A=√2+√6+√12+√20+√30+√42

A= 23.7579

B= 24

vậy => B > A

giúp mk với

A=√2+√6+√12+√20+√30+√42

A=23,75790715

Mà B=24

=>A<B

ban viet chu mau trang à?

A < B

\(\sqrt{2}+\sqrt{6}+\sqrt{12}+...+\sqrt{110}\)\(=\sqrt{1.2}+\sqrt{2.3}+\sqrt{3.4}+...+\sqrt{10.11}\)

\(< \frac{1+2}{2}+\frac{2+3}{2}+\frac{3+4}{2}+...+\frac{10+11}{2}\)\(=\frac{1}{2}\left[\left(1+2+3+...+10\right)+\left(2+3+4+...+11\right)\right]\)\(=\frac{1}{2}\left(\frac{11.10}{2}+\frac{13.10}{2}\right)=\frac{1}{2}\left(55+65\right)=60\)

Vậy \(\sqrt{2}+\sqrt{6}+\sqrt{12}+...+\sqrt{110}< 60.\)

chị ơi toán lớp 7 hả

\(\sqrt{2}+\sqrt{6}+\sqrt{12}+\sqrt{20}+\sqrt{30}+\sqrt{42}+\sqrt{56}+\sqrt{72}+\sqrt{90}+\sqrt{110}\) < 60 nha.

\(\sqrt{2}=\sqrt{1.2}< \dfrac{\left(1+2\right)}{2}=\dfrac{3}{2}\)

\(\sqrt{6}=\sqrt{2.3}< \dfrac{\left(2+3\right)}{2}=\dfrac{5}{2}\)

\(\sqrt{12}=\sqrt{3.4}< \dfrac{3+4}{2}=\dfrac{7}{2}\)

\(\sqrt{20}=\sqrt{4.5}< \dfrac{4+5}{2}=\dfrac{9}{2}\)

\(\sqrt{30}=\sqrt{5.6}< \dfrac{5+6}{2}=\dfrac{11}{2}\)

\(\sqrt{42}=\sqrt{6.7}< \dfrac{6+7}{2}=\dfrac{13}{2}\)

-----------------------------------

VT \(VT=A< \dfrac{3+5+7+9+11+13}{2}=\dfrac{48}{2}=24=VP=B\)

<

Ta có

\(\sqrt{2}\)=\(\sqrt{\dfrac{8}{4}}\)<\(\sqrt{\dfrac{9}{4}}\)=\(\dfrac{3}{2}\)

\(\sqrt{6}\)=\(\sqrt{\dfrac{24}{4}}\)<\(\sqrt{\dfrac{25}{4}}\)=\(\dfrac{5}{2}\)

\(\sqrt{12}\)=\(\sqrt{\dfrac{48}{4}}\)<\(\sqrt{\dfrac{49}{4}}\)=\(\dfrac{7}{2}\)

\(\sqrt{20}\)=\(\sqrt{\dfrac{80}{4}}\)<\(\sqrt{\dfrac{81}{4}}\)=\(\dfrac{9}{4}\)

\(\sqrt{30}\)=\(\sqrt{\dfrac{120}{4}}\)<\(\sqrt{\dfrac{121}{4}}\)=\(\dfrac{11}{2}\)

\(\sqrt{42}\)=\(\sqrt{\dfrac{168}{4}}\)<\(\sqrt{\dfrac{169}{4}}\)=\(\dfrac{13}{2}\)

Do đó A<\(\dfrac{3}{2}+\dfrac{5}{2}+\dfrac{7}{2}+\dfrac{9}{2}+\dfrac{11}{2}+\dfrac{13}{2}\)=24

Vậy A<24

Lời giải:

Với $a\neq b; a,b\geq 0$ ta luôn có: \(a+b>2\sqrt{ab}\Leftrightarrow 2(a+b)> (\sqrt{a}+\sqrt{b})^2\)

\(\Rightarrow \sqrt{2(a+b)}> \sqrt{a}+\sqrt{b}\).

Áp dụng BĐT trên:

\(\sqrt{2}+\sqrt{6}< \sqrt{2(2+6)}=4\)

\(\sqrt{12}+\sqrt{20}< \sqrt{2(12+20)}=8\)

\(\sqrt{30}+\sqrt{42}< \sqrt{2(30+42)}=12\)

Cộng theo vế:
\(\sqrt{2}+\sqrt{6}+\sqrt{12}+\sqrt{20}+\sqrt{30}+\sqrt{42}< 8+4+12=24\) (đpcm)

a) Có \(\sqrt{2}< \sqrt{2,25}=1,5\)

\(\sqrt{6}< \sqrt{6,25}=2,5\); 

\(\sqrt{12}< \sqrt{12,25}=3,5\); 

\(\sqrt{20}< \sqrt{20,25}=4,5\)

=> \(P=\sqrt{2}+\sqrt{6}+\sqrt{12}+\sqrt{20}< 1,5+2,5+3,5+4,5=12\)

Vậy P < 12

Answer:

ý a, tham khảo bài làm của @xyzquynhdi

\(\sqrt{2}+\sqrt{3}+\sqrt{5}\)

\(\sqrt{10+\sqrt{24}+\sqrt{40}+\sqrt{60}}\)

\(=\sqrt{10+2\sqrt{6}+2\sqrt{10}+2\sqrt{15}}\)

\(=\sqrt{\left(\sqrt{2}\right)^2+\left(\sqrt{3}\right)^2+\left(\sqrt{5}\right)^2+2\sqrt{2}\sqrt{3}+2\sqrt{2}\sqrt{5}+2\sqrt{3}\sqrt{5}}\)

\(=\sqrt{\left(\sqrt{2}+\sqrt{3}+\sqrt{5}\right)^2}=\sqrt{2}+\sqrt{3}+\sqrt{5}\)