Căn bậc hai - Hỏi đáp

\(\sqrt{13+\sqrt{30+\sqrt{2+\sqrt{9+4\sqrt{2}}}}}\)

\(=\sqrt{13+30\sqrt{3+2\sqrt{2}}}\)

\(=\sqrt{13+30\left(\sqrt{2}+1\right)}\)

\(=\sqrt{43+30\sqrt{2}}\)

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

Bạn tìm câu hỏi tương tự có bài giống như vậy nè : https://hoc24.vn/hoi-dap/question/411105.html

ta có : a) \(m>1\Leftrightarrow m^2>m\Leftrightarrow m^2>\left(\sqrt{m}\right)^2\Leftrightarrow m>\sqrt{m}\) (đpcm)

b) ta có \(m< 1\Leftrightarrow m^2< m\) (m là số dương ) \(\Leftrightarrow\) \(m^2< \left(\sqrt{m}\right)^2\)\(\Leftrightarrow\) \(m< \sqrt{m}\) (đpcm)

a) ta có m>1

<=> m.m>1.m

<=> m²>m

<=> m²>^{\(\left(\sqrt{m}\right)^2\)}

<=> m>\(\sqrt{m}\)

câu b tương tự

a)

m>1=>\(\sqrt{m}>\sqrt{1}\)

mà \(\sqrt{1}=1\)

=>\(\sqrt{m}>1\)

b) m<1=> \(\sqrt{m}< \sqrt{1}\)

mà \(\sqrt{1}=1\)

=>\(\sqrt{m}< 1\)

ta có : a) \(m>1\Leftrightarrow m^2>m\Leftrightarrow m^2>\left(\sqrt{m}\right)^2\Leftrightarrow m>\sqrt{m}\) rút căn m ra ta có : \(\sqrt{m}>1\) (đpcm)

b) ta có \(m< 1\Leftrightarrow m^2< m\) (m là số dương ) \(\Leftrightarrow\) \(m^2< \left(\sqrt{m}\right)^2\)\(\Leftrightarrow\) \(m< \sqrt{m}\) rút căn m ra ta có :\(\sqrt{m}< 1\)(đpcm)

\(\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+3}{x-9}\right):\left(\dfrac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)

\(=\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right):\left(\dfrac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\right)\)

\(=\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-3x-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}:\dfrac{\sqrt{x}+1}{\sqrt{x}-3}=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\dfrac{\sqrt{x}-3}{\sqrt{x}+1}=\dfrac{-3\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}+1\right)}=\dfrac{-3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}+1\right)}=-\dfrac{3}{\sqrt{x}+3}\)

ta có : \(\left(\sqrt{18}+\sqrt{19}\right)^2=\left(\sqrt{18}\right)^2+2\sqrt{18.19}+\left(\sqrt{19}\right)^2\)

\(=18+\sqrt{1368}+19=37+\sqrt{1368}\)

ta có : \(9^2=81=37+44=37+\sqrt{1936}\)

vì \(1936>1368\Rightarrow37+\sqrt{1368}< 37+\sqrt{1936}\)

\(\Rightarrow\sqrt{18}+\sqrt{19}< 9\)

\(\left(\sqrt{18}+\sqrt{19}\right)^2=37+2\sqrt{18}\sqrt{19}\)

\(9^2=81\)

Có: \(18< 22\Rightarrow\sqrt{18}< \sqrt{22}\) , \(19< 22\Rightarrow\sqrt{19}< \sqrt{22}\)

\(\Rightarrow\sqrt{18}\sqrt{19}< \sqrt{22}\sqrt{22}=22\)

\(\Leftrightarrow2\sqrt{18}\sqrt{19}< 2\cdot22=44\)

\(\Leftrightarrow37+2\sqrt{18}\sqrt{19}< 37+44=81\)

\(\Rightarrow\left(\sqrt{18}+\sqrt{19}\right)^2< 9^2\)

\(\Rightarrow\sqrt{18}+\sqrt{19}< 9\)

Ta có: \(16a^4+4=16a^4+2.4a^2.2+4-16a^2\)

\(=\left(4a+2\right)^2-16a^2\)

\(=\left(4a+2\right)^2-16a^2\)

\(=\left(4a^2-4a+2\right).\left(4a^2+4a+2\right)\)

\(=\left[\left(2a-1\right)^2+1\right].\left[\left(2a+1\right)^2+1\right]\) ( a \(\in\) N^* )

Do đó: \(16a^4+4=\left[\left(2a-1\right)^2+1\right].\left[\left(2a+1\right)^2+1\right]\) ( * )

Thay a lần lượt bằng 1, 2, 3, ..., 2014, ta có:

\(16.1^4+4=\left[\left(2.1-1\right)^2+1\right].\left[\left(2.1+1\right)^2+1\right]=\left(1^2+1\right).\left(3^2+1\right)\)

\(16.2^4+4=\left[\left(2.2-1\right)^2+1\right].\left[\left(2.2+1\right)^2+1\right]=\left(3^2+1\right).\left(5^2+1\right)\)

\(16.3^4+4=\left[\left(2.3-1\right)^2+1\right].\left[\left(2.3+1\right)^2+1\right]=\left(5^2+1\right).\left(7^2+1\right)\)

\(16.4^4+4=\left[\left(2.4-1\right)^2+1\right].\left[\left(2.4+1\right)^2+1\right]=\left(7^2+1\right).\left(9^2+1\right)\)

\(......\)

\(16.2005^4+4=\left[\left(2.2005-1\right)^2+1\right].\left[\left(2.2005+1\right)^2+1\right]=\left(4009^2+1\right).\left(4011^2+1\right)\)

\(16.2006^4+4=\left[\left(2.2006-1\right)^2+1\right].\left[\left(2.2006+1\right)^2+1\right]=\left(4011^2+1\right).\left(4013^2+1\right)\)

Đặt \(T=\dfrac{\left(1^4+\dfrac{1}{4}\right).\left(3^4+\dfrac{1}{4}\right)...\left(2005^4+\dfrac{1}{4}\right)}{\left(2^4+\dfrac{1}{4}\right).\left(4^4+\dfrac{1}{4}\right)...\left(2006^4+\dfrac{1}{4}\right)}\)

\(\Leftrightarrow T=\dfrac{16.\left(1^4+\dfrac{1}{4}\right).16\left(3^4+\dfrac{1}{4}\right)...16\left(2005^4+\dfrac{1}{4}\right)}{16.\left(2^4+\dfrac{1}{4}\right).16\left(4^4+\dfrac{1}{4}\right)...16\left(2006^4+\dfrac{1}{4}\right)}\)

\(\Leftrightarrow T=\dfrac{\left(16.1^4+4\right).\left(16.3^4+4\right)...\left(16.2005^4+4\right)}{\left(16.2^4+4\right).\left(16.4^4+4\right)...\left(16.2006^4+4\right)}\)

\(\Leftrightarrow T=\dfrac{\left(1^2+1\right).\left(3^2+1\right).\left(5^2+1\right)...\left(4009^2+1\right).\left(4011^2+1\right)}{\left(3^2+1\right).\left(5^2+1\right).\left(7^2+1\right)...\left(4011^2+1\right).\left(4013^2+1\right)}\)

\(\Leftrightarrow T=\dfrac{1^2+1}{4013^2+1}\)

\(\Leftrightarrow T=\dfrac{2}{4013^2+1}\)

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

==> \(\sqrt{x}\) \(=\sqrt{\left(\sqrt{5}-1\right)^2}=\sqrt{5}-1\)

A = \(\dfrac{4\sqrt{x}}{x+4}=\dfrac{4\left(\sqrt{5}-1\right)}{10-2\sqrt{5}}=\dfrac{4\left(\sqrt{5}-1\right)}{2\sqrt{5}\left(\sqrt{5}-1\right)}=\dfrac{2}{\sqrt{5}}=\dfrac{2\sqrt{5}}{5}\)

Giải

Áp dụng BĐT AM-MG :

\(\left\{{}\begin{matrix}a^2+b^2\ge2ab\\b^2+c^2\ge2bc\\c^2+a^2\ge2ac\end{matrix}\right.\Rightarrow a^2+b^2+c^2\ge ab+bc+ac\) \(\Leftrightarrow\left(a+b+c\right)^2\ge3.\left(ab+bc+ac\right)\Leftrightarrow ab+bc+ac\le\dfrac{1}{3}\)

Do đó \(\dfrac{1}{ab+bc+ac}\ge\dfrac{1}{\dfrac{1}{3}}=3\)

Vậy P_min = 3 \(\Leftrightarrow a=b=c\dfrac{1}{3}\)

Lời giải:

Áp dụng BĐT AM-GM:

\(\left\{\begin{matrix} a^2+b^2\geq 2ab\\ b^2+c^2\geq 2bc\\ c^2+a^2\geq 2ac\end{matrix}\right.\Rightarrow a^2+b^2+c^2\geq ab+bc+ac\)

\(\Leftrightarrow (a+b+c)^2\geq3(ab+bc+ac)\Leftrightarrow ab+bc+ac\leq \frac{1}{3}\)

(do \(a+b+c=1\) )

Do đó, \(\frac{1}{ab+bc+ac}\geq \frac{1}{\frac{1}{3}}=3\)

Vậy \(P_{\min}=3\Leftrightarrow a=b=c=\frac{1}{3}\)

Áp dụng BĐT \(\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{4}{x+y}\) ta có:

\(\dfrac{ab}{c+1}=\dfrac{ab}{\left(c+a\right)+\left(b+c\right)}\le\dfrac{1}{4}\left(\dfrac{ab}{a+c}+\dfrac{ab}{b+c}\right)\)

Tương tự cho 2 BĐT còn lại

\(\dfrac{bc}{a+1}\le\dfrac{1}{4}\left(\dfrac{bc}{a+b}+\dfrac{bc}{a+c}\right);\dfrac{ac}{b+1}\le\dfrac{1}{4}\left(\dfrac{ac}{a+b}+\dfrac{ac}{b+c}\right)\)

Cộng theo vế 3 BĐT trên ta có:

\(P\le\dfrac{1}{4}\left(\dfrac{ab}{a+c}+\dfrac{bc}{a+c}+\dfrac{bc}{a+b}+\dfrac{ac}{a+b}+\dfrac{ab}{b+c}+\dfrac{ac}{b+c}\right)\)

\(=\dfrac{1}{4}\left(\dfrac{ab+bc}{a+c}+\dfrac{bc+ac}{a+b}+\dfrac{ab+ac}{b+c}\right)\)

\(=\dfrac{1}{4}\left(\dfrac{b\left(a+c\right)}{a+c}+\dfrac{c\left(a+b\right)}{a+b}+\dfrac{a\left(b+c\right)}{b+c}\right)\)

\(=\dfrac{1}{4}\left(a+b+c\right)=\dfrac{1}{4}\cdot1=\dfrac{1}{4}\left(a+b+c=1\right)\)

Đẳng thức xảy ra khi \(a=b=c=\dfrac{1}{3}\)