bạn ấn vào đúng 0 sẽ ra đáp án mình giải 

\(A=\frac{10.11+50.55+70.77}{11.12+55.60+77.84}\)

\(=\frac{10.11+5.10.5.11+7.10.7.11}{11.12+11.5.12.5+11.7.12.7}\)

\(=\frac{10.11\left(1+25+49\right)}{11.12\left(1+25+49\right)}\)

\(=\frac{10.11}{11.12}=\frac{10}{12}=\frac{5}{6}\)

\(B=\frac{1\times3\times5\times7\times........\times49}{26\times27\times28\times...........\times50}\)

\(=\frac{\left(1\times3\times5\times7\times.........\times49\right).\left(2\times4\times6.........48\times50\right)}{\left(26\times27\times28\times.........\times50\right).\left(2\times4\times6\times...........\times48\times50\right)}\)

\(=\frac{1\times2\times3\times4\times..........\times50}{\left(26\times27\times28\times..............\times50\right)2^{25}\left(1\times2\times3\times4\times............\times25\right)}=\frac{1}{2^{25}}\)

\(C=\frac{1.2.6+2.4.12+4.8.24+7.14.42}{1.6.9+2.12.18+4.24.36+7.42.63}\)

\(=\frac{1.2.6\left(1+8+64+343\right)}{1.6.9\left(1+8+64+343\right)}\)

\(=\frac{1.2.6}{1.6.9}=\frac{2}{9}\)

\(A=\frac{5}{6}\)

\(B=\frac{1}{33554432}\)

\(C=\frac{28}{117}\)

??? [ :( ] ??? trúng môn Toán ???

Tính .. thế này ... :

B = \(\frac{6+96+768+4116}{54+432+3456+18522}\)

B = \(\frac{102+768+4116}{486+3456+18522}\)

B = \(\frac{870+4116}{3942+18522}\)

B = \(\frac{4986}{22464}\)

B = \(\frac{277}{1248}\)

Cái này tớ làm theo khả năng, cơ bản bình thường. Nếu có sai sót hay thiếu mong cậu thông cảm ... :(

Chơi ăn gian nha.

bạn nào trả lời được k cho trong hôm nay vì mk đang cần gấp( giaỉ ra)

Mình nghĩ nên sửa lại chỗ 9 thành 8 hoặc 16 thành 18

\(B=\frac{1.2.6+2.4.12}{1.6.9+2.12.18}=\frac{1.2.6.\left(1+2^3\right)}{1.6.9.\left(1+2^3\right)}\)\(=\frac{1.2.6}{1.6.9}=\frac{2}{9}\)

hoặc 

\(B=\frac{1.2.6+2.4.12}{1.6.8+2.12.16}=\frac{1.2.6.\left(1+2^3\right)}{1.6.8.\left(1+2^3\right)}\)\(=\frac{1.2.6}{1.6.8}=\frac{2}{8}=\frac{1}{4}\)

help me, hic

a, ĐK: \(a\ne0,b\ne0,a+b\ne0\)

\(A=\left[\frac{1}{a^2}+\left(\frac{1}{a}+\frac{1}{b}\right):\frac{a+b}{2}+\frac{1}{b^2}\right].\frac{a^2b^2}{a^3+b^3}:\left(a+b\right)\)

\(=\left[\frac{1}{a^2}+\frac{a+b}{ab}:\frac{a+b}{2}+\frac{1}{b^2}\right].\frac{a^2b^2}{a^3+b^3}:\left(a+b\right)\)

\(=\left[\frac{1}{a^2}+\frac{2}{ab}+\frac{1}{b^2}\right].\frac{a^2b^2}{a^3+b^3}:\left(a+b\right)\)

\(=\frac{\left(a+b\right)^2}{a^2b^2}.\frac{a^2b^2}{\left(a+b\right)\left(a^2-ab+b^2\right)}.\frac{1}{a+b}\)

\(=\frac{1}{a^2-ab+b^2}\)

b, \(a^2-ab+b^2=\left(a-\frac{1}{2}b\right)^2+\frac{3}{4}b^2>0\left(a,b\ne0\right)\)

\(\Rightarrow A=\frac{1}{a^2-ab+b^2}>0\forall a;b\)