So sánh: 37 - 14 v à 6 - 15
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a) 3200=(32)100=9100 ; 2300=(23)100=8100
=> 9100>8100 hay 3200>2300
b) 7150=(712)25=504125 ; 3775=(373)25=5065325
=> 504125<5065325 hay 7150<3775
c)rút gọn
2016014/2017015=2014/2015
2016016014/2017017015=2014/2015
=> 2014/2015 = 2014/2015
a: -15/37>-25/37
b: -13/21=-26/42
-9/14=-27/42
mà -26>-42
nên -13/21>-9/14
c: -49/-63=7/9
56/80=7/10
=>-49/-63>56/80
d: 3/14=1-11/14
4/15=1-11/15
mà 11/14>11/15
nên 3/14<4/15
Ta có \(6=\sqrt{36}\)
\(\sqrt{37}-\sqrt{14}=\sqrt{37}+\left(-\sqrt{14}\right)\)
\(6-\sqrt{15}=\sqrt{36}-\sqrt{15}=\sqrt{36}+\left(-\sqrt{15}\right)\)
Vì \(\sqrt{37}>\sqrt{36}\) và \(-\sqrt{14}>-\sqrt{15}\)
\(\Rightarrow\sqrt{37}+\left(-\sqrt{14}\right)>\sqrt{36}+\left(-\sqrt{15}\right)\)
\(\Rightarrow\sqrt{37}-\sqrt{14}>\sqrt{36}-\sqrt{15}\)
hay \(\sqrt{37}-\sqrt{14}>6-\sqrt{15}\)
Chúc bn học tốt
a) Ta có: \(\dfrac{15}{7}>1\) (tử lớn hơn mẫu)
\(\dfrac{9}{14}< 1\) (tử nhỏ hơn mẫu)
Vậy: \(\dfrac{15}{7}>\dfrac{9}{14}\)
b) Ta có:
\(\dfrac{899}{900}=1-\dfrac{1}{900}\)
\(\dfrac{1235}{1236}=1-\dfrac{1}{1236}\)
Mà: \(\dfrac{1}{900}>\dfrac{1}{1236}\)
Vậy: \(\dfrac{1235}{1236}>\dfrac{899}{900}\)
c) Ta có:
\(\dfrac{77}{75}=1+\dfrac{2}{75}\)
\(\dfrac{37}{35}=1+\dfrac{2}{35}\)
Mà: \(\dfrac{2}{75}< \dfrac{2}{35}\)
Vậy: \(\dfrac{37}{35}>\dfrac{77}{75}\)
\(\left\{{}\begin{matrix}\dfrac{15}{7}=\dfrac{30}{14}\\\dfrac{9}{14}< \dfrac{30}{14}\end{matrix}\right.\Rightarrow\dfrac{15}{7}>\dfrac{9}{14}\)
\(\left\{{}\begin{matrix}\dfrac{899}{900}=\dfrac{899.1236}{900.1236}=\dfrac{\text{1111164}}{900.1236}\\\dfrac{1235}{1236}=\dfrac{1235.900}{900.1236}=\dfrac{\text{1111500}}{900.1236}>\dfrac{\text{1111164}}{900.1236}\end{matrix}\right.\Rightarrow\dfrac{1235}{1236}>\dfrac{899}{900}\)
\(\left\{{}\begin{matrix}\dfrac{77}{75}=\dfrac{539}{525}\\\dfrac{37}{35}=\dfrac{555}{525}>\dfrac{539}{525}\end{matrix}\right.\Rightarrow\dfrac{77}{73}< \dfrac{37}{35}\)
\(\sqrt{37}>6\)
\(-\sqrt{14}>-\sqrt{15}\)
=> \(\sqrt{37}-\sqrt{14}>6-\sqrt{15}\)