So sánh: 1 3 51 v à 1 5 150
Những câu hỏi liên quan
Câu 1: Chứng minh:
\(31.82+125.48+21.43=125.67=1500\)
Câu 2: So sánh:
1,\(\sqrt{51}-\sqrt{5}v\text{à}\sqrt{20}-\sqrt{6}\)
2,\(\sqrt{2}+\sqrt{8}v\text{à}\sqrt{3}+3\)
3,\(\sqrt{37}-\sqrt{14}v\text{à}6-\sqrt{15}\)
4,\(\sqrt{5}+\sqrt{10}v\text{à}5,3\)
So sánh
\(a,2^{30}+3^{30}+4^{30}v\text{à}3^{20}+6^{20}+8^{20}\)
\(b,2^{30}+3^{30}+4^{30}v\text{à}3.24^{10}\)
\(c,2^0+2^1+2^2+...+2^{50}v\text{à}2^{51}\)
c) Đặt \(A=2^0+2^1+2^2+...+2^{50}\)
\(\Leftrightarrow2A=2^1+2^2+2^3...+2^{51}\)
\(\Leftrightarrow2A-A=2^1+2^2+2^3...+2^{51}\)\(-2^0-2^1-2^2-...-2^{50}\)
\(\Leftrightarrow A=2^{51}-2^0=2^{51}-1< 2^{51}\)
Vậy \(2^0+2^1+2^2+...+2^{50}< 2^{51}\)
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a)Ta có: \(\hept{\begin{cases}2^{30}=\left(2^3\right)^{10}=8^{10}\\3^{30}=\left(3^3\right)^{10}=27^{10}\\4^{30}=\left(2^2\right)^{30}=2^{60}\end{cases}}\)và \(\hept{\begin{cases}3^{20}=\left(3^2\right)^{10}=9^{10}\\6^{20}=\left(6^2\right)^{10}=36^{10}\\8^{20}=\left(2^3\right)^{20}=2^{60}\end{cases}}\)
Mà \(8^{10}< 9^{10}\); \(27^{10}< 36^{10}\);\(2^{60}=2^{60}\)nên
\(8^{10}+27^{10}+2^{60}< 9^{10}+36^{10}+2^{60}\)
hay \(2^{30}+3^{30}+4^{30}< 3^{20}+6^{20}+8^{20}\)
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b) Ta có: \(4^{30}=2^{30}.2^{30}=8^{10}.4^{15}\)
\(3.24^{10}=3.8^{10}.3^{10}=3^{11}.8^{10}\)
Vì \(4^{15}>3^{11}\) nên \(8^{10}.4^{15}>3^{11}.8^{10}\)
hay \(2^{30}+3^{30}+4^{30}>3.24^{10}\)
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So sánh:
\(\frac{3}{124},\frac{1}{41},\frac{2}{83},\frac{5}{207}\)
\(\frac{-2525}{2929}v\text{à}\frac{-217}{245}\)
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So sánh các số sau:
a) \(0,5\sqrt{100}-\sqrt{\frac{4}{25}}v\text{à}\left(\sqrt{1\frac{1}{9}}-\sqrt{\frac{9}{16}}\right):5\)
b) \(\sqrt{25+9}v\text{à}\sqrt{25}+\sqrt{9}\)
So sánh :
a,\(\frac{7}{23}v\text{à}\frac{11}{28}\)
b,\(\frac{2014}{2015}+\frac{2015}{2016}v\text{à}\frac{2014+2015}{2015+2016}\)
c,A=\(\frac{2^{10}+1}{2^{11}+1}v\text{à B=\frac{2^{11}+1}{2^{12}+1}}\)
a)7/23<11/28
b)2014/2015+2015/2016>2014+2015/2015+2016
c) A= gì vậy
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So sánh với n thuộc N*
\(\frac{n+1}{n+2}v\text{à}\text{ }\frac{n}{n-3}\)
Vì (n+1)/(n+2)<1;n/(n-3)>1
=>(n+1)/(n+2)<n/(n-3)
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so sánh a)\(\frac{10^{2014}-1}{10^{2015}-1}v\text{à}\frac{10^{2013}-1}{10^{2014}-1}\)
b) \(\frac{n+3}{n-2}v\text{à}\frac{n+5}{n-4}\)
So sánh với n thuộc N*
\(\frac{2003\cdot2004-1}{2003\cdot2004}v\text{à}\frac{2004\cdot2005-1}{2004\cdot2005}\)
\(\frac{3535\cdot232323}{353535\cdot2323};\frac{3535}{3534}v\text{à}\frac{2323}{2322}\)
So sánh A=\(\frac{17^{18}+1}{17^{19}+1}v\text{à}B=\frac{17^{17}+1}{17^{18}+1}\)
ta có A=\(\frac{17^{18}+1}{17^{19}+1}\)<\(\frac{17^{18}+1+16}{17^{19}+1+16}\) (nếu a/b<1 thì a+c/b+c>a/b)
A<\(\frac{17\left(17^{17}+1\right)}{17\left(17^{18}+1\right)}\)
A,<\(\frac{17^{17}+1}{17^{18}+1}\)=B
hay A<B
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\(A=\frac{17^{18}+1}{17^{19}+1}\) với \(B=\frac{17^{17}+1}{17^{18}+1}\)
Ta có :B=\(\frac{17^{17}+1}{17^{18}+1}=\frac{17^{18}+17}{17^{19}+17}\)
Ta có:1-B=\(1-\frac{17^{18}+17}{17^{19}+17}=\frac{17^{19}+17-17^{18}-17}{17^{19}+17}=\frac{17^{19}-17^{18}}{17^{19}+17}\)
1-A=1-\(\frac{17^{18}+1}{17^{19}+1}=\frac{17^{19}+1-17^{18}-1}{17^{19}+1}=\frac{17^{19}-17^{18}}{17^{19}+1}\)
Do \(17^{19}+1< 17^{19}+10\Rightarrow1-A>1-B\)
\(\Rightarrow A< B\)
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So sánh
a) 3√3 và √12
b) 7 và 3√5
c) 1/3√51 và 1/5√150
d) 1/2√6 và 6√1/2