rút gọn:
b, \(B=\frac{16\cdot64\cdot8^2}{4^3\cdot2^5\cdot16}\)
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tử số : 2.4 + 4.8 + 8.12 + 12.16 + 16.20
= 2.(1.2+2.4+4.6+6.8+8.10)
ta được 2. A=( 1.2+2.4+4.6+6.8+8.10) / ( 1.2+2.4+4.6+6.8+8.10)
=> A=2
\(A=\frac{1.2+2.4+3.6+4.8+5.10}{3.4+6.8+9.12+12.16+15.20}\)
\(A=\frac{1.2.\left(1+2^2+3^2+4^2+5^2\right)}{3.4.\left(1+2^2+3^2+4^2+5^2\right)}\)
\(A=\frac{1.2}{3.4}\)
\(A=\frac{1}{6}\)
Ta thấy : \(B=\frac{111111}{666665}>\frac{111111}{666666}=\frac{1}{6}\)
Vậy B > A
Theo đề bài, ta có:
\(A=\frac{1\times2+2\times4+3\times6+4\times8+5\times10}{3\times4+6\times8+9\times12+12\times16+15\times20}\)
\(A=\frac{1\times2\times\left(1+2^2+3^2+4^2+5^2\right)}{3\times4\times\left(1+2^2+3^2+4^2+5^2\right)}\)
\(A=\frac{1\times2}{3\times4}\)
\(A=\frac{1}{6}\)
Ta thấy rằng: \(B=\frac{111111}{666665}>\frac{111111}{666666}=\frac{1}{6}\)
Vậy \(B>A\)
\(\dfrac{1.2.4+2.4.8+4.8.16+8.16.32}{1.3.4+2.6.8+4.12.16+8.24.32}\)
\(=\dfrac{8.\left(1+8+4.16+16.32\right)}{12.\left(1+8+4.16+16.32\right)}\)
\(=\dfrac{8}{12}=\dfrac{2}{3}\)
a, Tự chép đề bài ((:
\(=\frac{1}{9}\cdot1+\left(-\frac{1}{243}\right)\cdot\frac{9}{2}\)
\(=\frac{1}{9}-\frac{1}{54}\)
\(=\frac{5}{54}\)
b, 1. \(\left(\frac{2^2\cdot2^3}{4^2\cdot16}\right)^{15}\)
\(=\left(\frac{2^5}{2^4\cdot2^4}\right)^5=\left(\frac{2^5}{2^8}\right)^5=\left(\frac{1}{2^3}\right)^5=\left(\frac{1}{8}\right)^5=\frac{1}{8^5}\)(Để vậy đi :v)
2. \(\left(\frac{2^6}{16^2}\right)^{10}\)
\(=\left(\frac{2^6}{2^8}\right)^{10}=\left(\frac{1}{2^2}\right)^{10}=\frac{1}{2^{20}}\)
c, \(\frac{2^{15}\cdot9^4}{6^6\cdot8^3}\)
\(=\frac{2^{15}\cdot\left(3^2\right)^4}{\left(2\cdot3\right)^6\cdot\left(2^3\right)^3}=\frac{2^{15}\cdot3^8}{2^6\cdot3^6\cdot2^9}=\frac{2^{15}\cdot3^8}{2^{15}\cdot3^6}=\frac{3^2}{1}=3^2=9\)
\(G=\frac{1}{1.2}+\frac{2}{2.4}+\frac{3}{4.7}+\frac{4}{7.11}+\frac{5}{11.16}\)
\(G=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}\)
\(G=1-\frac{1}{16}\)
\(G=\frac{15}{16}\)
\(G=\frac{1}{1.2}+\frac{2}{2.4}+\frac{3}{4.7}+\frac{4}{7.11}+\frac{5}{11.16}\)
\(G=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}\)
\(G=1-\frac{1}{16}\)
\(G=\frac{15}{16}\)
\(B=\frac{2.4+2.4.8+4.8.16+8.16.32}{3.4+2.6.8+4.12.16+8.24.32}\)
\(B=\frac{2.4+2.4.8+4.2.4.16+2.4.16.32}{3.4+2.2.3.2.4+4.3.4.16+2.4.8.3.32}\)
\(B=\frac{2.4.\left(1+8+4.16+16.32\right)}{3.4.\left(1+2.2.2+4.16+2.8.32\right)}\)
\(B=\frac{2.4.\left(1+8+4.16+16.32\right)}{3.4.\left(1+8+4.16+16.32\right)}\)
\(B=\frac{2}{3}\)
Chúc bn học tốt !!!!
câu hỏi hay......nhưng tui xin nhường cho các bn khác
Hãy tích đúng cho tui nha
THANKS
\(B=\frac{16\cdot64\cdot8^2}{4^3\cdot2^5\cdot16}=\frac{16\cdot64\cdot64}{64\cdot32\cdot16}=\frac{64}{32}=2\)