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Câu hỏi của 123456 - Toán lớp 6 - Học toán với OnlineMath

~Hok tốt~

Vì 3^10+1/3^9+1>1 =>3^10+1/3^9+1>3^10+1+2/3^10+1+2 =3.(3^9+1)/3.(3^8+1)

                                                                                               =3^9+1/3^8+1

=>3^10+1/3^9+1 > 3^9+1/3^8+1

Vậy 3^10+1/3^19+1 > 3^9+1/3^8+1

Ta có: \(A=\dfrac{3^{10}+1}{3^9+1}\)

\(\Leftrightarrow A=\dfrac{3^{10}+3-2}{3^9+1}\)

hay \(A=3-\dfrac{2}{3^9+1}\)

Ta có: \(B=\dfrac{3^9+1}{3^8+1}\)

\(\Leftrightarrow B=\dfrac{3^9+3-2}{3^8+1}\)

hay \(B=3-\dfrac{2}{3^8+1}\)

Ta có: \(3^9+1>3^8+1\)

\(\Leftrightarrow\dfrac{2}{3^9+1}< \dfrac{2}{3^8+1}\)

\(\Leftrightarrow-\dfrac{2}{3^9+1}>-\dfrac{2}{3^8+1}\)

\(\Leftrightarrow-\dfrac{2}{3^9+1}+3>-\dfrac{2}{3^8+1}+3\)

hay A>B

Có : A = 3.(3^9+1)-2/3^9+1 = 3 - 2/3^9+1

B = 3.(3^8+1)-2/3^8+1 = 3 - 2/3^8+1

Vì 3^9+1 > 3^8+1 => 2/3^9+1 < 2/3^8+1

=> -2/3^9+1 > -2/3^8+1

=> A > B

Tk mk nha

A=3.(3⁹+1)-2

1. \(\frac{1}{2}\)​​​ssss^ssssss

A= 1/3+1/6+1/10+...+1/561

  = 2. (1/6+1/12+1/20+...+1/1122)

  = 2. [1/(2.3) + 1/(3.4) + 1/(4.5) +...+1/(33.34)]

  = 2. ( 1/2 - 1/3 +1/3 - 1/4 + 1/4 - 1/5 +...+ 1/33 - 1/34 )

  =2. (1/2 - 1/34)

  =2. 8/17

  =16/17

Vì 16/17 > 16/18 = 8/9 -> A > 8/9

\(A=\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{561}\)

\(A=\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{1122}\)

\(A=\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{33.34}\)

\(A=2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{33.34}\right)\)

\(A=2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{33}-\frac{1}{34}\right)\)

\(A=2.\left(\frac{1}{2}-\frac{1}{34}\right)\)

\(A=2.\left(\frac{17-1}{34}\right)\)

\(A=2.\frac{8}{17}\)

\(A=\frac{16}{17}>\frac{16}{18}=\frac{8}{9}\)

\(\Rightarrow A>\frac{8}{9}\)