\(\sqrt{27}+\sqrt{6}+1< \sqrt{48}\)

Bạn chỉ mình cách làm đc k

a)\(\sqrt{8}+3< \sqrt{9}+3=3+3=6< 6+\sqrt{2}\)

b)\(14=\sqrt{196}>\sqrt{195}=\sqrt{13.15}=\sqrt{13}.\sqrt{15}\)

c) Ta có: \(\hept{\begin{cases}\sqrt{27}>\sqrt{25}=5\\\sqrt{6}>\sqrt{4}=2\end{cases}\Rightarrow\sqrt{27}+\sqrt{6}+1>5+2+1=8}\)

Mà \(\sqrt{48}< \sqrt{49}=7< 8\)

\(\Rightarrow\sqrt{27}+\sqrt{6}+1>\sqrt{48}\)

Tham khảo nhé~

\(\sqrt{27+6}=\sqrt{33}\)

\(\sqrt{33}< \sqrt{48}\) 

27>25>0

→\(\sqrt{27}\)>\(\sqrt{25}\)

\(\sqrt{27}\)>5

6>4>0

\(\sqrt{6}\)>\(\sqrt{4}\) 

\(\sqrt{6}\)>2

\(\sqrt{27}\)+\(\sqrt{6}\)>2+5→\(\sqrt{27}\)+\(\sqrt{6}\)>7

0<48<49→\(\sqrt{48}\)<\(\sqrt{49}\)→\(\sqrt{48}\)<7

Từ đó suy ra \(\sqrt{27}\)+\(\sqrt{6}\)>\(\sqrt{48}\)

\(\sqrt{6+2\sqrt{5-\sqrt{13+\sqrt{48}}}}\)

\(=\sqrt{6+2\sqrt{5-\sqrt{\left(\sqrt{12}+1\right)^2}}}\)

\(=\sqrt{6+2\sqrt{5-\left(\sqrt{12}+1\right)}}\)

\(=\sqrt{6+2\sqrt{4-2\sqrt{3}}}\)

\(=\sqrt{6+2\sqrt{\left(\sqrt{3}-1\right)^2}}\)

\(=\sqrt{6+2\left(\sqrt{3}-1\right)}\)

\(=\sqrt{4+2\sqrt{3}}=\sqrt{3}+1\)

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

Tính ra rồi so sánh

a,\(\sqrt{12}=2\sqrt{3}=\sqrt{3}+\sqrt{3}\)

ta có \(\sqrt{5}>\sqrt{3}\)và\(\sqrt{7}>\sqrt{3}\)=>\(\sqrt{5}+\sqrt{7}>\sqrt{12}\)

dap an: b,<

            c,<

            d,>

cách làm thì cũng gần giống với câu a

\(\sqrt{6+2\sqrt{5-\sqrt{13+\sqrt{48}}}}=\sqrt{6+2\sqrt{5-\sqrt{13+4\sqrt{3}}}}\)

\(=\sqrt{6+2\sqrt{5-\sqrt{\left(1+\sqrt{12}\right)^2}}}=\sqrt{6+2\sqrt{5-\left|1+\sqrt{12}\right|}}=\sqrt{6+2\sqrt{5-1-\sqrt{12}}}\)

\(=\sqrt{6+2\sqrt{4-\sqrt{12}}}=\sqrt{6+2\sqrt{\left(\sqrt{3}-1\right)^2}}=\sqrt{6+2.\left|\sqrt{3}-1\right|}=\sqrt{6+2.\left(\sqrt{3}-1\right)}\)\(=\sqrt{6+2\sqrt{3}-2}=\sqrt{4+2\sqrt{3}}=\sqrt{\left(\sqrt{3}+1\right)^2}=\left|\sqrt{3}+1\right|=\sqrt{3}+1\)

Vậy: \(\sqrt{6+2\sqrt{5-\sqrt{13+\sqrt{48}}}}=\sqrt{3}+1\)

Chúc các bạn học tốt và vote cho mình nhé vì đây là lần đầu tiên mình trả lời! Cảm ơn!

$\sqrt{6+2\sqrt{5-\sqrt{13+\sqrt{48}}}}=\sqrt{6+2\sqrt{5-\sqrt{13+4\sqrt{3}}}}=\sqrt{6+2\sqrt{5-\sqrt{\left(1+\sqrt{12}\right)^2}}}=\sqrt{6+2\sqrt{5-\left|1+\sqrt{12}\right|}=\sqrt{6+2\sqrt{5-1-\sqrt{12}}}=\sqrt{6+2\sqrt{4-\sqrt{12}}}=\sqrt{6+2\sqrt{\left(\sqrt{3}-1\right)^2}}=\sqrt{6+2.\left|\sqrt{3}-1\right|}}$$\sqrt{6+2.\left(\sqrt{3}-1\right)}=\sqrt{6+2\sqrt{3}-2}=\sqrt{4+2\sqrt{3}}=\sqrt{\left(1+\sqrt{3}\right)^2}=\left|1+\sqrt{3}\right|=1+\sqrt{3}$

Vậy √6+2√5−√13+√48 = √3+1

(2√2004)2=4.2004=4008+2.2004(22004)2=4.2004=4008+2.2004

(√2003+√2005)2=2003+2√2003.2005+2005(2003+2005)2=2003+22003.2005+2005

$=4008+2\sqrt{2003.2005}$

So sánh 2004 và $\sqrt{2003.2005}$

Ta có: 

√2003.2005=√(2004−1)(2004+1)=√20042−1<√200422003.2005=(2004−1)(2004+1)=20042−1<20042

Suy ra: 

$\begin{array}{c}2004>\sqrt{2003.2005}\\ \Rightarrow 2.2004>2.\sqrt{2003.2005}\end{array}$

$\Rightarrow 4008+2.2004>4008+2\sqrt{2003.2005}$

⇒(2√2004)2>(√2003+√2005)2⇒(22004)2>(2003+2005)2

Vậy $2\sqrt{2004}>\sqrt{2003}+\sqrt{2005}$

a)A= \(\sqrt{6+2\sqrt{5-\sqrt{12}-1}}\)=\(\sqrt{6+2\sqrt{3}+2}\)

=> A²=8+2\(\sqrt{3}\)

B=\(\sqrt{3}+1\)=> B²=10+2\(\sqrt{3}\)

=>A>B