![](https://rs.olm.vn/images/avt/0.png?1311)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\sqrt{35}< \sqrt{36}=6,\)
\(\sqrt{15}< \sqrt{16}=4\)
\(\Rightarrow\sqrt{35}+\sqrt{15}< 6+4=10\)
![](https://rs.olm.vn/images/avt/0.png?1311)
ta có ; \(\sqrt{35}=\sqrt{10}+\sqrt{15}+\)\(\sqrt{5}\)
mà : \(\sqrt{5}< \sqrt{10};\sqrt{10}< \sqrt{25};1< \sqrt{5}\)
\(\Rightarrow\sqrt{35}>\sqrt{5}+\sqrt{10}+1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left\{{}\begin{matrix}a=\dfrac{35}{49}=\dfrac{5}{7}\\b=\sqrt{\dfrac{5^2}{7^2}}=\dfrac{5}{7}\\c=\dfrac{\sqrt{5^2}+\sqrt{35^2}}{\sqrt{7^2}+\sqrt{49^2}}=\dfrac{5+35}{7+49}=\dfrac{5}{7}\\d=\dfrac{\sqrt{5^2}-\sqrt{35^2}}{\sqrt{7^2}-\sqrt{49^2}}=\dfrac{5-35}{7-49}=\dfrac{5}{7}\end{matrix}\right.\)
\(\Rightarrow a=b=c=d=\dfrac{5}{7}\)
\(a=\dfrac{35}{49};b=\dfrac{5}{7}\\ c,=\dfrac{5+35}{7+49}=\dfrac{12}{14}=\dfrac{6}{7}\\ d,=\dfrac{5-35}{7-49}\)
Áp dụng t/c dtsbn:
\(\dfrac{5}{7}=\dfrac{35}{49}=\dfrac{5+35}{7+49}=\dfrac{5-35}{7-49}\) hay \(a=b=c=d\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\sqrt[]{50}+\sqrt[]{65}\Rightarrow A^2=50+65+2\sqrt[]{50.65}=115+2\sqrt[]{5.10.5.13=}115+10\sqrt[]{130}\left(1\right)\)
\(B=\sqrt[]{15}+\sqrt[]{115}\Rightarrow B^2=15+115+2\sqrt[]{15.115}=15+115+2\sqrt[]{3.5.5.23}=15+115+10\sqrt[]{69}\left(2\right)\)Ta có \(10\sqrt[]{130}< 10\sqrt[]{69.2}=10\sqrt[]{2}\sqrt[]{69}< 15+10\sqrt[]{69}\left(3\right)\)
\(\left(1\right),\left(2\right),\left(3\right)\Rightarrow A^2< B^2\Rightarrow A< B\)
\(\Rightarrow\sqrt[]{50}+\sqrt[]{65}< \sqrt[]{15}+\sqrt[]{115}\)
So sánh gì thế em, em nhập đủ đề vào hi
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có : \(\sqrt{61-35}=\sqrt{26}>\sqrt{25}=5\)(1)
\(\sqrt{61}-\sqrt{35}< \sqrt{64}-\sqrt{36}=8-6=2\)(2)
Từ (1) và (2) ta được : \(\sqrt{61-35}>5>2>\sqrt{61}-\sqrt{35}\)
\(\Rightarrow\sqrt{61-35}>\sqrt{61}-\sqrt{35}\)
Ta có :
\(\sqrt{35}+\sqrt{15}< \sqrt{36}+\sqrt{16}\)
\(\Rightarrow\sqrt{35}+\sqrt{15}< 6+4=10\)
Vậy \(\sqrt{35}+\sqrt{15}< 10\)
Ta có: \(\sqrt{35}+\sqrt{15}< \sqrt{36}+\sqrt{16}=6+4=10\)
\(\Rightarrow\sqrt{35}+\sqrt{15}< 10\)