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Ta có: \(\frac{2000}{-2001}=-\frac{2000}{2001}=-\left(\frac{2001-1}{2001}\right)=-\left(\frac{2001}{2001}-\frac{1}{2001}\right)=-\left(1-\frac{1}{2001}\right)=-1+\frac{1}{2001}\)
\(-\frac{2003}{2002}=-\left(\frac{2002+1}{2002}\right)=-\left(\frac{2002}{2002}+\frac{1}{2002}\right)=-\left(1+\frac{1}{2002}\right)=-1-\frac{1}{2002}\)
Vì \(\frac{1}{2001}>-\frac{1}{2002}\) nên \(-1+\frac{1}{2001}>-1-\frac{1}{2002}\)
hay \(\frac{2000}{-2001}>-\frac{2003}{2002}\)
a) \(\dfrac{2x+3}{24}=\dfrac{3x-1}{32}\)
\(\Rightarrow32\left(2x+3\right)=24\left(3x-1\right)\)
\(\Rightarrow64x+96=72x-24\)
\(\Rightarrow8x=120\Rightarrow x=15\)
b) \(\dfrac{13x-2}{2x+5}=\dfrac{76}{17}\)
\(\Rightarrow17\left(13x-2\right)=76\left(2x+5\right)\)
\(\Rightarrow221x-34=152x+380\)
\(\Rightarrow69x=414\Rightarrow x=6\)
`@` `\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}`
\(2^{333}=\left(2^3\right)^{111}=8^{111}\\ 3^{222}=\left(3^2\right)^{111}=9^{111}\)
VÌ\(8^{111}< 9^{111}\Rightarrow2^{333}< 3^{222}\)
\(2^{24}=(2^3)^8=8^8\)
\(3^{16}=\left(3^2\right)^8=9^8\)
vì \(8^8< 9^8\Rightarrow2^{24}< 3^{16}\)
\(2^{24}=\left(2^3\right)^8=8^8\)
\(3^{16}=\left(3^2\right)^8=9^8\)
\(8< 9\)
\(\Rightarrow8^9< 9^9\)
\(\Rightarrow2^{24}< 3^{16}\)
\(\left(-2\right)^{190}=2^{190}>2^{182}=\left(2^7\right)^{26}=128^{26}\)
\(128^{26}>125^{26}=\left(5^3\right)^{26}=5^{78}>5^{76}\)
\(\Rightarrow\left(-2\right)^{190}>5^{76}\)
5⁷⁶ = (5²)³⁸ = 25³⁸
(-2)¹⁹⁰ = 2¹⁹⁰ = (2⁵)³⁸ = 32³⁸
Do 26 < 32 nên 25³⁸ < 32³⁸
⇒ 5⁷⁶ < (-2)¹⁹⁰