So sánh: \({\left( {{{15}^3}} \right)^2}\) và \({15^{3.2}}\).
So sánh:
\(\left(\dfrac{1}{10}\right)^{15}\)và \(\left(\dfrac{3}{10}\right)^{20}\)
Ta có:
\(\left(\dfrac{1}{10}\right)^{15}=\left(\left(\dfrac{1}{10}\right)^3\right)^5=\left(\dfrac{1}{1000}\right)^5\)
\(\left(\dfrac{3}{10}\right)^{20}=\left(\left(\dfrac{3}{10}\right)^4\right)^5=\left(\dfrac{81}{10000}\right)^5\)
Ta có: \(\left(\dfrac{1}{10}\right)^{15}=\left(\dfrac{1}{10}^3\right)^5=\left(\dfrac{1}{1000}\right)^5\)
\(\left(\dfrac{3}{10}\right)^{20}=\left(\dfrac{3}{10}^4\right)^5=\left(\dfrac{3}{10000}\right)^5\)
Vì \(\dfrac{1}{1000}>\dfrac{3}{10000}\) nên \(\left(\dfrac{1}{10}\right)^{15}>\left(\dfrac{3}{10}\right)^{20}\)
Bài 1 : So sánh
\(\left(\frac{1}{10}\right)^{15}\) và \(\left(\frac{3}{10}\right)^{20}\)
Bài 2 : So sánh
A = \(\left(\frac{13^{15}+1}{13^{16}+1}\right)\) và B = \(\left(\frac{13^{16}+1}{13^{17}+1}\right)\)
Bài 1:
Ta có:
\(\left(\frac{1}{10}\right)^{15}=\left(\frac{1}{5}\right)^{3.5}=\left(\frac{1}{125}\right)^5\)
\(\left(\frac{3}{10}\right)^{20}=\left(\frac{3}{10}\right)^{4.5}=\left(\frac{81}{10000}\right)^5\)
Lại có:
\(\frac{1}{125}=\frac{80}{10000}< \frac{81}{10000}\Rightarrow\left(\frac{1}{125}\right)^5< \left(\frac{81}{10000}\right)^5\)
\(\Rightarrow\left(\frac{1}{10}\right)^{15}< \left(\frac{3}{10}\right)^{20}\)
Bài 2:
Ta có:
\(A=\frac{13^{15}+1}{13^{16}+1}\Rightarrow13A=\frac{13^{16}+13}{13^{16}+1}=1+\frac{12}{13^{16}+1}\)
\(B=\frac{13^{16}+1}{13^{17}+1}\Rightarrow13B=\frac{13^{17}+13}{13^{17}+1}=1+\frac{12}{13^{17}+1}\)
Mà \(\frac{12}{13^{16}+1}>\frac{12}{13^{17}+1}\)
\(\Rightarrow1+\frac{12}{13^{16}+1}>1+\frac{12}{13^{17}+1}\)
\(\Rightarrow13A>13B\Rightarrow A>B\)
so sánh
\(\left(\frac{-1}{2}\right)^{5^{13}}\)và \(\left(\frac{-1}{3}\right)^{3^{15}}\)
1) Rút gọn
\(A=\sqrt{\frac{8+\sqrt{15}}{2}}+\sqrt{\frac{8-\sqrt{15}}{2}}\)
2) So sánh: \(A=\sqrt{4-\sqrt{15}}\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\)và \(\sqrt{3}\)
1) \(A^2=2+2.\frac{\sqrt{\left(8+\sqrt{15}\right)\left(8-\sqrt{15}\right)}}{2}\)
\(2+\sqrt{64-15}=2+\sqrt{49}=2+7=9\) mà A>0
=> A=3
2) \(A=\sqrt{4-\sqrt{15}}\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right).\)
\(A=\sqrt{\left(4-\sqrt{15}\right)\left(4+\sqrt{15}\right)}\sqrt{4+\sqrt{15}}\left(\sqrt{10}-\sqrt{6}\right).\)
\(A=\sqrt{4+\sqrt{15}}\left(\sqrt{10}-\sqrt{6}\right).\)
\(A^2=\left(4+\sqrt{15}\right)\left(16-4\sqrt{15}\right)\)
\(=4\left(4+\sqrt{15}\right)\left(4-\sqrt{15}\right)=4\)
Mà A >0
=> A=2
Mà 4>3
=> \(\sqrt{4}=2>\sqrt{3}\)
=> \(A>\sqrt{3}\)
Giúp mình với nhé:
Tính:
a)\(\left(\frac{1}{2^2-1}\right).\left(\frac{1}{3^2-1}\right).\left(\frac{1}{4^2-1}\right)...\left(\frac{1}{98^2-1}\right).\left(\frac{1}{99^2-1}\right)\)
b)\(4\frac{1}{2}-\frac{15}{10.9}-\frac{15}{9\cdot8}-\frac{15}{3.2}-\frac{15}{2.1}\)
Khó quá
\(\frac{3!\left(2^x\left(15\%\left(-8\frac{2\frac{2}{5}}{5}+8\frac{3\frac{3}{4}}{5}\right)\right)\right)}{24}=\frac{81}{x^3.2}\)
x = ?
Cho \(A=\frac{\left(3\frac{2}{15}+\frac{1}{15}\right):2\frac{1}{2}}{\left(5\frac{3}{7}-2\frac{1}{4}\right):4\frac{43}{56}}\)
\(B=\frac{1;2:\left(1\frac{1}{5}:1\frac{1}{4}\right)}{0,32+\frac{2}{25}}\)
So sánh A và B
1. So sánh hai số trong các trường hợp sau
a. 340 và 230
b. 17 x 215 và 3.218
c. 19920 và 200315
\(3^{40}\)>\(2^{30}\)
17*\(2^{15}\)>\(3.2^{18}\)
\(199^{20}\)<\(2003^{15}\)
so sánh\(\left(\frac{-1}{2}\right)^{5^{13}}\)với \(\left(\frac{-1}{3}\right)^{3^{15}}\)