Đặt \(A=\frac{1}{6^2}+\frac{1}{9^2}+\cdots+\frac{1}{30^2}\)

\(=\frac{1}{3^2}\left(\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{10^2}\right)\)

\(=\frac19\left(\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{10^2}\right)\)

Ta có: \(\frac{1}{2^2}<\frac{1}{1\cdot2}=1-\frac12\)

\(\frac{1}{3^2}<\frac{1}{2\cdot3}=\frac12-\frac13\)

...

\(\frac{1}{10^2}<\frac{1}{9\cdot10}=\frac19-\frac{1}{10}\)

Do đó: \(\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{10^2}<1-\frac12+\frac12-\frac13+\frac13-\frac14+\cdots+\frac19-\frac{1}{10}\)

=>\(\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{10^2}<1-\frac{1}{10^2}<1\)

=>\(\frac19\left(\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{10^2}\right)<\frac19\)

=>\(A<\frac19\)

\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{10^2}\)

\(< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\)

\(=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{10-9}{9.10}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)

\(=1-\frac{1}{10}< 1\)

\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{10^2}\\ A< \frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{9\times10}\\ A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}=1-\frac{1}{10}\\ A< \frac{9}{10}< 1\Rightarrow A< 1\)

1. 2².3².5:2².3-3²=3.5-3²=15-9=6
2. 2.5²-2².3²:3²=2.(5²-2)=2.(25-2)=46

3. 3³.19-3³.12=3³.(19-12)=3³.7=189

4. 3.5²-16:2²=3.5²-2⁴:2²=3.25-4=75-4=71

bạn muốn trả lời đúng không

ai trả lời giúp mình

1^3+2^3+4^3+5^3=(1+2+3+4+5)^2

co ai giai dc ko cau 60 % .....

$\textbf{Ta có:}$

$A=\dfrac{-5^2-5\cdot3^2}{5^3+5^2\cdot3^2}$

$=\dfrac{-25-45}{125+225}$

$=\dfrac{-70}{350}$

$=-\dfrac15.$

Và

$B=\dfrac{2^{12}\cdot3^{10}+6^9\cdot120}{2^{12}\cdot3^{12}-2^{11}\cdot3^{11}}$

$=\dfrac{2^{12}3^{10}+2^{12}3^{10}\cdot5}{2^{11}3^{11}(2\cdot3-1)}$

$=\dfrac{2^{12}3^{10}(1+5)}{2^{11}3^{11}\cdot5}$

$=\dfrac{2^{12}3^{10}\cdot6}{2^{11}3^{11}\cdot5}$

$=\dfrac{2^2}{5}$

$=\dfrac45.$

=> $M=B-A$$=\dfrac45-\left(-\dfrac15\right)$

$=\dfrac55$$=1.$

chịu

Chịu