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a) \(M=\left\{20;21;22;23;24;25;26\right\}\)
b) \(N=\left\{1;2;3;4;5;6;7\right\}\)
c) \(P=\left\{47;48\right\}\)
_Chúc bạn học tốt_
\(a/M=\left\{20;21;23;24;25;26\right\}\)
\(b/N=\left\{1;2;3;4;5;6;7\right\}\)
\(c/P=\left\{47;48\right\}\)
\(\frac{a}{b}>1\Rightarrow a>b\)
Ta có :
\(\frac{a+m}{b+m}< \frac{a}{b}\)
<=> \(b\left(a+m\right)< a\left(b+m\right)\)
<=> \(ab+bm< ab+am\)
<=> \(bm< am\)
<=> \(b< a\) (Đúng do giả thiết cho)
Vậy ......
Ta có: \(\frac{a}{b}=\frac{a\left(b+m\right)}{b\left(b+m\right)}=\frac{ab+am}{b^2+bm}\)
\(\frac{a+m}{b+m}=\frac{b\left(a+m\right)}{b\left(b+m\right)}=\frac{ab+bm}{b^2+bm}\)
\(\Rightarrow\frac{a}{b}>1\Rightarrow a>b\)
\(\Rightarrow ab+am>ab+bm\)
\(\Rightarrow\frac{a+m}{b+m}< \frac{a}{b}\)
a) Ta có : \(0< \left|x+1\right|\le3\)
\(\Rightarrow\left|x+1\right|\in\left\{1;2;3\right\}\)
\(\Rightarrow x+1\in\left\{-1;1;-2;2;-3;3\right\}\)
\(\Rightarrow x\in\left\{-2;0;-3;1;-4;2\right\}\)
b) Ta có : \(0< \left|x\right|< 3\)
\(\Rightarrow\left|x\right|\in\left\{1;2\right\}\)
\(\Rightarrow x\in\left\{\pm1;\pm2\right\}\)
c) Ta có : \(-3\le\left|x+1\right|\le3\)
\(\Rightarrow\left|x+1\right|\in\left\{0;1;2;3\right\}\)
\(\Rightarrow x+1\in\left\{0;-1;1;-2;2;-3;3\right\}\)
\(\Rightarrow x\in\left\{-1;-2;0;-3;1;-4;2\right\}\)
a) \(500< 2^{x+1}< 1000\Leftrightarrow2^8< 500< 2^{x+1}< 1000< 2^{10}\)
\(\Rightarrow8< x+1< 10\Rightarrow7< x< 9\)
Do x là số tự nhiên nên x = 8.
b) \(\frac{1}{16}.2^x.4^{x+2}=64\)
\(\Leftrightarrow2^x.2^{2x+4}=1024\Leftrightarrow2^{3x+4}=2^{10}\)
\(\Leftrightarrow3x+4=10\Leftrightarrow x=2\)
\(A=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{99}{100}\)
\(\Rightarrow A>\frac{1}{2}.\frac{2}{3}.\frac{4}{5}...\frac{98}{99}\)
\(\Rightarrow A^2>\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}...\frac{98}{99}.\frac{99}{100}\)
\(\Rightarrow A^2>\frac{1}{100}=\frac{1}{10^2}\)
Vậy \(A>\frac{1}{10}\)
\(A=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{9999}{10000}\)
\(\Rightarrow A>\frac{1}{2}.\frac{2}{3}.\frac{4}{5}...\frac{9998}{9999}\)
\(\Rightarrow A^2>\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}...\frac{9998}{9999}.\frac{9999}{10000}\)
\(\Rightarrow A^2>\frac{1}{10000}=\frac{1}{100^2}\)
\(VayA>\frac{1}{100}=B\)
Đáp án cần chọn là: B