Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\frac{a+5}{a-5}=\frac{b+6}{b-6}=>\left(a+5\right)\left(b-6\right)=\left(a-5\right)\left(b+6\right)\)
\(=>a\left(b-6\right)+5\left(b-6\right)=a\left(b+6\right)-5\left(b+6\right)\)
\(=>ab-6a+5b-30=ab+6a-5b-30=>-6a+5b=6a-5b=>6a-\left(-6a\right)=5b-\left(-5b\right)\)
\(=>12a=10b=>\frac{a}{b}=\frac{10}{12}=\frac{5}{6}\) (đpcm)
a)Ta có:A:B=\(\left(\frac{1}{4}.\frac{3}{6}.\frac{5}{8}....\frac{43}{46}.\frac{45}{48}\right):\left(\frac{2}{5}.\frac{4}{7}.\frac{6}{9}....\frac{44}{47}.\frac{46}{49}\right)=\frac{\left(1.3.5...45\right).\left(2.4.6...46\right)}{\left(4.6.8...48\right)\left(5.7.9...49\right)}=\frac{3.2}{47.48.49}<1\)
=>A<B
b)Do A có tử nhỏ hơn mẫu nên A<1<133
Vậy A<133
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 \(\frac{1}{2x3}\)
\(\frac{a+5}{a-5}=\frac{b+6}{b-6}\)
\(\Leftrightarrow\left(a+5\right)\left(b-6\right)=\left(b+6\right)\left(a-5\right)\)
\(\Leftrightarrow ab-6a+5b-30=ab-5b+6a-30\)
\(\Leftrightarrow-6a+5b=6a-5b\)
\(\Leftrightarrow5b+5b=6a+6a\)
\(\Leftrightarrow10b=12a\)
\(\Leftrightarrow\frac{a}{b}=\frac{10}{12}=\frac{5}{6}\)