Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(A=-\frac{\frac{-6}{5}+\frac{6}{19}-\frac{6}{23}}{\frac{9}{5}-\frac{9}{19}+\frac{9}{23}}\)
\(=\frac{-6.\left(\frac{1}{5}-\frac{1}{19}+\frac{1}{23}\right)}{9.\left(\frac{1}{5}-\frac{1}{19}+\frac{1}{23}\right)}\)
\(=-\frac{6}{9}=-\frac{2}{3}\)
1. sai dấu nhé
2.a, \(\frac{45^{10}.5^{20}}{75^{15}}=\frac{\left(3^2.5\right)^{10}.5^{20}}{\left(5^2.3\right)^{15}}=\frac{3^{20}.5^{30}}{5^{30}.3^{15}}=3^5=243\)
b, \(\frac{\left(0,8\right)^5}{\left(0,4\right)^6}=\frac{\left(\frac{4}{5}\right)^5}{\left(\frac{2}{5}\right)^6}=\frac{\left(\frac{2}{5}\cdot2\right)^5}{\left(\frac{2}{5}\right)^6}=\frac{\left(\frac{2}{5}\right)^5\cdot2^5}{\left(\frac{2}{5}\right)^5\cdot\frac{2}{5}}=2^5\div\frac{2}{5}=32\cdot\frac{5}{2}=80\)
c, \(\frac{2^{15}.9^4}{6^6.8^3}=\frac{2^{15}.3^8}{2^6.3^6.2^9}=\frac{2^{15}.3^2}{2^{15}}=3^2=9\)
Có : a/ab+a+1 = a/ab+a+abc = 1/b+1+bc = 1/bc+b+1
c/ca+c+1 = bc/abc+bc+b = b/1+bc+b = b/bc+b+1
=> A = 1+bc+b/bc+b+1 = 1
Tk mk nha
BÀI 1:
\(\frac{a}{ab+a+1}+\frac{b}{bc+b+1}+\frac{c}{ca+c+1}\)
\(=\frac{a}{ab+a+1}+\frac{ab}{a\left(bc+b+1\right)}+\frac{abc}{ab\left(ca+c+1\right)}\)
\(=\frac{a}{ab+a+1}+\frac{ab}{abc+ab+a} +\frac{abc}{a^2bc+abc+ab}\)
\(=\frac{a}{ab+a+1}+\frac{ab}{ab+a+1}+\frac{1}{ab+a+1}\) (thay abc = 1)
\(=\frac{a+ab+1}{a+ab+1}=1\)
Từ a+b+c=2010
\(\Rightarrow\)a= 2010-(b+c)
\(\Rightarrow\)b= 2010-(c+a)
\(\Rightarrow\)c= 2010-(a+b)
Thay vào A, ta được:
A=\(\frac{2010-\left(b+c\right)}{b+c}\)+ \(\frac{2010-\left(c+a\right)}{c+a}\) + \(\frac{2010-\left(a+b\right)}{a+b}\)
A= \(\frac{2010}{b+c}\)+ \(\frac{2010}{c+a}\)+\(\frac{2010}{a+b}\)- 3
A= 2010( \(\frac{1}{b+c}\)+\(\frac{1}{c+a}\)+\(\frac{1}{a+b}\) ) -3
A= 2010. \(\frac{1}{10}\)-3
A=201-3
A= 198
Vậy A=198
`Answer:`
\(C=\left(1-\frac{1}{3}\right)\left(1-\frac{1}{6}\right)\left(1-\frac{1}{10}\right)\left(1-\frac{1}{15}\right)...\left(1-\frac{1}{210}\right)\)
\(=\left(\frac{3}{3}-\frac{1}{3}\right)\left(\frac{6}{6}-\frac{1}{6}\right)\left(\frac{10}{10}-\frac{1}{10}\right)\left(\frac{15}{15}-\frac{1}{15}\right)...\left(\frac{210}{210}-\frac{1}{210}\right)\)
\(=\frac{2}{3}.\frac{5}{6}.\frac{9}{10}.\frac{14}{15}...\frac{209}{210}\)
\(=\frac{2.2}{3.2}.\frac{5.2}{6.2}.\frac{9.2}{10.2}...\frac{209.2}{210.2}\)
\(=\frac{4}{6}.\frac{10}{12}.\frac{18}{20}...\frac{418}{420}\)
\(=\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}...\frac{19.22}{20.21}\)
\(=\frac{1.4.2.5.3.6...19.22}{2.3.3.4.4.5...20.21}\)
\(=\frac{\left(1.2.3...19\right)\left(4.5.6...22\right)}{\left(2.3.4...20\right)\left(3.4.5...21\right)}\)
\(=\frac{11}{30}\)