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Ta có : \(\left\{{}\begin{matrix}2^{24}=\left(2^3\right)^8=8^8\\3^{16}=\left(3^2\right)^8=9^8\end{matrix}\right.\)
Thấy : \(9>8\)
\(\Leftrightarrow9^8>8^8\)
\(\Leftrightarrow3^{16}>2^{24}\)
Vậy ...
Siêu tốc thần sầu
\(VP=2^{30}+3^{30}+4^{30}\ge3\sqrt[3]{\left(2.3.4\right)^{3.10}}=3.24^{10}=VT\)
VP=230+330+430
VP= 230+330+430\(\ge\)\(3^3\sqrt{\left(2.3.4\right)^{3.10}}\)=\(3\cdot24^{10}\)
VP=VT
\(\Rightarrow\)230+330+430\(\ge\)\(3\cdot24^{10}\)
\(\left(-2\right)^{190}=2^{190}>2^{182}=\left(2^7\right)^{26}=128^{26}\)
\(128^{26}>125^{26}=\left(5^3\right)^{26}=5^{78}>5^{76}\)
\(\Rightarrow\left(-2\right)^{190}>5^{76}\)
5⁷⁶ = (5²)³⁸ = 25³⁸
(-2)¹⁹⁰ = 2¹⁹⁰ = (2⁵)³⁸ = 32³⁸
Do 26 < 32 nên 25³⁸ < 32³⁸
⇒ 5⁷⁶ < (-2)¹⁹⁰
\(2^{24}=(2^3)^8=8^8\)
\(3^{16}=\left(3^2\right)^8=9^8\)
vì \(8^8< 9^8\Rightarrow2^{24}< 3^{16}\)
\(2^{24}=\left(2^3\right)^8=8^8\)
\(3^{16}=\left(3^2\right)^8=9^8\)
\(8< 9\)
\(\Rightarrow8^9< 9^9\)
\(\Rightarrow2^{24}< 3^{16}\)
Áp dụng t/c dãy tỉ số bằng nhau:
a.
\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{2x}{6}=\dfrac{4y}{20}=\dfrac{2x+4y}{6+20}=\dfrac{28}{26}=\dfrac{14}{13}\)
\(\Rightarrow\left\{{}\begin{matrix}x=3.\dfrac{14}{13}=\dfrac{52}{13}\\y=5.\dfrac{14}{13}=\dfrac{70}{13}\end{matrix}\right.\)
(Em có nhầm đề 26 thành 28 ko nhỉ, số xấu quá)
b.
\(4x=5y\Rightarrow\dfrac{x}{5}=\dfrac{y}{4}=\dfrac{3x}{15}=\dfrac{-2y}{-8}=\dfrac{3x-2y}{15-8}=\dfrac{35}{7}=5\)
\(\Rightarrow\left\{{}\begin{matrix}x=5.5=25\\y=4.2=20\end{matrix}\right.\)
c.
\(\dfrac{x}{-3}=\dfrac{y}{-7}=\dfrac{2x}{-6}=\dfrac{4y}{-28}=\dfrac{2x+4y}{-6-28}=\dfrac{68}{-34}=-2\)
\(\Rightarrow\left\{{}\begin{matrix}x=-3.\left(-2\right)=6\\y=-7.\left(-2\right)=14\end{matrix}\right.\)
d.
\(\dfrac{x}{2}=\dfrac{y}{-3}=\dfrac{z}{4}=\dfrac{4x}{8}=\dfrac{-3y}{9}=\dfrac{-2z}{-8}=\dfrac{4x-3y-2z}{8+9-8}=\dfrac{16}{9}\)
\(\Rightarrow\left\{{}\begin{matrix}x=2.\dfrac{16}{9}=\dfrac{32}{9}\\y=-3.\dfrac{16}{9}=-\dfrac{48}{9}\\z=4.\dfrac{16}{9}=\dfrac{64}{9}\end{matrix}\right.\)
`@` `\text {Ans}`
`\downarrow`
Ta có:
\(5^{333}=\left(5^3\right)^{111}=125^{111}\)
\(11^{222}=\left(11^2\right)^{111}=121^{111}\)
Vì `125 > 121 =>`\(125^{111}>121^{111}\)
`=>`\(5^{333}>11^{222}\)
Vậy, \(5^{333}>11^{222}\)
_____
`@` So sánh lũy thừa cùng cơ số:
Nếu `m > n =>`\(a^m>a^n\left(m,n\ne0,a>1\right)\)
`@` So sánh lũy thừa cùng số mũ:
Nếu `a > b =>`\(a^m>b^m\left(a,b>1,m\ne0\right)\)
`@` `\text {Kaizuu lv uuu}`
ta có
2^24=(2^3)^8=8^8 (1)
5^16=(5^2)^8=25^8 (2)
từ (1);(2)suy ra 5^16>2^24
ta có: 2^24=8^8
5^16=10^8
=>8^8<10^8=>2^24<5^16