\(\Leftrightarrow B=\left(x^2-3x^2+3x-1\right)+1010=\left(x-1\right)^2+1010=10^3+1010=100+1010=1110\)

y=13

Trả lời:

\(G< H\)

Hok tốt!

_ _ _ 

ghi rõ lời giải bạn ơi ! giúp mình với tối nay mk phải đi học rồi

G = 1.(2.3) + 2.(4.6) + 4.(8.12)

H = 1.(3.5) + 2.(6.10) + 4.(12.20)

suy ra : 1.(3.5)>1.(2.3)

             2.(6.10)>2.(4.6)

            4.(12.20)>4.(8.12)

suy ra : G<H

VẬY .....

Chúc bạn học tốt tch mình nhé !

Bài 5:

a) Ta có: \(A=1\cdot2+2\cdot3+3\cdot4+...+9\cdot10\)

\(\Leftrightarrow3\cdot A=3\cdot\left(1\cdot2+2\cdot3+3\cdot4+...+9\cdot10\right)\)

\(\Leftrightarrow3A=1\cdot2\cdot\left(3-0\right)+2\cdot3\cdot\left(4-1\right)+3\cdot4\cdot\left(5-2\right)+...+9\cdot10\cdot\left(11-8\right)\)

\(\Leftrightarrow3A=1\cdot2\cdot3+2\cdot3\cdot4-1\cdot2\cdot3+3\cdot4\cdot5-2\cdot3\cdot4+...+8\cdot9\cdot10-8\cdot9\cdot10+9\cdot10\cdot11\)

\(\Leftrightarrow3\cdot A=9\cdot10\cdot11=90\cdot11=990\)

hay A=330

Vậy: A=330

a) \(A=\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+...+\frac{2}{98\cdot99\cdot100}\)

\(A=\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+...+\frac{1}{98\cdot99}-\frac{1}{99\cdot100}\)

\(A=\frac{1}{2}-\frac{1}{99\cdot100}=\frac{1}{2}-\frac{1}{9900}=\frac{4949}{9900}\)

b) \(B=\frac{17}{1\cdot3\cdot5}+\frac{17}{3\cdot5\cdot7}+\frac{17}{5\cdot7\cdot9}+...+\frac{17}{47\cdot49\cdot51}\)

\(B=\frac{17}{4}\left(\frac{4}{1\cdot3\cdot5}+\frac{4}{3\cdot5\cdot7}+\frac{4}{5\cdot7\cdot9}+...+\frac{4}{47\cdot49\cdot51}\right)\)

\(B=\frac{17}{4}\left(\frac{1}{1\cdot3}-\frac{1}{3\cdot5}+\frac{1}{3\cdot5}-\frac{1}{5\cdot7}+...+\frac{1}{47\cdot49}-\frac{1}{49\cdot51}\right)\)

\(B=\frac{17}{4}\left(\frac{1}{3}-\frac{1}{2499}\right)=\frac{17}{4}\cdot\frac{832}{2499}=\frac{208}{147}\)