A=1/2²+1/3²+...+1/9²

Ta có:1/2²>1/2.3,1/3²>1/3.4,...,1/9²>1/9.10

⇒A>1/2.3+1/3.4+...+1/9.10

A>1/2-1/3+1/3-1/4+...+1/9-1/10

A>1/2-1/10

A>2/5(đpcm)

Ta có: A = 1/4 + 1/9 + 1/16 + 1/25 +1/36 + 1/49 + 1/64 + 1/81

Vì 1/2²>1/2.3,1/3²>1/3.4,...,1/9²>1/9.10

=>A>1/2.3+1/3.4+...+1/9.10

=>A>1/2-1/3+1/3-1/4+...+1/9-1/10

=>A>1/2-1/10

=>A>2/5

Giải:

\(A=\dfrac{1}{4}+\dfrac{1}{9}+\dfrac{1}{16}+\dfrac{1}{25}+\dfrac{1}{36}+\dfrac{1}{49}+\dfrac{1}{64}+\dfrac{1}{81}\) 

\(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2} +\dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+\dfrac{1}{8^2}+\dfrac{1}{9^2}\) 

Ta có:

\(\dfrac{1}{2^2}=\dfrac{1}{2.2}>\dfrac{1}{2.3}\) 

\(\dfrac{1}{3^2}=\dfrac{1}{3.3}>\dfrac{1}{3.4}\) 

\(\dfrac{1}{4^2}=\dfrac{1}{4.4}>\dfrac{1}{4.5}\) 

\(\dfrac{1}{5^2}=\dfrac{1}{5.5}>\dfrac{1}{5.6}\) 

\(\dfrac{1}{6^2}=\dfrac{1}{6.6}>\dfrac{1}{6.7}\) 

\(\dfrac{1}{7^2}=\dfrac{1}{7.7}>\dfrac{1}{7.8}\) 

\(\dfrac{1}{8^2}=\dfrac{1}{8.8}>\dfrac{1}{8.9}\) 

\(\dfrac{1}{9^2}=\dfrac{1}{9.9}>\dfrac{1}{9.10}\) 

\(\Rightarrow A>\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}\) 

\(\Rightarrow A>\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\)

\(\Rightarrow A>\dfrac{1}{2}-\dfrac{1}{10}\) 

\(\Rightarrow A>\dfrac{2}{5}\left(đpcm\right)\) 

Chúc bạn học tốt!

\(=\dfrac{1}{15}+\dfrac{2}{15}+\dfrac{3}{15}+...+\dfrac{9}{15}\)

\(=\dfrac{1+2+3+...+9}{15}\)

\(=\dfrac{45}{15}=3\)

$A=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{99}{100}$

$\Rightarrow A<\frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{100}{101}$

$\Rightarrow {}^{}<\frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{100}{101}.\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{99}{100}$

$\Rightarrow {}^{}<\frac{1}{101}<\frac{1}{100}=\frac{1}{{}^{}}$

$\iff A<$

$A=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{99}{100}$

$\Rightarrow A<\frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{100}{101}$

$\Rightarrow {}^{}<\frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{100}{101}.\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{99}{100}$

$\Rightarrow {}^{}<\frac{1}{101}<\frac{1}{100}=\frac{1}{{}^{}}$

$\iff A<$

    A = 1 + 4 + 9 + 16 +....+ 64 + 81 + 100

=>A = 1 + 1 + 3 + 1 + 8 +...+ 1 + 63 + 1 + 80 + 1 + 99

=>A = 1 + 1 + ..... + 1 + 3 + 8 + 14 +....+80 + 99

Bạn tự tìm kết quả nhé

Hok tốt

A=(1+9)+(4+16)+(9+81)+(16+64)+(25+36+49+100)

A=10+20+90+80+(61+49+100)

A=10+20+90+80+110+100

A=(100+10)+(20+90)+80+110

A=110+110+110+80

A=(110.3)+80

A=330+80

A=410

t i c k cho mình nha

bạnnnnnnnnnnnnnnnnnnnnn

a,234.4+234.5+234=234(4+5+1)=234.10=2340

b,135.16-135.2-135.4=135(16-4-2)=135.10=1350

c,1/2+1/4+1/8+1/16+1/32+1/64=1-1/2+1/2-1/4+1/4-1/8+1/8-1/16+1/16-1/32=1-1/32=31/32

a) 234 x4 + 234 x 5 + 234 = 234 x ( 1+4+5)=2340

b)135 x16 -135 x2 -4 x135 =135 x (16-2-4) =1350

a) 234.(4+5+1)=234.10=2340

b)135.(16-2-4)=135.10=1350

a,=16.(64+17)+81.84
   =16.81+81.84
   =81(16+84)
   =81.100=8100
b,=32.19+32
   =32.(19+1)
   =32.20=640
c,=12.19+12.37+44.12
   =12.(19+37+44)
   =12.100=1200
d, Khoảng cách là:
        5-1=4;9-5=4
    Số số hạng là:
        (81-1):4+1=21(số)
    Tổng dãy số là:
        (81+1).21:2=861

a

\(\text{=16.(64+17)+81.84}\)
\(\text{=16.81+81.84}\)
\(\text{ =81.(16+84)}\)
 \(\text{=81.100=8100}\)

b

\(\text{=32.19+32}\)
\(\text{ =32.(19+1)}\)
\(\text{ =32.20=640}\)
c

\(\text{=12.19+12.37+44.12}\)
\(\text{ =12.(19+37+44)}\)
\(\text{ =12.100}\)

\(=1200\)

d
Có tất cả số hạng là

\(\text{( 81 - 1 ) : 4 + 1 = 21 (số )}\)

Tổng là

\(\text{( 81 + 1 ) x 21 : 2 = 861}\)