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

Đặt T = 1 + 2 x 2 + ... + 10 x 2 = 385

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

9x9=81

8x8=64

7x7=49

6x6=36

5x5=25

TK nha!

81,64,49 và 36 nha bn

9 x 9 = 81

8 x 8 = 64

7 x 7 = 49

6 x 6 = 36

5 x 5 = 25

~~~~~hok tốt ~~~~~~!!!!###$$$$>>>>33333

lấy máy tính mà tính ko rảnh

mún thì vào hoc24.vn đăg

7x7-8+8x8-65+9x9-81=49-8+64-65+81-81=41+64-65=105-65=40

máy tính 

Tick mình nha !

1,4,9,16

ai đúng mình tích

1x1=1

2x2=4

3x3=9

4x4=16

5x5=25

6x6=36

7x7=49

8x8=64

9x9=81

10x10=100

11x11=121

12x12=144

13x13=169

14x14=196

15x15=225

16x16=256

17x17=289

18x18=324

19x19=361

20x20=400

7x7=49

8x8=64

9x9=81

k mình nha

7 x 7 = 49

8 x 8 = 64

9 x 9 = 81 nha bạn nhaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa

7 x 7 = 49

8 x 8 = 64

9 x 9 = 81

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

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

\(B=\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+\frac{1}{5.5}+\frac{1}{6.6.}+\frac{1}{7.7}+\frac{1}{8.8}\)\(=\)\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+\frac{1}{8^2}\)

\(B=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}\)

\(B=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{7}-\frac{1}{8}\)

\(B=1-\frac{1}{8}\)

\(\Rightarrow B< 1\left(ĐPCM\right)\)