Các bài toán Dạng SO SÁNH SỐ HỮU TỈ:
1, \(3^{800}\)và \(5^{500}\)
2, \(\left(-2\right)^{3000}\)và \(\left(-3\right)^{2000}\)
3, \(\left(\frac{1}{2^5}\right)^{25}\)và \(\left(\frac{1}{3^{25}}\right)^5\)
4, \(6^{1000}\)và \(2^{2500}\)
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Ta có
\(A=\frac{\left(3\frac{2}{5}+\frac{1}{5}\right):2\frac{1}{2}}{\left(5\frac{3}{7}-2\frac{1}{4}\right):4\frac{43}{56}}\) \(B=\frac{1,2:\left(1\frac{1}{5}-1\frac{1}{4}\right)}{0,32+\frac{2}{25}}\)
\(\Leftrightarrow A=\frac{\left(\frac{17}{5}+\frac{1}{5}\right):\frac{5}{2}}{\left(\frac{38}{7}-\frac{9}{4}\right):\frac{276}{56}}\) \(\Leftrightarrow B=\frac{\frac{6}{5}:\left(\frac{6}{5}-\frac{5}{4}\right)}{\frac{8}{25}+\frac{2}{25}}\)
\(\Leftrightarrow A=\frac{\frac{18}{5}:\frac{5}{2}}{\frac{89}{28}:\frac{276}{56}}\) \(\Leftrightarrow B=\frac{\frac{6}{5}:\left(-\frac{1}{20}\right)}{\frac{2}{5}}\)
\(\Leftrightarrow A=\frac{\frac{36}{25}}{\frac{89}{138}}\) \(\Leftrightarrow B=\frac{\frac{5}{4}}{\frac{2}{5}}\)
\(\Leftrightarrow A=\frac{4968}{2225}\) \(\Leftrightarrow B=\frac{25}{8}\)
\(\Leftrightarrow A=\frac{39744}{17800}\) \(\Leftrightarrow B=\frac{55625}{17800}\)
Ta có: 39744<55625
\(\Rightarrow A< B\)
Vậy A<B
A =\(\frac{\left(\frac{17}{5}+\frac{1}{5}\right).\frac{2}{5}}{\left(\frac{38}{7}-\frac{9}{4}\right).\frac{56}{267}}\)
A=\(\frac{36}{25}\).\(\frac{3}{2}\)=\(\frac{54}{25}\)=2,16
B=\(\frac{1,2:\left(\frac{6}{5}-\frac{5}{4}\right)}{0,32+\frac{2}{25}}\)=-24.\(\frac{5}{2}\)=-60
vì 2,16 > -60 Vậy A>B
Câu 1 :
a) \(4.\left(\frac{1}{32}\right)^{-2}:\left(2^3.\frac{1}{16}\right)\)
\(=2^2.32^2:\left(\frac{1}{8}.16\right)=\left(2.32\right)^2:2=64^2:2\)
\(=2048=2^{11}\)
b) \(5^2.3^5.\left(\frac{3}{5}\right)^2\)
\(=\left(5.\frac{3}{5}\right)^2.3^5=3^2.3^5=3^7\)
VIẾT CÁC BIỂU THỨC DƯỚI DẠNG LUỸ THỪA CỦA 1 SỐ HỮU TỈ
\(a,4\cdot\left(\frac{1}{32}\right)^{-2}:\left(2^3\cdot\frac{1}{16}\right)\\ =4\cdot1024:\left(8\cdot\frac{1}{16}\right)\\ =4\cdot1024:\frac{1}{2}\\ =2\cdot1024\\ =2\cdot2^{10}\\ =2^{11}\)
\(b,5^2\cdot3^5\cdot\left(\frac{3}{5}\right)^2\\ =5^2\cdot\left(\frac{3}{5}\right)^2\cdot3^5\\ =3^2\cdot3^5\\ =3^7\)
2 SO SÁNH
\(a,10^{20}\text{ và }9^{10}\)
Có: \(9^{10}=\left(3^2\right)^{10}=3^{20}\)
\(\Rightarrow10^{20}>3^{20}\\ \text{hay}\text{ }10^{20}>9^{10}\)
\(b,\left(-5\right)^3\text{ và }\left(-3\right)^{50}\)
Có: \(\left(-3\right)^{50}=3^{50}\)
\(\Rightarrow\left(-5\right)^3< 3^{50}\\ \text{hay }\left(-5\right)^3< \left(-3\right)^{50}\)
\(c,64^3\text{ và }16^{12}\)
Có: \(64^3=\left(4^3\right)^3=4^9;16^{12}=\left(4^2\right)^{12}=4^{24}\)
\(\Rightarrow4^9< 4^{24}\\ hay\text{ }64^3< 16^{12}\)
\(d,\left(\frac{1}{16}\right)^{10}\text{ và }\left(\frac{1}{2}\right)^{50}\)
Có: \(\left(\frac{1}{2}\right)^{50}=\left(\frac{1}{2}\right)^{5\cdot10}=\left[\left(\frac{1}{2}\right)^5\right]^{10}=\left(\frac{1}{32}\right)^{10}\)
\(\Rightarrow\left(\frac{1}{16}\right)^{10}>\left(\frac{1}{32}\right)^{10}\\ \text{hay }\left(\frac{1}{16}\right)^{10}>\left(\frac{1}{2}\right)^{50}\)
B1:Ta có:
2800=(28)100=256100(1)
3600=(36)100=729100(2)
Từ (1),(2)
Ta có: 256100<729100(Vì 256<729)
=>2800<3600
Mình làm bài 1 thôi nhé!
Ta có :2^800=2^2.400=(2^2)^400=4^400=4^4.100=(4^4)^100=256^100
3^600=3^2.300=(3^2)^300=9^300=9^3.100=(9^3)^100=729^100
Vì 256^100<729^100 nên 2^800<3^600
^ là mũ nhé!
3. \(\left(\frac{1}{2^5}\right)^{25}=\left(\frac{1^5}{2^5}\right)^{25}=\left[\left(\frac{1}{2}\right)^5\right]^{25}=\left(\frac{1}{2}\right)^{125}\)
\(\left(\frac{1}{3^{25}}\right)^5=\left(\frac{1^{25}}{3^{25}}\right)^5=\left[\left(\frac{1}{3}\right)^{25}\right]^5=\left(\frac{1}{3}\right)^{125}\)
Vì \(\frac{1}{2}>\frac{1}{3}\Rightarrow\left(\frac{1}{2^5}\right)^{25}>\left(\frac{1}{3^{25}}\right)^5\)
1. \(3^{800}=\left(3^8\right)^{100}=6561^{100}\)
\(5^{500}=\left(5^5\right)^{100}=3125^{100}\)
Vì \(6561>3125\Rightarrow3^{800}>5^{500}\)
2. \(\left(-2\right)^{3000}=\left[\left(-2\right)^3\right]^{1000}=\left(-8\right)^{1000}\)
\(\left(-3\right)^{2000}=\left[\left(-3\right)^2\right]^{1000}=9^{1000}\)
Vì \(-8< 9\Rightarrow\left(-2\right)^{3000}< \left(-3\right)^{2000}\)