\(\left(\sqrt{24}+\sqrt{26}\right)^2=50+8\sqrt{39}\)

\(10^2=100=50+50\)

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) <

b) <

c) >

d) <

      a <

            b <

                           c >

                   d <

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: \(\dfrac{\sqrt[3]{-8}}{\sqrt[3]{27}}=\sqrt[3]{-\dfrac{8}{27}}\)

ko biết

Em mới học lớp 6 thôi để em thử àm xem nó ra sao:

a)<

b)<

c)<

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m ko bt đâu

\(\sqrt{27}>\sqrt{25}=5.\)

\(\sqrt{26}>\sqrt{25}=5.\)

\(\sqrt{27}+\sqrt{26}+1>5+5+1=11.\)

\(\sqrt{99}< \sqrt{100}=10\)

\(\sqrt{27}+\sqrt{26}+1>\sqrt{99}\)

ta có : \(\sqrt{27}+\sqrt{26}+1\approx11,29\)

                \(\sqrt{99}\approx9,94\)

\(\Rightarrow\sqrt{27}+\sqrt{26}+1>\sqrt{99}\)

\(\sqrt{27}>\sqrt{25}=5\)(1)

\(\sqrt{26}>\sqrt{25}=5\)(2)

\(\sqrt{99}< \sqrt{100}=10\)(3)

Từ (1) và (2) và (3) ta có \(:\)\(\sqrt{27}+\sqrt{26}>5+5=10>\sqrt{99}\)

                        \(\Rightarrow\sqrt{27}+\sqrt{26}+1>\sqrt{99}\)

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 \(\sqrt{144}< \sqrt{37}+\sqrt{26}+1\)

Ta có \(\sqrt{144}\)=12=6+5+1=\(\sqrt{36}+\sqrt{25}+\sqrt{1}\)

   Vì 0<25<26=>\(\sqrt{25}< \sqrt{26}\)(1)

    Vì 0<36<37=>\(\sqrt{36}< \sqrt{37}\)(2)

Từ (1) và (2), ta có \(\sqrt{36}+\sqrt{25}< \sqrt{37}+\sqrt{26}\)

=>\(\sqrt{36}+\sqrt{25}+\sqrt{1}< \sqrt{37}+\sqrt{26}+\sqrt{1}\)

Hay 12<\(\sqrt{37}+\sqrt{26}+1\)

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

1.Ta có:

\(A=\)\(\sqrt{13}+\sqrt{20}=\sqrt{13}+2\sqrt{5}\)

\(B=\)\(\sqrt{24}+\sqrt{19}=\sqrt{19}+2\sqrt{6}\)

So sánh ta thấy:

\(\sqrt{13}<\sqrt{19}\) ; \(2\sqrt{5}<2\sqrt{6}\)

Vậy A < B