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\(A=\frac{0,87.7+0,8^2}{\left(1,25.7-\frac{4}{5}.1,25\right)+31,64}\)
\(A=\frac{0,87\left(7+0,8\right)}{\left[1,25\left(7-\frac{4}{5}\right)\right]+31,64}\)
\(A=\frac{6,786}{7,75+31,64}\)
\(A=\frac{87}{505}\)
\(B=\frac{\left(1,09-0,29\right).\frac{5}{4}}{\left(18,9-16,65\right).\frac{8}{9}}\)
\(B=\frac{0,8.\frac{5}{4}}{2,25.\frac{8}{9}}\)
\(B=\frac{1}{2}\)
A gấp B số lần là:\(\frac{87}{505}:\frac{1}{2}=\frac{174}{505}\)
a,
\(-\frac{13}{38}=-1--\frac{25}{38}=-1+\frac{25}{38}\)
\(\frac{29}{-88}=-\frac{29}{88}=-1--\frac{59}{88}=-1+\frac{59}{88}\)
Vì \(\frac{25}{38}< \frac{59}{88}\Rightarrow-\frac{13}{38}< \frac{29}{-88}\)
b,
Ta có:
3301 > 3300 = [33]100 = 27100
5199 < 5200 = [52]100 = 25100
Mà 27100 > 25100 => 3301 > 5199
c,
\(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{\left[2n+1\right]\left[2n+3\right]}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2n+1}-\frac{1}{2n+3}\)
\(=1-\frac{1}{2n+3}< 1\)
Vậy P < 1
\(5^{199}=\left(5^{\frac{199}{301}}\right)^{301}\)
\(5^{\frac{199}{301}}< 3^1\)
\(\Leftrightarrow5^{199}< 3^{301}\)
\(2.THPT\)
\(A=\frac{9}{1.2}+\frac{9}{2.3}+\frac{9}{3.4}+...+\frac{9}{98.99}+\frac{9}{99.100}\)
\(A=9\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)\)
\(A=9\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(A=9\left(1-\frac{1}{100}\right)\)
\(A=9.\frac{99}{100}\)
\(A=\frac{891}{100}\)
\(B=\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+...+\frac{2}{93.95}\)
\(B=\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{93}-\frac{1}{95}\)
\(B=\frac{1}{5}-\frac{1}{95}\)
\(B=\frac{18}{95}\)
\(D=\frac{5}{2.7}+\frac{4}{7.11}+\frac{3}{11.14}+\frac{1}{14.15}+\frac{13}{15.28}\)
\(D=\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{15}+\frac{1}{15}-\frac{1}{28}\)
\(D=\frac{1}{2}-\frac{1}{28}\)
\(D=\frac{13}{28}\)
a ) Ta có :
\(5^{36}=\left(5^3\right)^{12}=125^{12}\)
\(11^{24}=\left(11^2\right)^{12}=121^{12}\)
Do \(125^{12}>121^{12}\Rightarrow5^{36}>11^{24}\)
b ) \(3^{2n}=\left(3^2\right)^n=9^n\)
\(2^{3n}=\left(2^3\right)^n=8^n\)
Do \(9^n>8^n\)
\(\Rightarrow3^{2n}>2^{3n}\)
Chúc bạn học tốt !!!
a) 536 = ( 53 )12 = 12512 < 1 >
1124 = ( 112 )12 = 12112 < 2 >
Từ < 1 > và < 2 > => 536 = 12512 > 12112 = 1124
=> 536 > 1124.
Vậy 536 > 1124.
b) 32n = 9n < 1 >
23n = 8n < 2 >
Từ < 1 > và < 2 > => 32n = 9n > 8n = 23n.
=> 32n > 23n.
Vậy 32n > 23n.