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Áp dụng BĐT Bunhiacôpxki:
\(123^2=\left(m\sqrt{123-n^2}+n\sqrt{123-m^2}\right)^2\)
\(\Rightarrow123^2\le\left(m^2+n^2\right)\left(123-n^2+123-m^2\right)\)
\(\Leftrightarrow123^2\le\left(m^2+n^2\right)\left(2.123-m^2-n^2\right)\)
Đặt \(m^2+n^2=x\)
\(\Rightarrow123^2\le x\left(2.123-x\right)\)
\(\Leftrightarrow x^2-2.x.123+123^2\le0\)
\(\Leftrightarrow\left(x-123\right)^2\le0\)
\(\Leftrightarrow x-123=0\Rightarrow x=123\)
1) \(M=\dfrac{10}{\sqrt{x}+2};M_{\left(16\right)}=\dfrac{10}{\sqrt{16}+2}=\dfrac{10}{6}=\dfrac{5}{3}\)
2)\(N=\dfrac{2\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}-18}{x-4}=2+\dfrac{4}{\sqrt{x}-2}+\dfrac{\sqrt{x}-18}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=2+\dfrac{4\sqrt{x}+8+\sqrt{x}-18}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)\(N=2+\dfrac{5}{\sqrt{x}+2}=\dfrac{2\sqrt{x}+9}{\sqrt{x}+2}\)
N khác 0 mọi x thuộc đk
\(\dfrac{M}{N}=M.\dfrac{1}{N}=\dfrac{10}{\sqrt{x}+2}.\dfrac{\sqrt{x}+2}{\left(2\sqrt{x}+9\right)}=\dfrac{10}{2\sqrt{x}+9}\)
\(\dfrac{M}{N}=\dfrac{12-\sqrt{x}}{13}=\dfrac{10}{2\sqrt{x}+9}\)
\(\Leftrightarrow\left(12-\sqrt{x}\right)\left(2\sqrt{x}+9\right)=130\)
\(15\sqrt{x}+12.9-2x=130\)
\(2x-15\sqrt{x}+22=0\)
\(\Delta_{\sqrt{x}}=15^2-4.2.22=137\)
\(\sqrt{x}=\dfrac{15+-\sqrt{137}}{4}\)
\(\left[{}\begin{matrix}x_1=\dfrac{181-15.\sqrt{137}}{8}\\x_2=\dfrac{181+15.\sqrt{137}}{8}\end{matrix}\right.\) tự kiểm tra số liểu (nhẩm tính có thể nhầm; thấy lẻ quá)
a)\(2\sqrt{27}=\sqrt{4\cdot27}=\sqrt{108}< \sqrt{147}\)
b)\(-3\sqrt{5}=-\sqrt{9\cdot5}=-\sqrt{45}>-\sqrt{75}=-\sqrt{25\cdot3}=-5\sqrt{3}\)
c) ta có
\(21=\sqrt{21\cdot21}=\sqrt{441}\\ 2\sqrt{7}=\sqrt{28}\\ 15\sqrt{3}=\sqrt{\left(15\cdot15\right)\cdot3}=\sqrt{675}\\ -\sqrt{123}\)
=> thứ tự lần lượt là:
\(-\sqrt{123};2\sqrt{7};21;15\sqrt{3}\)
d)\(2\sqrt{15}=\sqrt{60}>\sqrt{59}\)
e)\(2\sqrt{2}=\sqrt{8}-1< \sqrt{9}-1=3-1=2\)
f)\(6=\sqrt{36}< \sqrt{41}\)
Ta có
\(\sqrt{123-22\sqrt{2}}=11-\sqrt{2}\)
\(\sqrt[3]{77\sqrt{2}-115}=\sqrt{2}-5\)
\(\Rightarrow\sqrt{123-22\sqrt{2}}+\sqrt[3]{77\sqrt{2}-115}=11-\sqrt{2}+\sqrt{2}-5=6\)
\(\sqrt{227-30\sqrt{2}}+\sqrt{123+22\sqrt{2}}\)
=\(\sqrt{225+2.15.\sqrt{2}+2}+\sqrt{121+2.11\sqrt{2}+2}\)
=\(\sqrt{\left(15+\sqrt{2}\right)^2}+\sqrt{\left(11+\sqrt{2}\right)^2}\)
=\(15+\sqrt{2}+11+\sqrt{2}\)
=\(26+2\sqrt{2}\)
\(\left\{{}\begin{matrix}\sqrt{a}=x\ge0\\\sqrt{b}=y\ge0\end{matrix}\right.\) \(\Rightarrow x+y=1\)
\(P=x^3+y^3=\left(x+y\right)\left(x^2+y^2-xy\right)=x^2+y^2-xy\)
\(P=\left(x+y\right)^2-3xy=1-3xy\)
Do \(\left\{{}\begin{matrix}x\ge0\\y\ge0\end{matrix}\right.\) \(\Rightarrow xy\ge0\Rightarrow P\le1\Rightarrow P_{max}=1\) khi \(\left(x;y\right)=\left(1;0\right);\left(0;1\right)\) hay \(\left(a;b\right)=\left(1;0\right);\left(0;1\right)\)
Lại có \(xy\le\frac{\left(x+y\right)^2}{4}=\frac{1}{4}\Rightarrow P\ge1-3.\frac{1}{4}=\frac{1}{4}\)
\(\Rightarrow P_{min}=\frac{1}{4}\) khi \(x=y=\frac{1}{2}\) hay \(a=b=\frac{1}{4}\)
\(\Rightarrow m^2+n^2=1+\frac{1}{16}=\frac{17}{16}\)
Lời giải:
Áp dụng BĐT Bunhiacopxky và Cauchy ngược dấu ta có:
\((m\sqrt{123-n^2}+n\sqrt{123-m^2})^2\leq (m^2+n^2)(123-n^2+123-m^2)\leq \left(\frac{m^2+n^2+123-n^2+123-m^2}{2}\right)^2\)
\(\Leftrightarrow (m\sqrt{123-n^2}+n\sqrt{123-m^2})^2\leq 123^2\)
\(\Rightarrow m\sqrt{123-n^2}+n\sqrt{123-m^2}\leq 123\)
Dấu "=" xảy ra khi \(\left\{\begin{matrix} \frac{m}{\sqrt{123-n^2}}=\frac{n}{\sqrt{123-m^2}}\\ m^2+n^2=123-n^2+123-m^2(1)\end{matrix}\right.\)
Từ (1) \(\Rightarrow m^2+n^2=123\)