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\(I=\dfrac{a-\sqrt{ab}+b+3\sqrt{ab}}{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}\cdot\left[\left(\dfrac{a+\sqrt{ab}+b+3\sqrt{ab}}{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}\right)\cdot\dfrac{a+\sqrt{ab}+b}{a-b}\right]\)
\(=\dfrac{a+2\sqrt{ab}+b}{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}\cdot\left(\dfrac{a+4\sqrt{ab}+b}{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}\cdot\dfrac{a+\sqrt{ab}+b}{a-b}\right)\)
\(=\dfrac{\sqrt{a}+\sqrt{b}}{a-\sqrt{ab}+b}\cdot\dfrac{a+4\sqrt{ab}+b}{\left(\sqrt{a}-\sqrt{b}\right)\left(a-b\right)}\)
\(=\dfrac{a+4\sqrt{ab}+b}{\left(\sqrt{a}-\sqrt{b}\right)^2\cdot\left(a-\sqrt{ab}+b\right)}\)
Khi a=16 và b=4 thì \(I=\dfrac{16+4+4\cdot\sqrt{16\cdot4}}{\left(4-2\right)^2\cdot\left(16-\sqrt{16\cdot4}+4\right)}=\dfrac{20+4\cdot8}{4\cdot12}\)
\(=\dfrac{20+32}{48}=\dfrac{52}{48}=\dfrac{13}{12}\)
Áp dụng BĐT Cô-si:
\(A\le\dfrac{a+b}{2\sqrt{c+ab}}+\dfrac{b+c}{2\sqrt{a+bc}}+\dfrac{c+a}{2\sqrt{b+ac}}\)\(\le\dfrac{a+b}{2\sqrt{2\sqrt{abc}}}+\dfrac{b+c}{2\sqrt{2\sqrt{abc}}}+\dfrac{c+a}{2\sqrt{2\sqrt{abc}}}\)\(=\dfrac{a+b+c}{\sqrt[4]{4abc}}=\dfrac{1}{\sqrt[4]{4abc}}\ge\dfrac{1}{\sqrt{\left(a+b+c\right).\dfrac{2}{3}}}\)(BĐT Cô-si)\(=\dfrac{1}{\sqrt{\dfrac{2}{3}}}=\dfrac{\sqrt{6}}{2}\)
Vậy Amin=\(\dfrac{\sqrt{6}}{2}\Leftrightarrow a=b=c=\dfrac{1}{3}\)
a: Ta có: \(A=\left(1-\dfrac{2\sqrt{x}-2}{x-1}\right):\left(\dfrac{1}{\sqrt{x}+1}-\dfrac{\sqrt{x}}{x\sqrt{x}+1}\right)\)
\(=\dfrac{x-1-2\sqrt{x}+2}{x-1}:\dfrac{x-\sqrt{x}+1-\sqrt{x}}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)
\(=\dfrac{x-\sqrt{x}+1}{\sqrt{x}-1}\)
a) \(A=\dfrac{3x}{x\sqrt{x}+1}-\dfrac{\sqrt{x}-1}{x-\sqrt{x}+1}-\dfrac{1}{1+\sqrt{x}}\)
\(=\dfrac{3x}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}-\dfrac{\sqrt{x}-1}{x-\sqrt{x}+1}-\dfrac{1}{1+\sqrt{x}}\)
\(=\dfrac{3x-\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)-x+\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)
\(\dfrac{x+\sqrt{x}}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}}{x-\sqrt{x}+1}\)
a: ĐKXĐ: \(\left\{{}\begin{matrix}a>0\\b>0\\a< >b\end{matrix}\right.\)
b: \(M=\dfrac{\sqrt{a}+\sqrt{b}-1}{a+\sqrt{ab}}+\dfrac{\sqrt{a}-\sqrt{b}}{2\sqrt{ab}}\cdot\left(\dfrac{b}{a-\sqrt{ab}}+\dfrac{\sqrt{b}}{a+\sqrt{ab}}\right)\)
\(=\dfrac{\sqrt{a}+\sqrt{b}-1}{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}+\dfrac{\sqrt{a}-\sqrt{b}}{2\sqrt{ab}}\cdot\dfrac{b\left(a+\sqrt{ab}\right)+\sqrt{b}\left(a-\sqrt{ab}\right)}{a^2-ab}\)
\(=\dfrac{\sqrt{a}+\sqrt{b}-1}{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}+\dfrac{\sqrt{a}-\sqrt{b}}{a\left(a-b\right)}\cdot\dfrac{ab+b\sqrt{ab}+a\sqrt{b}-b\sqrt{a}}{2\sqrt{ab}}\)
\(=\dfrac{\sqrt{a}+\sqrt{b}-1}{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}+\dfrac{\sqrt{a}-\sqrt{b}}{a\left(\sqrt{a}-\sqrt{b}\right)\cdot\left(\sqrt{a}+\sqrt{b}\right)}\cdot\dfrac{\sqrt{ab}\left(\sqrt{ab}+b+\sqrt{a}-\sqrt{b}\right)}{2\sqrt{ab}}\)
\(=\dfrac{\sqrt{a}+\sqrt{b}-1}{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}+\dfrac{1}{a\left(\sqrt{a}+\sqrt{b}\right)}\cdot\dfrac{\sqrt{ab}+b+\sqrt{a}-\sqrt{b}}{2}\)
\(=\dfrac{2\sqrt{a}\left(\sqrt{a}+\sqrt{b}-1\right)+\sqrt{ab}+b+\sqrt{a}-\sqrt{b}}{2a\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\dfrac{2a+2\sqrt{ab}-2\sqrt{a}+\sqrt{ab}+b+\sqrt{a}-\sqrt{b}}{2a\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\dfrac{2a+3\sqrt{ab}-\sqrt{a}+b-\sqrt{b}}{2a\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\dfrac{2a+3\sqrt{ab}+b-\left(\sqrt{a}+\sqrt{b}\right)}{2a\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\dfrac{\left(2\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)-\left(\sqrt{a}+\sqrt{b}\right)}{2a\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\dfrac{2\sqrt{a}+\sqrt{b}-1}{2a}\)
Giả sử như a=0,1 và b=0,11 thì M<0 nha bạn
=>Đề này sai rồia: ĐKXĐ:
b: \(M=\dfrac{\sqrt{a}+\sqrt{b}-1}{a+\sqrt{ab}}+\dfrac{\sqrt{a}-\sqrt{b}}{2\sqrt{ab}}\cdot\left(\dfrac{b}{a-\sqrt{ab}}+\dfrac{\sqrt{b}}{a+\sqrt{ab}}\right)\)
\(=\dfrac{\sqrt{a}+\sqrt{b}-1}{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}+\dfrac{\sqrt{a}-\sqrt{b}}{2\sqrt{ab}}\cdot\dfrac{b\left(a+\sqrt{ab}\right)+\sqrt{b}\left(a-\sqrt{ab}\right)}{a^2-ab}\)
\(=\dfrac{\sqrt{a}+\sqrt{b}-1}{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}+\dfrac{\sqrt{a}-\sqrt{b}}{a\left(a-b\right)}\cdot\dfrac{ab+b\sqrt{ab}+a\sqrt{b}-b\sqrt{a}}{2\sqrt{ab}}\)
\(=\dfrac{\sqrt{a}+\sqrt{b}-1}{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}+\dfrac{\sqrt{a}-\sqrt{b}}{a\left(\sqrt{a}-\sqrt{b}\right)\cdot\left(\sqrt{a}+\sqrt{b}\right)}\cdot\dfrac{\sqrt{ab}\left(\sqrt{ab}+b+\sqrt{a}-\sqrt{b}\right)}{2\sqrt{ab}}\)
\(=\dfrac{\sqrt{a}+\sqrt{b}-1}{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}+\dfrac{1}{a\left(\sqrt{a}+\sqrt{b}\right)}\cdot\dfrac{\sqrt{ab}+b+\sqrt{a}-\sqrt{b}}{2}\)
\(=\dfrac{2\sqrt{a}\left(\sqrt{a}+\sqrt{b}-1\right)+\sqrt{ab}+b+\sqrt{a}-\sqrt{b}}{2a\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\dfrac{2a+2\sqrt{ab}-2\sqrt{a}+\sqrt{ab}+b+\sqrt{a}-\sqrt{b}}{2a\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\dfrac{2a+3\sqrt{ab}-\sqrt{a}+b-\sqrt{b}}{2a\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\dfrac{2a+3\sqrt{ab}+b-\left(\sqrt{a}+\sqrt{b}\right)}{2a\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\dfrac{\left(2\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)-\left(\sqrt{a}+\sqrt{b}\right)}{2a\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\dfrac{2\sqrt{a}+\sqrt{b}-1}{2a}\)
Giả sử như a=0,1 và b=0,11 thì M<0 nha bạn
=>Đề này sai rồi
a) ĐKXĐ: \(\left\{{}\begin{matrix}a>0\\b>0\\a\ne b\end{matrix}\right.\)
P = \(\dfrac{a-\sqrt{ab}+b+3\sqrt{ab}}{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}.\left[\left(\dfrac{a+\sqrt{ab}+b-3\sqrt{ab}}{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}\right):\dfrac{a-b}{a+\sqrt{ab}+b}\right]\)= \(\dfrac{\left(\sqrt{a}+\sqrt{b}\right)^2}{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}.\left[\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}.\dfrac{a+\sqrt{ab}+b}{a-b}\right]\)
= \(\dfrac{\sqrt{a}+\sqrt{b}}{a-\sqrt{ab}+b}.\dfrac{\sqrt{a}-\sqrt{b}}{a-b}\)
= \(\dfrac{1}{a-\sqrt{ab}+b}\)
b) có a = 16 và b = 4 (thoả mãn ĐKXĐ)
Thay a = 16, b =4 vào P có:
P = \(\dfrac{1}{16-\sqrt{16.4}+4}\)= \(\dfrac{1}{12}\)
Vậy tại a =16, b = 4 thì P = \(\dfrac{1}{12}\)
ĐKXĐ : \(\left\{{}\begin{matrix}a,b\ge0\\a.b\ne1\end{matrix}\right.\)
a ) \(P=\left(\dfrac{\sqrt{a}+1}{\sqrt{ab}+1}+\dfrac{\sqrt{ab}+\sqrt{a}}{\sqrt{ab}-1}-1\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{ab}+1}-\dfrac{\sqrt{ab}+\sqrt{a}}{\sqrt{ab}-1}+1\right)\)
\(=\dfrac{\left(\sqrt{a}+1\right)\left(\sqrt{ab}-1\right)+\left(\sqrt{ab}+\sqrt{a}\right)\left(\sqrt{ab}+1\right)-\left(\sqrt{ab}+1\right)\left(\sqrt{ab}-1\right)}{\left(\sqrt{ab}+1\right)\left(\sqrt{ab}-1\right)}:\dfrac{\left(\sqrt{a}+1\right)\left(\sqrt{ab}-1\right)-\left(\sqrt{ab}+\sqrt{a}\right)\left(\sqrt{ab}+1\right)+\left(\sqrt{ab}+1\right)\left(\sqrt{ab}-1\right)}{\left(\sqrt{ab}+1\right)\left(\sqrt{ab}-1\right)}\)
\(=\dfrac{a\sqrt{b}-\sqrt{a}+\sqrt{ab}-1+ab+\sqrt{ab}+a\sqrt{b}+\sqrt{a}-ab+1}{\left(\sqrt{ab}+1\right)\left(\sqrt{ab}-1\right)}.\dfrac{\left(\sqrt{ab}+1\right)\left(\sqrt{ab}-1\right)}{a\sqrt{b}-\sqrt{a}+\sqrt{ab}-1-ab-\sqrt{ab}-a\sqrt{b}-\sqrt{a}+ab-1}\)
\(=\dfrac{2a\sqrt{b}+2\sqrt{ab}}{-2\sqrt{a}-2}=-\dfrac{2\sqrt{ab}\left(\sqrt{a}+1\right)}{2\left(\sqrt{a}+1\right)}=-\sqrt{ab}\)
Câu b : Ta có : \(b=\dfrac{\sqrt{3}-1}{1+\sqrt{3}}=\dfrac{\left(\sqrt{3}-1\right)\left(\sqrt{3}-1\right)}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}=\dfrac{3-2\sqrt{3}+1}{2}=2-\sqrt{3}\)
\(P=-\sqrt{ab}=-\sqrt{\left(2-\sqrt{3}\right)^2}=-\left|2-\sqrt{3}\right|=\sqrt{3}-2\)
Câu c : \(\sqrt{a}+\sqrt{b}=4\Rightarrow\sqrt{a}=4-\sqrt{b}\)
\(P=-\sqrt{ab}=-\left(4-\sqrt{b}\right)\sqrt{b}=b-4\sqrt{b}=\left(\sqrt{b}-2\right)^2-4\ge-4\)
Vậy GTNN của P là -4 . Dấu bằng xảy ra khi \(a=b=4\)
cho em với
ĐKXĐ \(a\ge0,b\ge0\)
\(\Rightarrow\left(\dfrac{\left(\sqrt{a}+1\right)\left(\sqrt{ab}-1\right)+\left(\sqrt{ab}+\sqrt{a}\right)\left(\sqrt{ab}+1\right)-\left(\sqrt{ab}+1\right)\left(\sqrt{ab}-1\right)}{\left(\sqrt{ab}+1\right)\left(\sqrt{ab}-1\right)}\right):\left(\dfrac{\left(\sqrt{a}+1\right)\left(\sqrt{ab}-1\right)-\left(\sqrt{ab}+\sqrt{a}\right)\left(\sqrt{ab}+1\right)+\left(\sqrt{ab}+1\right)\left(\sqrt{ab}-1\right)}{\left(\sqrt{ab}-1\right)\left(\sqrt{ab}+1\right)}\right)\)
=\(\left(\dfrac{a\sqrt{b}-\sqrt{a}+\sqrt{ab}-1+ab+\sqrt{ab}+a\sqrt{b}+\sqrt{a}-ab+1}{\left(\sqrt{ab}+1\right)\left(\sqrt{ab}-1\right)}\right):\left(\dfrac{a\sqrt{b}-\sqrt{a}+\sqrt{ab}-1-ab-\sqrt{ab}-a\sqrt{b}-\sqrt{a}+ab-1}{\left(\sqrt{ab}-1\right)\left(\sqrt{ab}+1\right)}\right)\)
=\(\left(\dfrac{2a\sqrt{b}+2\sqrt{ab}}{\left(\sqrt{ab}-1\right)\left(\sqrt{ab}+1\right)}\right):\left(\dfrac{-2\sqrt{a}-2}{\left(\sqrt{ab}-1\right)\left(\sqrt{ab}+1\right)}\right)\)
= \(\dfrac{2\sqrt{ab}\left(\sqrt{a}+1\right)}{-2\left(\sqrt{a}+1\right)}\) = \(-\sqrt{ab}\)
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