So sánh 9/15 va 6/10
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![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(4\frac{7}{10}< 6\frac{7}{10}\)(4 < 6)
b) \(3\frac{4}{15}< 3\frac{11}{15}\)(4/15 < 11/15)
c) \(5\frac{1}{9}>2\frac{2}{5}\)(5 > 2)
d) \(2\frac{2}{3}=2\frac{10}{15}\)(10/15 = 2/3)
![](https://rs.olm.vn/images/avt/0.png?1311)
1520=(3.5)20 = 320. 520
910.521= (32)10 .521 =320. 521
Ta thấy 320 .520 < 320. 521
=> 1520 < 910. 520
Chúc học tốt nha
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Ta có:
\(\begin{array}{l}\frac{6}{{10}} = \frac{{6:2}}{{10:2}} = \frac{3}{5};\\\frac{9}{{15}} = \frac{{9:3}}{{15:3}} = \frac{3}{5}\end{array}\)
\(\begin{array}{l}\frac{{6 + 9}}{{10 + 15}} = \frac{{15}}{{25}} = \frac{{15:5}}{{25:5}} = \frac{3}{5};\\\frac{{6 - 9}}{{10 - 15}} = \frac{{ - 3}}{{ - 5}} = \frac{3}{5}\end{array}\)
Ta được: \(\frac{{6 + 9}}{{10 + 15}} = \frac{{6 - 9}}{{10 - 15}} = \frac{6}{{10}} = \frac{9}{{15}}\)
b) - Vì \(k = \frac{a}{b} \Rightarrow a = k.b\)
Vì \(k = \frac{c}{d} \Rightarrow c = k.d\)
- Ta có:
\(\begin{array}{l}\frac{{a + c}}{{b + d}} = \frac{{k.b + k.d}}{{b + d}} = \frac{{k.(b + d)}}{{b + d}} = k;\\\frac{{a - c}}{{b - d}} = \frac{{k.b - k.d}}{{b - d}} = \frac{{k.(b - d)}}{{b - d}} = k\end{array}\)
- Như vậy, \(\frac{{a + c}}{{b + d}}\) =\(\frac{{a - c}}{{b - d}}\) = \(\frac{a}{b}\) =\(\frac{c}{d}\)( = k)
a: \(\dfrac{6+9}{10+15}=\dfrac{15}{25}=\dfrac{3}{5};\dfrac{6-9}{10-15}=\dfrac{-3}{-5}=\dfrac{3}{5}\)
=>Bằng nhau
b: a/b=c/d=k
=>a=bk; c=dk
\(\dfrac{a+c}{b+d}=\dfrac{bk+dk}{b+d}=k;\dfrac{a-c}{b-d}=\dfrac{bk-dk}{b-d}=k\)
=>\(\dfrac{a+c}{b+d}=\dfrac{a-c}{b-d}=\dfrac{a}{b}=\dfrac{c}{d}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: 200410+20049=20049(2004+1)=20049.2005
Mà 20049<20059 => 20049.2005<20059.2005 hay 200410+20049<200510
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có \(A=\frac{10}{2^7}+\frac{10}{2^6}=\frac{5}{2^6}+\frac{10}{2^6}=\frac{15}{2^6}\)
Lại có B = \(\frac{11}{2^7}+\frac{9}{2^6}=\frac{5,5}{2^6}+\frac{9}{2^6}=\frac{14,5}{2^6}\)
Vì \(\frac{15}{2^6}>\frac{14,5}{2^6}\Rightarrow A>B\)
b) Ta có : \(A=\frac{-7}{10^{2005}}+\frac{-15}{10^{2006}}=\frac{-70}{10^{2006}}+\frac{-15}{10^{2006}}=\frac{-85}{10^{2006}}\)
Lại có B = \(\frac{-15}{10^{2005}}+\frac{-7}{10^{2006}}=\frac{-150}{10^{2006}}+\frac{-7}{10^{2006}}=\frac{-157}{10^{2006}}\)
Vì \(\frac{-85}{10^{2006}}>\frac{-157}{10^{2006}}\Rightarrow A< B\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) (-4). (-8) > 0
b) (-23). 6 < (-4). (-15)
c) (+12) .(+7) < (-9). (-10)
9/15 = 6/10
9/15=3/5
6/10=3/5
mà 3/5=3/5
=>9/15=6/10