Question

Chứng tỏ rằng : B=\(\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}

Answer 1

Ta có : \(B=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{8^2}\)

Mà \(\frac{1}{2^2}<\frac{1}{1.2};\frac{1}{3^2}<\frac{1}{2.3};...;\frac{1}{8^2}<\frac{1}{7.8}\)

\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{8^2}<\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{7.8}=1-\frac{1}{8}<1\)

Vậy B < 1

Answer 2

Answer 3

Ta có:\(\frac{1}{2^2}<\frac{1}{1.2};\frac{1}{3^2}<\frac{1}{2.3};......;\frac{1}{8^2}<\frac{1}{7.8}\) 

<=> B<\(\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{7.8}\) 

<=> B<\(\frac{1}{1}-\frac{1}{2}+.......+\frac{1}{7}-\frac{1}{8}\) 

<=> B<\(1-\frac{1}{8}\) 

<=> B<\(\frac{7}{8}\) <1

Answer 4

ghi lại biểu thức

B<1/1*2+1/2*3+1/3*4+...+1/7*8

B<1/1-1/2+1/2-1/3+1/3-1/4+...+1/7-1/8(công thức)

B<1-1/8<1

nên B<1

 

Answer 5

\(\frac{1}{2^2}+\frac{1}{3^2}+....+\frac{1}{8^2}<\frac{1^{ }}{1\cdot2}+\frac{1}{2.3}+...+\frac{1}{7\cdot8}=1-\frac{1}{2}+\frac{1}{2}-...-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}=1-\frac{1}{8}=>B<1_{^{ }}\)

Answer 6

vì 1/2^2<1/1x2,1/3^2<1/2x3.....1/8^2<1/7x8

=>1/2^2+1/3^2+.....+1/8^2<1/1x2+1/2x3+...+1/7x8=﴾tự tính﴿=1

=>1/2^2+1/3^2+.....+1/8^2<1﴾đpcm﴿ 

vậy B < 1

Answer 7

a. 8 2/3 : - 10= -8

b. x+ 30%x = -13

c. 3/10 - x = 25%

d. x . 9/17 = 17/5

GIÚP MK VS, ĐỀ KIỂM TRA THI ĐẤY. KO GIÚP LẦN NÀY MK RỚT ĐIỂM, LÀM ƠN..