So sánh 2 biểu thức sau:
a)A = 10^8+2/10^8-1 và B = 10^8/10^8-3
b)C=17^203+1/17^204+1 và D = 17^202+1/17^203+1
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Giải:
a) A=1718+1/1719+1
17A=1719+17/1719+1
17A=1719+1+16/1719+1
17A=1+16/1719+1
Tương tự:
B=1717+1/1718+1
17B=1718+17/1718+1
17B=1718+1+16/1718+1
17B=1+16/1718+1
Vì 16/1719+1<16/1718+1 nên 17A<17B
⇒A<B
b) A=108-2/108+2
A=108+2-4/108+2
A=1+-4/108+2
Tương tự:
B=108/108+4
B=108+4-4/108+1
B=1+-4/108+1
Vì -4/108+2>-4/108+1 nên A>B
c)A=2010+1/2010-1
A=2010-1+2/2010-1
A=1+2/2010-1
Tương tự:
B=2010-1/2010-3
B=2010-3+2/2010-3
B=1+2/2010-3
Vì 2/2010-3>2/2010-1 nên B>A
⇒A<B
Chúc bạn học tốt!
17A=1719+1+16/1719+1
17A=1+16/1719+1
phần in nghiêng mình không hiểu lắm, bn giải thích cho mình được ko?
a) Ta có: \(\frac{2}{{ - 5}} = \frac{{ - 16}}{{40}}\) và \(\frac{{ - 3}}{8} = \frac{{ - 15}}{{40}}\)
Do \(\frac{{ - 16}}{{40}} < \frac{{ - 15}}{{40}}\,\, \Rightarrow \,\frac{2}{{ - 5}} < \frac{{ - 3}}{8}\).
b) Ta có: \( - 0,85 = \frac{{ - 85}}{{100}} = \frac{{ - 17}}{{20}}\). Vậy \( - 0,85\)=\(\frac{{ - 17}}{{20}}\).
c) Ta có: \(\frac{{37}}{{ - 25}} = \frac{{ - 296}}{{200}}\)
Do \(\frac{{ - 137}}{{200}} > \frac{{ - 296}}{{200}}\) nên \(\frac{{ - 137}}{{200}}\) > \(\frac{{37}}{{ - 25}}\) .
d) Ta có: \( - 1\frac{3}{{10}}=\frac{-13}{10}\) ;
\(-\left( {\frac{{ - 13}}{{ - 10}}} \right) = \frac{{-13}}{{10}}\).
Vậy \(- 1\frac{3}{{10}} =-(\frac{{-13}}{{-10}})\,\).
a, \(\sqrt{15}+\sqrt{8}< \sqrt{16}+\sqrt{9}=4+3=7\)
\(\Rightarrow\sqrt{15}+\sqrt{8}< 7\)
b, \(\sqrt{10}+\sqrt{17}+1>\sqrt{9}+\sqrt{16}+1=3+4+1=8\)
\(\sqrt{61}< \sqrt{64}=8\)
\(\Rightarrow\sqrt{10}+\sqrt{17}+1>\sqrt{61}\)
c, \(\sqrt{10}+\sqrt{5}+1>\sqrt{9}+\sqrt{4}+1=3+2+1=6\)
\(\sqrt{35}< \sqrt{36}=6\)
\(\Rightarrow\sqrt{10}+\sqrt{5}+1>\sqrt{35}\)
Ta có : \(A=\frac{10^{17}+5}{10^{17}-8}=\frac{10^{17}-8+13}{10^{17}-8}=1+\frac{13}{10^{17}-8}\)
Lại có B = \(\frac{10^{17}-13+13}{10^{17}-13}=1+\frac{13}{10^{17}-13}\)
Nhận thấy 1017 - 8 > 1017 - 13
=> \(\frac{13}{10^{17}-8}< \frac{13}{10^{17}-13}\)
=> \(1+\frac{13}{10^{17}-8}< 1+\frac{13}{10^{17}-13}\)
=> A < B
7 = 3 + 4 = √9 + √16
Do 10 > 9 nên √10 > √9
17 > 16 nên √17 > √16
⇒ √10 + √17 > √9 + √16
Vậy √10 + √17 > 7
--------
(1/8)²³ = 1/(2³)²³ = 1/2⁶⁹
(1/32)¹⁶ = 1/(2⁵)¹⁶ = 1/2⁸⁰
Do 69 < 80 nên 2⁶⁹ < 2⁸⁰
⇒ 1/2⁶⁹ > 1/2⁸⁰
Vậy (1/8)²³ > (1/³²)¹⁶
--------
5 = √25
Do 27 > 25 nên √27 > √25
Vậy √27 > 5
\(A=\frac{10^{17}+5}{10^{17}-8}=\frac{10^{17}-8+13}{10^{17}-8}=\frac{10^{17}-8}{10^{17}-8}+\frac{13}{10^{17}-8}=1+\frac{13}{10^{17}-8}\)
\(B=\frac{10^{17}}{10^{17}-3}=\frac{10^{17}-3+13}{10^{17}-3}=\frac{10^{17}-3}{10^{17}-3}+\frac{13}{10^{17}-3}=1+\frac{13}{10^{17}-3}\)
Nhận xét: \(10^{17}-8\frac{13}{10^{17}-3}\Rightarrow1+\frac{13}{10^{17}-8}>1+\frac{13}{10^{17}-3}\Rightarrow A>B\)
\(A=\frac{10^{17}+5}{10^{17}-8}=\frac{10^{17}-8+13}{10^{17}-8}=\frac{10^{17}-8}{10^{17}-8}+\frac{13}{10^{17}-8}=2+\frac{3}{10^{17}-8}\)
\(B=\frac{10^{17}}{10^{17}-3}=\frac{10^{17}-3+3}{10^{17}-3}=\frac{10^{17}-3}{10^{17}-3}+\frac{3}{10^{17}-3}=1+\frac{3}{10^{17}-3}\)
Do \(2+\frac{3}{10^{17}-8}>1+\frac{3}{10^{17}-3}\)n\(A>B\)
a) Ta có: \(\dfrac{-5}{7}\left(\dfrac{14}{5}-\dfrac{7}{10}\right):\left|-\dfrac{2}{3}\right|-\dfrac{3}{4}\left(\dfrac{8}{9}+\dfrac{16}{3}\right)+\dfrac{10}{3}\left(\dfrac{1}{3}+\dfrac{1}{5}\right)\)
\(=\dfrac{-5}{7}\cdot\dfrac{3}{2}\cdot\dfrac{21}{10}-\dfrac{3}{4}\cdot\dfrac{56}{3}+\dfrac{10}{3}\cdot\dfrac{8}{15}\)
\(=\dfrac{-9}{4}-14+\dfrac{16}{9}\)
\(=\dfrac{-1621}{126}\)
b) Ta có: \(\dfrac{17}{-26}\cdot\left(\dfrac{1}{6}-\dfrac{5}{3}\right):\dfrac{17}{13}-\dfrac{20}{3}\left(\dfrac{2}{5}-\dfrac{1}{4}\right)+\dfrac{2}{3}\left(\dfrac{6}{5}-\dfrac{9}{2}\right)\)
\(=\dfrac{-17}{26}\cdot\dfrac{13}{17}\cdot\dfrac{-3}{2}-\dfrac{20}{3}\cdot\dfrac{3}{20}+\dfrac{2}{3}\cdot\dfrac{-33}{10}\)
\(=\dfrac{3}{4}-1-\dfrac{11}{5}\)
\(=-\dfrac{49}{20}\)
A=\(\frac{10^8+2}{10^8-1}=1+\frac{3}{10^8-1}\)
\(B=\frac{10^8}{10^8-3}=1+\frac{3}{10^8-3}\)
Vì\(10^8-1>10^8-3\)
\(\Rightarrow\frac{3}{10^8-1}< \frac{3}{10^8-3}\)
\(\Rightarrow1+\frac{3}{10^8-1}< 1+\frac{3}{10^8-3}\)
Vậy \(A< B\)
2:
a: =4+3/8+5+2/3
=9+3/8+2/3
=216/24+9/24+16/24
=216/24+25/24
=241/24
b; =2+3/8+1+1/4+3+6/7
=6+3/8+1/4+6/7
=6+5/8+6/7
=419/56
c: \(=2+\dfrac{3}{8}-1-\dfrac{1}{4}+5+\dfrac{1}{3}\)
=6+3/8-1/4+1/3
=6+1/8+1/3
=6+11/24
=155/24
d: \(=3+\dfrac{5}{6}+6\cdot\dfrac{13}{6}\)
=3+13+5/6
=16+5/6
=101/6
e: =3+1/2+4+5/7-5-5/14
=3+4-5+1/2+5/7-5/14
=2+7/14+10/14-5/14
=2+12/14
=2+6/7=20/7
f: =9/2+1/2:11/2
=9/2+1/11
=99/22+2/22=101/22
\(A=\frac{10^8+2}{10^8-1}=\frac{10^8-1+3}{10^8-1}=1+\frac{3}{10^8-1}\)
\(B=\frac{10^8}{10^8-3}=\frac{10^8-3+3}{10^8-3}=1+\frac{3}{10^8-3}\)
Nhận thầy 108 - 1 > 108 - 3
=> \(\frac{3}{10^8-1}< \frac{3}{10^8-3}\)
=> \(1+\frac{3}{10^8-1}< \frac{3}{10^8-3}+1\)
=> A < B
b) 17C = \(\frac{17\left(17^{203}+1\right)}{17^{204}+1}=\frac{17^{204}+1+16}{17^{204}+1}=1+\frac{16}{17^{204}+1}\)
17D = \(\frac{17\left(17^{202}+1\right)}{17^{203}+1}=\frac{17^{203}+1+16}{17^{203}+1}=1+\frac{16}{17^{203}+1}\)
Nhận thầy 17203 + 1 < 17204 + 1
=> \(\frac{16}{17^{203}+1}>\frac{16}{17^{204}+1}\)
=> \(\frac{16}{17^{203}+1}+1>\frac{16}{17^{204}+1}+1\Rightarrow17C>17D\Rightarrow C>D\)