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a) \(A=\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)
\(\Rightarrow A< \frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}\)
\(\Rightarrow A< \frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(\Rightarrow A< \frac{1}{2}-\frac{1}{100}< \frac{1}{2}\)
b) b = a - c => b + c = a
\(\left\{{}\begin{matrix}\frac{a}{b}\cdot\frac{a}{c}=\frac{a^2}{bc}\\\frac{a}{b}+\frac{a}{c}=\frac{ac+ab}{bc}=\frac{a\left(b+c\right)}{bc}=\frac{a^2}{bc}\end{matrix}\right.\)
\(\Rightarrow\frac{a}{b}\cdot\frac{a}{c}=\frac{a}{b}+\frac{a}{c}\)
Bước 2 bạn sai rồi. Vd: \(\frac{1}{3x3}\) đâu bằng hay nhỏ hơn \(\frac{1}{2x3}\)
\(\frac{1}{a}+\frac{1}{b}=\frac{1}{c}\Rightarrow\frac{bc+ac}{abc}=\frac{ab}{abc}\Rightarrow bc+ac=ab\)
\(\Rightarrow ab-ac-bc=0\Rightarrow a\left(b-c\right)-c\left(b-c\right)=c^2\)
\(\Rightarrow\left(b-c\right)\left(a-c\right)=c^2\Rightarrow\frac{a-c}{c}=\frac{c}{b-c}\)
a) Ta có: \(\frac{a}{b}+\frac{a}{c}=\frac{ac+ab}{bc}=\frac{a\left(b+c\right)}{bc}=\frac{a.a}{bc}\) (thay b+c = a) (1)
\(\frac{a}{b}\times\frac{a}{c}=\frac{a.a}{bc}\) (2)
Từ (1) và (2) suy ra: \(\frac{a}{b}+\frac{a}{c}=\frac{a}{b}\times\frac{a}{c}\) (đpcm)
b) \(c=a+b\)\(\Rightarrow\)\(a=c-b\)
Ta có: \(\frac{a}{b}-\frac{a}{c}=\frac{ac-ab}{bc}=\frac{a\left(c-b\right)}{bc}=\frac{a^2}{bc}\) (thay c-b = a) (3)
\(\frac{a}{b}\times\frac{a}{c}=\frac{a^2}{bc}\) (4)
Từ (3) và (4) suy ra: \(\frac{a}{b}-\frac{a}{c}=\frac{a}{b}\times\frac{a}{c}\) (đpcm)
\(\frac{a}{b}=\frac{b}{3c}=\frac{c}{9a}\)
\(\Leftrightarrow\left(\frac{a}{b}\right)^3=\left(\frac{b}{3c}\right)^3=\left(\frac{c}{9a}\right)^3=\left(\frac{a.b.c}{b.3c.c.9a}\right)=\frac{1}{27}=k^3\)
\(\Leftrightarrow k=\left(\frac{1}{27}\div\frac{1}{27}\right)\div3=\frac{1}{3}\)
\(\Leftrightarrow\frac{b}{3c}=\frac{1}{3}\)
Vậy \(\Rightarrow b=c\left(đpcm\right)\)