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* Cách 1 :
Ta có :
\(16A=\frac{4^{17}+16}{4^{17}+1}=\frac{4^{17}+1+15}{4^{17}+1}=\frac{4^{17}+1}{4^{17}+1}+\frac{15}{4^{17}+1}=1+\frac{15}{4^{17}+1}\)
\(16B=\frac{4^{14}+16}{4^{14}+1}=\frac{4^{14}+1+15}{4^{14}+1}=\frac{4^{14}+1}{4^{14}+1}+\frac{15}{4^{14}+1}=1+\frac{15}{4^{14}+1}\)
Vì \(\frac{15}{4^{17}+1}< \frac{15}{4^{14}+1}\) nên \(1+\frac{15}{4^{17}+1}< 1+\frac{15}{4^{14}+1}\)
\(\Rightarrow\)\(16A< 16B\) hay \(A< B\)
Vậy \(A< B\)
Chúc bạn học tốt ~
\(4^2.A=\frac{4^2\left(4^{15}+1\right)}{4^{17}+1}\); \(4^2.B=\frac{4^2\left(4^{12}+1\right)}{4^{14}+1}\)
=> \(4^2.A=\frac{4^{17}+4^2}{4^{17}+1}\);\(4^2.B=\frac{4^{14}+4^2}{4^{14}+1}\)
=> \(4^2.A=\frac{4^{17}+1+4^2-1}{4^{17}+1}\); \(4^2.B=\frac{4^{14}+1+4^2-1}{4^{14}+1}\)
=> \(4^2.A=\frac{4^{17}+1}{4^{17}+1}+\frac{4^2-1}{4^{17}+1}\); \(4^2.B=\frac{4^{14}+1}{4^{14}+1}+\frac{4^2-1}{4^{14}+1}\)
=> \(4^2.A=1+\frac{4^2-1}{4^{17}+1}\); \(4^2.B=1+\frac{4^2-1}{4^{14}+1}\)
Mà \(4^{17}>4^{14}\)
=> \(4^{17}+1>4^{14}+1\)
=> \(\frac{4^2-1}{4^{17}+1}< \frac{4^2-1}{4^{14}+1}\)
=> \(1+\frac{4^2-1}{4^{17}+1}< 1+\frac{4^2-1}{4^{14}+1}\)
=> \(4^2.A< 4^2.B\)
=> \(A< B\)
a) \(\dfrac{-1}{-4}\)=\(\dfrac{1}{4}>0\)
\(\dfrac{3}{-4}< 0\)
\(\Rightarrow\dfrac{1}{4}>\dfrac{3}{-4}hay\dfrac{-1}{-4}>\dfrac{3}{-4}\)
b) Ta có:
\(\dfrac{15}{17}=1-\dfrac{2}{17}\\ \)
\(\dfrac{25}{27}=1-\dfrac{2}{27}\\ \\ \)
Mà \(\dfrac{2}{17}>\dfrac{2}{27}\left(17< 27\right)\)
\(\Rightarrow1-\dfrac{2}{17}< 1-\dfrac{2}{27}\)hay \(\dfrac{15}{17}< \dfrac{25}{27}\)
Câu1:
a: \(=2008^2-\left(2008-2\right)\left(2008+2\right)\)
\(=2008^2-\left(2008^2-4\right)\)
=4
b: \(=\dfrac{23\cdot29\cdot10101}{23\cdot29\cdot10101}=1\)
c: \(=\dfrac{\left(2^{17}+5^{17}\right)\left(3^{14}-5^{12}\right)\cdot\left(16-16\right)}{15^2+5^3+67^7}\)
=0
a) \(\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{6}\right)\left(1-\dfrac{1}{10}\right)...\left(1-\dfrac{1}{780}\right)\)
\(=\dfrac{2}{3}.\dfrac{5}{6}.\dfrac{9}{10}.....\dfrac{779}{780}\)\(=\)
c) E = \(\dfrac{4116-14}{10290-35}\) và K = \(\dfrac{2929-101}{2.1919+404}\)
E = \(\dfrac{4116-14}{10290-35}\)
E = \(\dfrac{14.\left(294-1\right)}{35.\left(294-1\right)}\)
E = \(\dfrac{14}{35}\)
K = \(\dfrac{2929-101}{2.1919+404}\)
K = \(\dfrac{101.\left(29-1\right)}{101.\left(38+4\right)}\)
K = \(\dfrac{29-1}{34+8}\)
K = \(\dfrac{28}{42}\) = \(\dfrac{2}{3}\)
Ta có : E = \(\dfrac{14}{35}\) và K = \(\dfrac{2}{3}\)
\(\dfrac{14}{35}\) = \(\dfrac{42}{105}\)
\(\dfrac{2}{3}\) = \(\dfrac{70}{105}\)
Vậy E < K
Các câu còn lại tương tự
A = \(\dfrac{17}{15}.\dfrac{-31}{125}.\dfrac{1}{2}.\dfrac{10}{17}.\dfrac{-1}{8}\)
= \(\dfrac{17.\left(-31\right).1.10.\left(-1\right)}{15.125.2.17.8}\)
= \(\dfrac{17.\left[\left(-31\right).\left(-1\right)\right].1.2.5}{5.3.125.17.4.2}\)
= \(\dfrac{31.1}{3.125.4}\)
= \(\dfrac{31}{1500}\)
B = \(\left(\dfrac{11}{4}.\dfrac{-5}{9}-\dfrac{4}{9}.\dfrac{11}{4}\right).\dfrac{8}{33}\)
= \(\left[\dfrac{11}{4}.\left(\dfrac{-5}{9}-\dfrac{4}{9}\right)\right].\dfrac{8}{33}\)
= \(\left(\dfrac{11}{4}.\dfrac{-9}{9}\right).\dfrac{8}{33}\)
= \(\left[\dfrac{11}{4}.\left(-1\right)\right].\dfrac{4.2}{\left(-11\right).\left(-3\right)}\)
= \(\dfrac{-11}{4}.\dfrac{4.2}{\left(-11\right).\left(-3\right)}\)
= \(\dfrac{\left(-11\right).4.2}{4.\left(-11\right)\left(-3\right)}\)
= \(\dfrac{2}{-3}\)
Ok nhá!
Dấu " / " là phân số nhé
a) 5/-4 . 16/25 + -5/4 . 9/25
= -5/4 . 16/25 + -5/4 . 9/25
= -5/4 . ( 16/25 + 9/25 )
= -5/4 . 1
= -5/4
b) 4 11/23 - 9/14 + 2 12/23 - 5/4
= 103/23 - 9/14 + 58/23 - 5/4
= 103/23 + 58/23 - 9/14 - 5/4
= 7 - 9/14 - 5/4
= 143/28
c) 2 13/27 - 7/15 + 3 14/27 - 8/15
= 67/27 - 7/15 + 95/27 - 8/15
= 67/27 + 95/27 - 7/15 - 8/15
= 6 - 7/15 - 8/15
= 5
Ta có :
\(\dfrac{1}{11}>\dfrac{1}{20}\\ \dfrac{1}{12}>\dfrac{1}{20}\\ ..........\\ \dfrac{1}{20}=\dfrac{1}{20}\)
\(\Rightarrow\dfrac{1}{11}+\dfrac{1}{12}+\dfrac{1}{13}+...+\dfrac{1}{20}>\dfrac{1}{20}+\dfrac{1}{20}+...+\dfrac{1}{20}\\ \Rightarrow S>\dfrac{10}{20}\\ \Rightarrow S>\dfrac{1}{2}\)
4 mũ 15+1/4 mũ 17 +1= 1/16+1
4 mũ 12+1/ 4 mũ 14+1= 1/16+1
suy ra 1/17=1/17
suy ra A=B
nhớ tích cho tớ nhé