So sánh
\(5+\sqrt{27}\)và \(\sqrt{24}+\sqrt{26}\)
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\(\left(\sqrt{24}+\sqrt{26}\right)^2=50+8\sqrt{39}\)
\(10^2=100=50+50\)
mà \(8\sqrt{39}< 50\)
nên \(\sqrt{24}+\sqrt{26}< 10\)
a)
Ta có:
\(\left(\sqrt{26}+\sqrt{5}\right)^2=26+2\sqrt{26}\sqrt{5}+5\)
\(=31+2\sqrt{130}\)(1)
Mặt khác: \(\left(\sqrt{7}\right)^2=7\) (2)
Từ (1) và (2) =>\(\sqrt{26}+\sqrt{5}>\sqrt{7}\)
a) \(\sqrt{26}+\sqrt{5}< \sqrt{25}+\sqrt{4}=5+2=7\)
b) \(\sqrt{8}+\sqrt{24}< \sqrt{9}+\sqrt{25}=3+5=8\)
\(\sqrt{65}>\sqrt{64}=8\)
\(\Rightarrow\sqrt{8}+\sqrt{24}< \sqrt{65}\)
a: \(\sqrt[3]{-8}\cdot\sqrt[3]{27}=-2\cdot3=-6\)
\(\sqrt[3]{\left(-8\right)\cdot27}=\sqrt[3]{-216}=-6\)
Do đó: \(\sqrt[3]{-8}\cdot\sqrt[3]{27}=\sqrt[3]{\left(-8\right)\cdot27}\)
b: \(\dfrac{\sqrt[3]{-8}}{\sqrt[3]{27}}=-\dfrac{2}{3}\)
\(\sqrt[3]{-\dfrac{8}{27}}=-\dfrac{2}{3}\)
Do đó: \(\dfrac{\sqrt[3]{-8}}{\sqrt[3]{27}}=\sqrt[3]{-\dfrac{8}{27}}\)
\(\sqrt{27}>\sqrt{25}=5.\)
\(\sqrt{26}>\sqrt{25}=5.\)
\(\sqrt{27}+\sqrt{26}+1>5+5+1=11.\)
\(\sqrt{99}< \sqrt{100}=10\)
\(\sqrt{27}+\sqrt{26}+1>\sqrt{99}\)
ta có : \(\sqrt{27}+\sqrt{26}+1\approx11,29\)
\(\sqrt{99}\approx9,94\)
\(\Rightarrow\sqrt{27}+\sqrt{26}+1>\sqrt{99}\)
Dễ mà:vvv
Ta có: \(\left\{{}\begin{matrix}\sqrt{37}>\sqrt{36}=6\\\sqrt{26}>\sqrt{25}=5\end{matrix}\right.\)
=> \(\sqrt{37}+\sqrt{26}+1>\sqrt{36}+\sqrt{25}+1=6+5+1=12\)
Mà \(\sqrt{144}=12\)
=> \(\sqrt{37}+\sqrt{26}+1>\sqrt{144}\)
Ta có: \(\sqrt{37}>\sqrt{36}=6\)
\(\sqrt{26}>\sqrt{25}=5\)
Do đó: \(\sqrt{37}+\sqrt{26}>6+5=11\)
\(\Leftrightarrow\sqrt{37}+\sqrt{26}+1>12\)
hay \(\sqrt{144}< \sqrt{37}+\sqrt{26}+1\)
Bài 2:
\(A=\sqrt{26}+\sqrt{10}>\sqrt{25}+\sqrt{9}=5+3=8\)
\(B=\sqrt{64}=8\)
Do đó: A>B
\(\text{Ta có : }\hept{\begin{cases}5>\sqrt{24}\left(\sqrt{25}>\sqrt{24}\right)\\\sqrt{27}>\sqrt{26}\left(\text{luôn đúng}\right)\end{cases}}\)
\(\Rightarrow5+\sqrt{27}>\sqrt{24}+\sqrt{26}\)
\(\text{Vậy }\)\(5+\sqrt{27}>\sqrt{24}+\sqrt{26}\)
Vì 5=căn 25>căn 24
căn 27>căn 26
=>5+ căn 27>căn 24+ căn 26