Lời giải:

$(\frac{1}{2^2}-1)(\frac{1}{3^2}-1)...(\frac{1}{100^2}-1)$

$=\frac{(1-2^2)(1-3^2)...(1-100^2)}{2^2.3^2....100^2}$

$=\frac{-(2^2-1)(3^2-1)...(100^2-1)}{2^2.3^2...100^2}$

$=\frac{-[(2-1)(3-1)...(100-1)][(2+1)(3+1)...(100+1)]}{(2.3...100)(2.3..100)}$

$=\frac{-(1.2.3....99)(3.4....101)}{(2.3...100)(2.3...100)}$

$=-\frac{1.2.3...99}{2.3...100}.\frac{3.4....101}{2.3...100}$

$=-\frac{1}{100}.\frac{101}{2}=\frac{-101}{200}$

Số lớn là:

42 : \(\dfrac{3}{11}\) = 154

Kết luận số lớn là 154 

số lớn là 

42 : 3 x 11 = 154

đs 154

nhớ tích cho mình nha mình cảm ơn rất nhìu  =)))

T = \(\dfrac{1}{2x^2}\) \(\times\) ( -4\(x^3\))

T = \(\dfrac{-4}{2}\) \(\times\) \(\dfrac{x^3}{x^2}\)

T = - 2\(x\)

T(2) = -2 \(\times\) ( 2) = -4

a, T=\(\dfrac{1}{2X^2}\) x (-4X³) = -2X

b, M=-2X =>M=-4

Lời giải:
Gọi tổng trên là $A$
$A=\frac{1}{\frac{3.4}{2}}+\frac{1}{\frac{4.5}{2}}+....+\frac{1}{\frac{2023.2024}{2}}$

$=\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{2023.2024}$

$=2(\frac{4-3}{3.4}+\frac{5-4}{4.5}+...+\frac{2024-2023}{2023.2024})$

$=2(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{2023}-\frac{1}{2024})$

$=2(\frac{1}{3}-\frac{1}{2024})=\frac{2021}{3036}$

$A=\genfrac{}{}{0ex}{}{1}{\genfrac{}{}{0ex}{}{3.4}{2}}+\genfrac{}{}{0ex}{}{1}{\genfrac{}{}{0ex}{}{4.5}{2}}+....+\genfrac{}{}{0ex}{}{1}{\genfrac{}{}{0ex}{}{2023.2024}{2}}$

$=\genfrac{}{}{0ex}{}{2}{3.4}+\genfrac{}{}{0ex}{}{2}{4.5}+...+\genfrac{}{}{0ex}{}{2}{2023.2024}$

$=2(\genfrac{}{}{0ex}{}{4-3}{3.4}+\genfrac{}{}{0ex}{}{5-4}{4.5}+...+\genfrac{}{}{0ex}{}{2024-2023}{2023.2024})$

$=2(\genfrac{}{}{0ex}{}{1}{3}-\genfrac{}{}{0ex}{}{1}{4}+\genfrac{}{}{0ex}{}{1}{4}-\genfrac{}{}{0ex}{}{1}{5}+....+\genfrac{}{}{0ex}{}{1}{2023}-\genfrac{}{}{0ex}{}{1}{2024})$

$=2(\genfrac{}{}{0ex}{}{1}{3}-\genfrac{}{}{0ex}{}{1}{2024})=\genfrac{}{}{0ex}{}{2021}{3036}$