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Ta có (x-1)2>0
(y+2)2>0
=>(x-1)2+(y+2)2>0 mà theo bài ra (x-1)2+(y+2)2<0
=>(x-1)2+(y+2)2=0
=>x-1=0=>x=1;y+2=0=>y=-2
Vậy x=1;y=-2
Ta có A = 1/2+2/22+3/23+4/24+...+100/2100
<=> A = 1/2+2/4+3/9+4/16+...+100/2100
\(\frac{1}{7^2}A=\frac{1}{7^2}\left(\frac{1}{7^2}-\frac{1}{7^4}+\frac{1}{7^6}-\frac{1}{7^8}+...+\frac{1}{7^{98}}-\frac{1}{7^{100}}\right)\)
\(\Leftrightarrow\frac{1}{7^2}A=\frac{1}{7^4}-\frac{1}{7^6}+\frac{1}{7^8}-\frac{1}{7^{10}}+...+\frac{1}{7^{100}}-\frac{1}{7^{102}}\)
\(\Leftrightarrow A+\frac{1}{7^2}A=\frac{1}{49}-\frac{1}{7^{102}}\Rightarrow\frac{50}{49}A=\frac{1}{49}-\frac{1}{7^{102}}\)
\(\Rightarrow A=\left(\frac{1}{49}-\frac{1}{7^{102}}\right)\cdot\frac{49}{50}< \frac{1}{50}\left(đpcm\right)\)
\(b^2=ac\Rightarrow\frac{a}{b}=\frac{b}{c}\left(1\right)\)
\(c^2=bd\Rightarrow\frac{c}{d}=\frac{b}{c}\left(2\right)\)
Từ (1);(2) dễ dàng suy ra:
\(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}\)
\(\Rightarrow\frac{a^3}{b^3}=\frac{b^3}{c^3}=\frac{c^3}{d^3}=\frac{a\cdot b\cdot c}{b\cdot c\cdot d}\)
\(=\frac{a^3+b^3+c^3}{b^3+c^3+d^3}=\frac{a}{d}\left(đpcm\right)\)
\(1^2+2^2+3^2+...+10^2=385\)
Mà \(1^2.2=2^2\), \(2^2.2=4^2\)
\(\Rightarrow\left(1+2^2+3^2+...+10^2\right).2=S\)
\(\Rightarrow S=385.2=770\)
a)
\(2.16\ge2^n>4\)
\(\Rightarrow32\ge2^n>2^2\)
\(\Rightarrow2^5\ge2^n>2^2\)
\(\Rightarrow n\in\left\{3;4;5\right\}\)
b)
\(9.27\le3^n\le243\)
\(\Rightarrow3^2.3^3\le3^n\le3^5\)
\(\Rightarrow3^5\le3^n\le3^5\)
\(\Rightarrow n=5\)
1
a) Ta có\(\frac{31}{40}=\frac{31.6}{40.6}=\frac{186}{240}\)
Vì \(240< 241\)
nên\(\frac{286}{240}>\frac{286}{241}\)
Vậy\(\frac{31}{40}>\frac{286}{240}\)
b)Ta có\(\frac{411}{911}=\frac{911-500}{911}=1-\frac{500}{911}\)
\(\frac{41}{91}=\frac{91-50}{91}=1-\frac{50}{91}=1-\frac{500}{910}\)
Vì \(\frac{500}{911}< \frac{500}{910}\)nên\(1-\frac{500}{911}>1-\frac{500}{910}\)
Vậy \(\frac{411}{911}>\frac{41}{91}\)
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