ahihi Dồ     ahihi đồ chó

bn có bị j ko z

\(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\)

\(3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}\)

\(3A-A=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\right)\)

\(2A=1-\frac{1}{3^{99}}\)

\(A=\frac{1-\frac{1}{3^{99}}}{2}\)

Ta đặt \(C=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)

Ta so sánh giữa A và C.

\(\frac{1}{3}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};\frac{1}{3^3}< \frac{1}{3.4};....;\frac{1}{3^{99}}< \frac{1}{99.100}\Leftrightarrow A< C\)( 1 )

 \(C=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}=\frac{1}{1}-\frac{1}{100}=\frac{99}{100}\)

Mà \(\frac{99}{100}< \frac{1}{2}\Rightarrow C< B\)( 2 )

Từ ( 1 ) và ( 2 )

 \(\Rightarrow A< C< B\Leftrightarrow A< B\)

lp 6-7 hay 7-8 z bn

\(=\frac{bzx-cxy}{ax}=\frac{cxy-ayz}{by}=\frac{ayz-bzx}{cz}=\frac{bzx-cxy+cxy-ayz+ayz-bzx}{ax+by+cz}=0\)

=>bz-cy=0;cx-az=0;ay-bx=0

\(\Rightarrow\frac{x}{a}=\frac{y}{b}=\frac{z}{c}\left(đpcm\right)\)

\(\frac{a}{b}=\frac{b}{c}\Rightarrow\frac{a}{b}.\frac{b}{c}=\frac{a}{b}.\frac{a}{b}=\frac{b}{c}.\frac{b}{c}\Rightarrow\frac{a}{c}=\frac{a^2}{b^2}=\frac{b^2}{c^2}=\frac{a^2+b^2}{b^2+c^2}\)

\(\Leftrightarrow\frac{a^2}{b^2}=\frac{b^2}{c^2}=\frac{a}{b}.\frac{b}{c}=\frac{a}{c}=\frac{a^2+b^2}{b^2+c^2}\)

Ta có :

5²⁷  = ( 5³ )⁹ = 125⁹ < 128⁹ = ( 2⁷ )⁹ = 2⁶³

=> 5²⁷ < 2⁶³

Mà 2⁶³ < 2⁶⁴ = ( 2¹⁶ )⁴ = 65536⁴ < 5²⁸ = ( 5⁷ )⁴ = 78125⁴ 

=> 5²⁷ < 2⁶³ < 5²⁸

\(5^{27}< 5^{28}< 5^{63}\)