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\(3x^2y^4\)-\(5xy^3\)-\(\dfrac{3}{2}x^2y^4\)+\(3xy^3\)+\(2xy^3\)+1=1,5\(x^2y^4\)+1>0
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1. a, Ta có: \(2^{24}=2^{3^8}=8^8\)
Lại có: \(3^{16}=3^{2^8}=9^8\)
Vì \(8^8< 9^8\Rightarrow2^{24}< 3^{16}\)
b, Ta có: \(5^{300}=5^{3^{100}}=125^{100}\)
Lại có: \(3^{500}=3^{5^{100}}=243^{100}\)
Vì \(125^{100}< 243^{100}\Rightarrow5^{300}< 3^{500}\)
c, Ta có: \(2^{700}=2^{7^{100}}=128^{100}\)
Lại có: \(5^{300}=5^{3^{100}}=125^{100}\)
Vì \(128^{100}>125^{100}\Rightarrow2^{700}>5^{300}\)
d, Ta có: \(2^{400}=2^{2^{200}}=4^{200}\)
\(\Rightarrow2^{400}=4^{200}\)
e, Ta có: \(99^{20}=99^{2^{10}}=9801^{10}\)
Vì \(9801^{10}< 9999^{10}\Rightarrow99^{20}< 9999^{10}\)
Bài 1:
a) Ta có: 224 = (23)8 = 88 ; 316 = (32)8 = 98
Vì 8 < 9 nên 88 < 98
Vậy 224 < 316.
b) Ta có: 5300 = (53)100 =125100 ; 3500 = (35)100 = 243100
Vì 125 < 243 nên 125100 < 243100
Vậy 5300 < 3500.
c) Ta có: 2700 = (27)100 = 128100; 5300 = (53)100 = 125100
Vì 128 > 125 nên 128100 > 125100
Vậy 2700 > 5300.
d) (làm tương tự)
Vậy 2400 = 4200.
e) (tương tự)
Vậy 9920 < 999910.
f) Ta có: 321 = 320. 3 = 910. 3 ; 231 = 230. 3 = 810. 2
Vì 910 > 810 ; 3 > 2
Nên 910. 3 > 810. 2
Vậy 321 > 231.
Bài 2: phương trình dễ ợt :v
!!!!!!!!!
\(A=\dfrac{4^2}{1.3}+\dfrac{4^2}{3.5}+\dfrac{4^2}{5.8}+...+\dfrac{4^2}{45.47}.\dfrac{1-3-5-...-49}{8}\)
\(A=4\left(\dfrac{4}{1.3}+\dfrac{4}{3.5}+\dfrac{4}{5.8}+...+\dfrac{4}{45.47}\right).\dfrac{1-3-5-...-49}{8}\)\(A=4\left[2\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{45}-\dfrac{1}{47}\right)\right].\dfrac{1-3-5-...-49}{8}\)\(A=8\left(1-\dfrac{1}{47}\right).\dfrac{1-3-5-...-49}{8}\)
\(A=8\left(1-\dfrac{1}{47}\right).\dfrac{-623}{8}\)
\(A=\dfrac{368}{47}.\dfrac{-623}{8}=\dfrac{-28658}{47}\)
\(\left\{{}\begin{matrix}P\left(x\right)=x+x^2-x^3+2x^3+2=x^3+x^2+x+2\\Q\left(x\right)=1+3x-x^2-4x+x^3=x^3-x^2-x+1\end{matrix}\right.\)
\(\left\{{}\begin{matrix}P\left(x\right)+Q\left(x\right)=2x^3+3\\P\left(x\right)-Q\left(x\right)=2x^2+2x+1\end{matrix}\right.\)
\(\dfrac{4^5\cdot9^4-2\cdot6^9}{2^{10}\cdot3^8+6^8\cdot20}\)=\(\dfrac{\left(2^2\right)^5\cdot\left(3^2\right)^4-2\cdot6^9}{2^{10}\cdot3^8+6^8\cdot2\cdot10}=\dfrac{2^{10}\cdot3^8-2\cdot6^9}{2^{10}\cdot3^8+6^8\cdot2\cdot10}=\dfrac{6}{10}=\dfrac{3}{5}\)
Ta có: \(1^2+2^2+3^2+...+10^2=358\)
\(S=2^2+4^2+6^2+...+20^2\)
\(=\left(1.2\right)^2+\left(2.2\right)^2+\left(2.3\right)^2+...+\left(2.10\right)^2\)
\(=1^2.2^2+2^2.2^2+3^2.2^2+...+10^2.2^2\)
\(=2^2\left(1^2+2^2+3^2+...+10^3\right)\)
\(=2^2.385\)
\(=4.385=1540\)
\(S=2^2+4^2+6^2+....+20^2\)
\(S=\left(1.2\right)^2+\left(2.2\right)^4+\left(2.3\right)^2+...+\left(2.10\right)^2\)
\(S=1^2+2^2+2^2+2^2+2^2+3^2+...+2^2+10^2\)
\(S=2^2.\left(1^2+2^2+3^2+...+10^2\right)\)
\(S=2^2.385\)
\(S=4.385\)
\(\Rightarrow S=1540\)
Vậy...