kho lm giup minh voi

Bạn tham khảo nhé 

A=14 +116 +136 +164 +1100 +1144 +1196 =122 +142 +162 +182 +1102 +1122 +1142 

2A=222 +242 +262 +282 +2102 +2122 +2142 

2A<12 +22.4 +24.6 +26.8 +28.10 +210.12 +212.14 

2A<12 +12 −14 +14 −1

Đặt \(A\)\(=\frac{1}{2^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)

Ta có: \(A< \frac{1}{2^2-1}+\frac{1}{4^2-1}+...+\frac{1}{100^2-1}\)

 \(A< \frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{99.101}\)

\(A< \frac{1}{2}.\left(1-\frac{1}{101}\right)\)

\(A< \frac{1}{2}.1\)( VÌ \(1-\frac{1}{101}< 1\))

\(A< \frac{1}{2}\)

đặt A=1/2-1/4+1/8-1/16+1/32-1/64

2A=1-1/2+1/4-1/8+1/16-1/32

2A-A=1-1/64

A=63/64

Vì 63/64<1/3

nên 1/2-1/4+1/8-1/16+1/32-1/64<1/3

Vậy 1/2-1/4+1/8-1/16+1/32-1/64<1/3

nhầm!

chỗ A dòng 4 là 21/64

Tự mình hỏi mà tự mình trả lời he!..he!

S=1/4+1/9+1/16+1/25+1/36+1/49+1/64+1/81=1-1/81=1/81

80/81 là đúng

vô lí vì 2/2 = 1 mà 8/9 < 1

a) Ta có 

\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{8^2}=\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{8.8}\)

Mà \(\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{8.8}<\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{7.8}\)

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{7.8}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{7}-\frac{1}{8}\)

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

                                                       \(=\frac{7}{8}<1\)

Vì \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{8^2}=\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{8.8}<\frac{7}{8}<1\)

nên \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{8^2}<1\)

Ta có: \(VT=\dfrac{1}{2^2}+\dfrac{1}{4^2}+\dfrac{1}{6^2}+...+\dfrac{1}{100^2}\)

\(4VT=\dfrac{1}{2^2:2^2}+\dfrac{1}{4^2:2^2}+\dfrac{1}{6^2:2^2}+...+\dfrac{1}{100^2:2^2}\)

\(4VT=\dfrac{1}{1^2}+\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{50^2}\)

Lại có: \(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)

\(\dfrac{1}{3^2}< \dfrac{1}{2.3}\)

\(...\)

\(\dfrac{1}{4^2}< \dfrac{1}{3.4}\)

\(\Rightarrow4VT-1< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{49.50}\)(*)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\)

\(=1-\dfrac{1}{50}\) (**)

Từ (*) và (**) \(\Rightarrow4VT< 2-\dfrac{1}{50}\)

\(\Rightarrow VT< \dfrac{1}{2}-\dfrac{1}{200}< VP\Rightarrow\) đpcm

b) Ta có: \(2VT=1-\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{8}+\dfrac{1}{16}-\dfrac{1}{32}\)

\(2VT+VT=\left(1-\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{8}+\dfrac{1}{16}-\dfrac{1}{32}\right)+\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16}+\dfrac{1}{32}-\dfrac{1}{64}\right)\)

\(3VT=1-\dfrac{1}{64}< 1\)

\(\Rightarrow VT< \dfrac{1}{3}\) (đpcm)

mk ko bjt có đúng ko

=1/(2+18+14+16+36+64)

=1/(20+30+100)

=1/150

=1/150

=tự tính nhé

256 - 128 - 64 - 36 - 16 - 8 - 4 - 2 - 1

= (256 - 16) - (128 - 8) - (64 - 4) - (36 - 16) - (16 - 4 - 2) - 1

= 240 - 120 - 60 - 20 - 10 - 1

= 29 

sửa tí

= (256 - 16) - (128 - 8) - (64 - 4) - (36 - 16) - 8 - 2 - 1

= 240 - 120 - 60 - 20 - 8 - 2 - 1

= -3

sửa lại đáp án

= 29

cm A<1/2

và B<3/2 thì có thể nhưng bất đẳng thức thì ko có đâu