 = 16
5 x 5 = 25
6 x 6 = 36
7 x 7 = 49
8 x 8 = 64
9 x 9 = 81
10 x 10 = 100
\(1x1=1\)
\(2x2=4\)
\(3x3=9\)
\(4x4=16\)
\(5x5=25\)
\(6x6=36\)
\(7x7=49\)
\(8x8=64\)
\(9x9=81\)
\(10x10=100\)
1 x 1 = 1
2 x 2 = 4
3 x 3 = 9
4 x 4 = 16
5 x 5 = 25
6 x 6 = 36
7 x 7 = 49
8 x 8 = 64
9 x 9 = 81
10 x 10 = 100
So sánh số sau: 1/2x2+1/3x3+1/4x4+1/5x5+1/6x6+1/7x7+1/8x8 với 1.
đặt \(A=\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+\frac{1}{5.5}+\frac{1}{6.6}+\frac{1}{7.7}+\frac{1}{8.8}=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+\frac{1}{8^2}\)
\(A
CM rằng
\(\frac{1}{6}< \frac{1}{5x5}+\frac{1}{6x6}+\frac{1}{7x7}+............+\frac{1}{100x100}< \frac{1}{4}\)
Ta có:
1/5×5 < 1/4×5
1/6×6 < 1/5×6
1/7×7 < 1/6×7
.........
1/100×100 < 1/99×100
=> 1/5×5 + 1/6×6 + 1/7×7 +.....+ 1/100×100 < 1/4×5 + 1/5×6 + 1/6×7 +.....+ 1/99×100
= 1/4-1/5 + 1/5-1/6 + 1/6-1/7 +......+ 1/99-1/100
= 1/4-1/100 < 1/4
=> 1/5×5 + 1/6×6+1/7×7 +...+1/100×100<1/4 (1)
Lại có:
1/5×5 > 1/6×7
1/6×6 > 1/7×8
1/7×7 > 1/8×9
........
1/100×100 > 1/101×102
=> 1/5×5 + 1/6×6 + 1/7×7 +.....+ 1/100×100 > 1/5×6 + 1/6×7 + 1/7×8 +.....+1/100×101
= 1/5-1/6 + 1/6-1/7 + 1/7-1/8 +.....+ 1/100 - 1/101
= 1/5 - 1/101 > 1/5 - 1/30 = 1/6
=> 1/5×5 + 1/6×6 +1/7×7 +.....+ 1/100×100>1/6 (2)
Từ (1) và (2)
=> 1/6 < 1/5×5 +1/6×6+ 1/7×7 +...+1/100×100<1/4
Đặt \(A=\frac{1}{5.5}+\frac{1}{6.6}+...+\frac{1}{100.100}\)
Có \(\frac{1}{5.5}< \frac{1}{4.5};\frac{1}{6.6}< \frac{1}{5.6};...;\frac{1}{100.100}< \frac{1}{99.100}\)
\(\Rightarrow A< \frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{99.100}=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}=\frac{1}{4}-\frac{1}{100}< \frac{1}{4}\)(1)
Lại có :\(\frac{1}{5.5}>\frac{1}{5.6};\frac{1}{6.6}>\frac{1}{6.7};...;\frac{1}{100.100}>\frac{1}{100.101}\)
\(\Rightarrow A>\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{100.101}=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{101}=\frac{1}{5}-\frac{1}{101}=\frac{96}{505}>\frac{1}{6}\left(2\right)\)
Từ (1) và (2) \(\RightarrowĐCCM\)
Đặt \(X=\frac{1}{5.5}+\frac{1}{6.6}+...+\frac{1}{100.100}\)
\(\Rightarrow X< \frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{99.100}\)
\(\Rightarrow X=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)
\(\Rightarrow X< \frac{1}{4}-\frac{1}{100}< \frac{1}{4}\)
\(\Rightarrow X< \frac{1}{4}\left(1\right)\)
\(\Rightarrow X>\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{100.101}\)
\(\Rightarrow X>\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{101}\)
\(\Rightarrow X>\frac{1}{5}-\frac{1}{101}>\frac{1}{5}-\frac{1}{30}=\frac{1}{6}\)
\(\Rightarrow X>\frac{1}{6}\left(2\right)\)
Từ \(\left(1\right);\left(2\right)\Rightarrow\frac{1}{6}< X< \frac{1}{4}\)
Chứng tỏ rằng
B=\(\frac{1}{2x2}+\frac{1}{3x3}+\frac{1}{4x4}+\frac{1}{5x5}+\frac{1}{6x6}+\frac{1}{7x7}+\frac{1}{8x8}< 1\)
Ta thấy:
1/2*2<1/1*2)vì 2*2>1*2).
1/3*3<1/2*3(vì 3*3>2*3).
...
1/8*8<1/7*8(vì 8*8>7*8).
=>1/2*2+1/3*3+1/4*4+...+1/8*8<1/1*2+1/2*3+1/3*4+...+1/7*8.
=>B<1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8.
=>B<1-1/8.
=>B<7/8.
Mà 7/8<1.
=>B<1.
Vậy B<1(đpcm).
\(< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}\)
\(\Rightarrow1-\frac{1}{8}< 1\)
=>B<1
\(B=\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+\frac{1}{5.5}+\frac{1}{6.6.}+\frac{1}{7.7}+\frac{1}{8.8}\)\(=\)\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+\frac{1}{8^2}\)
\(B=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}\)
\(B=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{7}-\frac{1}{8}\)
\(B=1-\frac{1}{8}\)
\(\Rightarrow B< 1\left(ĐPCM\right)\)