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\(12^8.9^{12}\) và \(18^{18}.\)
Ta có:
\(12^8.9^{12}\)
\(=\left(2^2.3\right)^8.\left(3^2\right)^{12}\)
\(=2^{16}.3^8.3^{24}\)
\(=2^{16}.3^{32}\)
\(=2^{16}.\left(3^2\right)^{16}\)
\(=2^{16}.9^{16}\)
\(=\left(2.9\right)^{16}\)
\(=18^{16}.\)
Vì \(18^{16}< 18^{18}.\)
\(\Rightarrow12^8.9^{12}< 18^{18}.\)
Chúc bạn học tốt!
Giải:
\(C=\left(1-\dfrac{2}{2.3}\right)\left(1-\dfrac{2}{3.4}\right)\left(1-\dfrac{2}{4.5}\right)...\left(1-\dfrac{2}{n\left(n+1\right)}\right)\)
Đk: \(n\ne0;n\ne-1\)
\(C=\left(1-\dfrac{2}{2.3}\right)\left(1-\dfrac{2}{3.4}\right)\left(1-\dfrac{2}{4.5}\right)...\left(1-\dfrac{2}{n\left(n+1\right)}\right)\)
\(\Leftrightarrow C=\left(\dfrac{2.3-2}{2.3}\right)\left(\dfrac{3.4-2}{3.4}\right)\left(\dfrac{4.5-2}{4.5}\right)...\left(\dfrac{n\left(n-1\right)-2}{n\left(n+1\right)}\right)\)
\(\Leftrightarrow C=\dfrac{4}{2.3}.\dfrac{10}{3.4}.\dfrac{18}{4.5}...\left(\dfrac{n\left(n-1\right)-2}{n\left(n+1\right)}\right)\)
\(\Leftrightarrow C=\dfrac{1.4}{2.3}.\dfrac{2.5}{3.4}.\dfrac{3.6}{4.5}...\left(\dfrac{\left(n-1\right)\left(n+2\right)}{n\left(n+1\right)}\right)\)
\(\Leftrightarrow C=\dfrac{1.4.2.5.3.6...\left(n-1\right)\left(n+2\right)}{2.3.3.4.4.5.n\left(n+1\right)}\)
\(\Leftrightarrow C=\dfrac{\left[1.2.3...\left(n-1\right)\right]\left[4.5.6\left(n+2\right)\right]}{\left(2.3.4...n\right)\left[3.4.5....\left(n+1\right)\right]}\)
\(\Leftrightarrow C=\dfrac{n+2}{3n}\)
Vì \(\dfrac{n+2}{3n}< \dfrac{2n+2}{3n}\)
\(\Leftrightarrow C< \dfrac{2n+2}{3n}\)
Vậy ...
Giải:
\(C=\left(1-\dfrac{2}{2.3}\right)\left(1-\dfrac{2}{3.4}\right)\left(1-\dfrac{2}{4.5}\right)...\left(1-\dfrac{2}{n\left(n+1\right)}\right)\)
Đk: \(n\ne0;n\ne-1\)
\(C=\left(1-\dfrac{2}{2.3}\right)\left(1-\dfrac{2}{3.4}\right)\left(1-\dfrac{2}{4.5}\right)...\left(1-\dfrac{2}{n\left(n+1\right)}\right)\)
\(\Leftrightarrow C=\left(\dfrac{2.3-2}{2.3}\right)\left(\dfrac{3.4-2}{3.4}\right)\left(\dfrac{4.5-2}{4.5}\right)...\left(\dfrac{n\left(n-1\right)-2}{n\left(n+1\right)}\right)\)
\(\Leftrightarrow C=\dfrac{4}{2.3}.\dfrac{10}{3.4}.\dfrac{18}{4.5}...\left(\dfrac{n\left(n-1\right)-2}{n\left(n+1\right)}\right)\)
\(\Leftrightarrow C=\dfrac{1.4}{2.3}.\dfrac{2.5}{3.4}.\dfrac{3.6}{4.5}...\left(\dfrac{\left(n-1\right)\left(n+2\right)}{n\left(n+1\right)}\right)\)
\(\Leftrightarrow C=\dfrac{1.4.2.5.3.6...\left(n-1\right)\left(n+2\right)}{2.3.3.4.4.5.n\left(n+1\right)}\)
\(\Leftrightarrow C=\dfrac{\left[1.2.3...\left(n-1\right)\right]\left[4.5.6\left(n+2\right)\right]}{\left(2.3.4...n\right)\left[3.4.5....\left(n+1\right)\right]}\)
\(\Leftrightarrow C=\dfrac{n+2}{3n}\)
Vì \(\dfrac{n+2}{3n}< \dfrac{2n+2}{3n}\)
\(\Leftrightarrow C< \dfrac{2n+2}{3n}\)
Vậy ...
Gọi 3 máy là a,b,c. Ta có:
a+b = 1 giờ 20 phút = 80 (phút)
b+c = 1 giờ 30 phút = 90 (phút)
a+c = 2 giờ 24 phút = 144 (phút)
mà a+b+b+c+a+c = 80+90+144 = 314 ( phút)
hay 2(a+b+c) = 314 => a+b+c = 314/2 = 157
- c = (a+b+c) - (a+b) = 157 - 80 = 77 (phút)
- a = (a+b+c) - (b+c) = 157 - 90 = 67 (phút)
- b = (a+b+c) - (a+c) = 157 - 144 = 13 (phút)
Ta có : \(1=3-2=\sqrt{9}-\sqrt{4}\)
Vì \(\left\{{}\begin{matrix}\sqrt{9}>\sqrt{8}\\\sqrt{4}< \sqrt{5}\end{matrix}\right.\Rightarrow}\left\{{}\sqrt{8}-\sqrt{5}< \sqrt{9}-\sqrt{4}=1}\)
Ta có: \(\sqrt{2}>1\)
\(\Rightarrow1+\sqrt{2}>1+1\)
\(\Rightarrow1+\sqrt{2}>2\)
Ta có:\(\sqrt{2}>\sqrt{1}\)
\(\Leftrightarrow1+\sqrt{2}>1+\sqrt{1}=2\)