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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 \(m^2+n^2=x\)
\(\Rightarrow123^2\le x\left(2.123-x\right)\)
\(\Leftrightarrow x^2-2.x.123+123^2\le0\)
\(\Leftrightarrow\left(x-123\right)^2\le0\)
\(\Leftrightarrow x-123=0\Rightarrow x=123\)
\(\sqrt{227-30\sqrt{2}}+\sqrt{123+22\sqrt{2}}\)
=\(\sqrt{225+2.15.\sqrt{2}+2}+\sqrt{121+2.11\sqrt{2}+2}\)
=\(\sqrt{\left(15+\sqrt{2}\right)^2}+\sqrt{\left(11+\sqrt{2}\right)^2}\)
=\(15+\sqrt{2}+11+\sqrt{2}\)
=\(26+2\sqrt{2}\)
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)
\(\sqrt{227-30\sqrt{2}}+\sqrt{123+22\sqrt{2}}=\sqrt{225-30\sqrt{2}+2}+\sqrt{121+22\sqrt{2}+2}=\sqrt{15^2-15.2.\sqrt{2}+\left(\sqrt{2}\right)^2}+\sqrt{11^2+11.2\sqrt{2}+\left(\sqrt{2}\right)^2}=\sqrt{\left(15-\sqrt{2}\right)^2}+\sqrt{\left(11+\sqrt{2}\right)^2}=15-\sqrt{2}+11+\sqrt{2}\left(do:15-\sqrt{2}>0;11+\sqrt{2}>0\right)=26\)
a) \(ĐKXĐ:\hept{\begin{cases}x\ge0\\x\ne9\end{cases}}\)
\(M=\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}+1}{\sqrt{x}-3}+\frac{3-11\sqrt{x}}{9-x}\)
\(=\frac{2\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\frac{11\sqrt{x}-3}{x-9}\)
\(=\frac{2x-6\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\frac{x+4\sqrt{x}+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\frac{11\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{2x-6\sqrt{x}+x+4\sqrt{x}+3+11\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{3x+9\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\frac{3\sqrt{x}.\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\frac{3\sqrt{x}}{\sqrt{x}-3}\)
b) Ta có: \(x=\sqrt{\sqrt{3}-\sqrt{4-2\sqrt{3}}}=\sqrt{\sqrt{3}-\sqrt{3-2\sqrt{3}+1}}\)
\(=\sqrt{\sqrt{3}-\sqrt{\left(\sqrt{3}-1\right)^2}}=\sqrt{\sqrt{3}-\left|\sqrt{3}-1\right|}\)
\(=\sqrt{\sqrt{3}-\sqrt{3}+1}=\sqrt{1}=1\)( thỏa mãn ĐKXĐ )
Thay \(x=1\)vào M ta được:
\(M=\frac{3\sqrt{1}}{\sqrt{1}-3}=\frac{3}{1-3}=\frac{-3}{2}\)
c) \(M=\frac{3\sqrt{x}}{\sqrt{x}-3}=\frac{3\sqrt{x}-9+9}{\sqrt{x}-3}=\frac{3\left(\sqrt{x}-3\right)+9}{\sqrt{x}-3}=3+\frac{9}{\sqrt{x}-3}\)
Vì \(x\inℕ\)\(\Rightarrow\)Để M là số tự nhiên thì \(\frac{9}{\sqrt{x}-3}\inℕ\)
\(\Rightarrow9⋮\left(\sqrt{x}-3\right)\)\(\Rightarrow\sqrt{x}-3\inƯ\left(9\right)\)(1)
Vì \(x\ge0\)\(\Rightarrow\sqrt{x}\ge0\)\(\Rightarrow\sqrt{x}-3\ge-3\)(2)
Từ (1) và (2) \(\Rightarrow\sqrt{x}-3\in\left\{-3;-1;1;3;9\right\}\)
\(\Rightarrow\sqrt{x}\in\left\{0;2;4;6;12\right\}\)\(\Rightarrow x\in\left\{0;4;16;36;144\right\}\)( thỏa mãn ĐKXĐ )
Thử lại với \(x=4\)ta thấy M không là số tự nhiên
Vậy \(x\in\left\{0;16;36;144\right\}\)
ĐKXĐ:...
\(M=\frac{3x+3\sqrt{x}-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}-\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}+\frac{\sqrt{x}-2}{\sqrt{x}}\left(\frac{\sqrt{x}}{1-\sqrt{x}}\right)\)
\(=\frac{3x+3\sqrt{x}-3-x+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}-\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{2x+3\sqrt{x}-2-x+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}=\frac{x+3\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}=\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(x=4+2\sqrt{3}\Rightarrow\sqrt{x}=\sqrt{4+2\sqrt{3}}=\sqrt{3}+1\)
\(\Rightarrow P=\frac{\sqrt{3}+1+1}{\sqrt{3}+1-1}=\frac{2+\sqrt{3}}{\sqrt{3}}=\frac{3+2\sqrt{3}}{3}\)
Để \(M=\sqrt{x}\Leftrightarrow\sqrt{x}=\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(\Leftrightarrow x-2\sqrt{x}-1=0\Rightarrow\left[{}\begin{matrix}\sqrt{x}=1+\sqrt{2}\\\sqrt{x}=1-\sqrt{2}< 0\left(l\right)\end{matrix}\right.\) \(\Rightarrow x=3+2\sqrt{2}\)
Để \(M>\frac{1}{2}\Rightarrow\frac{\sqrt{x}+1}{\sqrt{x}-1}-\frac{1}{2}>0\Rightarrow\frac{\sqrt{x}+3}{2\left(\sqrt{x}-1\right)}>0\)
\(\Rightarrow\sqrt{x}-1>0\Rightarrow x>1\)
Ta có: \(M=\frac{\sqrt{x}+1}{\sqrt{x}-1}=1+\frac{2}{\sqrt{x}-1}\)
Để M nguyên \(\Rightarrow\sqrt{x}-1=Ư\left(2\right)=\left\{-2;-1;1;2\right\}\)
\(\Rightarrow\sqrt{x}=\left\{2;3\right\}\Rightarrow x=\left\{4;9\right\}\)
a) \(\sqrt{13+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}\\ =\sqrt{13+30\sqrt{2+\sqrt{\left(2\sqrt{2}+1\right)^2}}}\\ =\sqrt{13+30\sqrt{2+2\sqrt{2}+1}}\\ =\sqrt{13+30\sqrt{\left(\sqrt{2}+1\right)^2}}\\ =\sqrt{13+30\left(\sqrt{2}+1\right)}\\ =\sqrt{13+30\sqrt{2}+30}\\ =\sqrt{43+30\sqrt{2}}\\ =\sqrt{\left(3\sqrt{2}+5\right)^2}\\ =3\sqrt{2}+5\)
b) \(\sqrt{227-30\sqrt{2}}+\sqrt{123+22\sqrt{2}}\\ =\sqrt{225-2\cdot15\sqrt{2}+2}+\sqrt{121+2\cdot11\sqrt{2}+2}\\ =\sqrt{\left(15-\sqrt{2}\right)^2}+\sqrt{\left(11+\sqrt{2}\right)^2}\\ =15-\sqrt{2}+11+\sqrt{2}\\ =26\)
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) \(\Rightarrow m^2+n^2=123\)