\(=\left(1^2+2^2+...+100^2\right)\cdot0=0\)

Bài 3: 

a: Ta có: 60-3(x-2)=51

\(\Leftrightarrow x-2=3\)

hay x=5

b: Ta có: \(4x-20=25:2^2\)

\(\Leftrightarrow4x=\dfrac{25}{4}+20=\dfrac{105}{4}\)

hay \(x=\dfrac{105}{16}\)

c: Ta có: \(8\cdot6+288:\left(x-3\right)^2=50\)

\(\Leftrightarrow288:\left(x-3\right)^2=50-48=2\)

\(\Leftrightarrow\left(x-3\right)^2=144\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=12\\x-3=-12\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=15\\x=-9\end{matrix}\right.\)

a) (-124) + 24 = -100
b) 37. 78 + 37. 22
=  37 . ( 78 + 22 )
=  37 . 100
= 3700

tk cho mik nha

\(\left(2^3\cdot9^4+9^3+45\right):\left(9^2\cdot10-9^2\right)\)

\(=\dfrac{9^3\cdot\left(2^3\cdot9+1\right)+45}{9^3}\)

\(=\dfrac{9^3\cdot73+45}{9^3}=\dfrac{5918}{81}\)

   \(3^{^{ }2}\) . 5 + 2³.10  - 81: 3

= 9.5 + 8.10 - 27

= 45 + 80 - 27

= 125 - 27

= 98 

\(3^2\cdot5+2^3\cdot10-81:3\)

\(=9\cdot5+8\cdot10-3^4:3\)

\(=40+80-3^3\)

\(=120-27\)

\(=93\)

\(=64\cdot\dfrac{-1}{8}+100\cdot\dfrac{1}{100}\)

=-8+1

=-7

\(A=-\dfrac{1}{6}-\left[\dfrac{9}{16}-\dfrac{9}{8}+\dfrac{3}{4}\right]\)

\(=\dfrac{-1}{6}-\dfrac{9-18+12}{16}=\dfrac{-1}{6}-\dfrac{3}{16}\)

\(=\dfrac{-8}{48}-\dfrac{9}{48}=-\dfrac{17}{48}\)

\(B=\left(-\dfrac{3}{4}+\dfrac{3}{4}\right)^2+\dfrac{3}{4}\cdot\dfrac{-5}{9}+7^3\)

=-15/36+343

=343-5/12=4111/12

e: =5/7(5/11+2/11-14/11)

=5/7(-7/11)=-5/11

f: =-5/4(2/11+9/11)+1+3/4

=-5/4+7/4

=2/4=1/2

g: =-5/7(3/13+10/13)+1+5/7

=-5/7+1+5/7

=1

h: =7/4(29/5-9/5)+41/13

=-7+41/13=-50/13

\(\dfrac{\dfrac{5}{3}-\dfrac{5}{7}+\dfrac{5}{9}}{\dfrac{10}{3}-\dfrac{10}{7}+\dfrac{10}{9}}=\dfrac{5\left(\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{9}\right)}{10\left(\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{9}\right)}=\dfrac{1}{2}\)

thank you