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Bài 1 :
Ta có : \(\dfrac{1}{3a^2+b^2}+\dfrac{2}{b^2+3ab}=\dfrac{1}{3a^2+b^2}+\dfrac{4}{2b^2+6ab}\)
Theo BĐT Cô - Si dưới dạng engel ta có :
\(\dfrac{1}{3a^2+b^2}+\dfrac{4}{2b^2+6ab}\ge\dfrac{\left(1+2\right)^2}{3a^2+6ab+3b^2}=\dfrac{9}{3\left(a+b\right)^2}=\dfrac{9}{3.1}=3\)
Dấu \("="\) xảy ra khi : \(a=b=\dfrac{1}{2}\)
\(M=4x^2-3x+\dfrac{1}{4x}+2017\)
\(=\left(4x^2-4x+1\right)+\left(x+\dfrac{1}{4x}\right)+2016\ge2017\)
1.
a, ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne4\end{matrix}\right.\)
b,
\(M=(\dfrac{\sqrt{x}}{\sqrt{x}-2}\times\dfrac{\sqrt{x}}{\sqrt{x}+2})\times\dfrac{x-4}{\sqrt{4x}}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)+\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\times\dfrac{x-4}{2\sqrt{x}}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+2+\sqrt{x}-2\right)}{x-4}\times\dfrac{x-4}{2\sqrt{x}}\)
\(=(\sqrt{x}\times2\sqrt{x})\times\dfrac{1}{2\sqrt{x}}\)
\(=\sqrt{x}\)
c,
\(M>3\Leftrightarrow\sqrt{x}>3\Leftrightarrow x>9\)
Bài 2:
a: \(A=\dfrac{3+\sqrt{1-a^2}}{\sqrt{1+a}}:\dfrac{3+\sqrt{1-a^2}}{\sqrt{1-a^2}}=\sqrt{\dfrac{1-a^2}{1+a}}=\sqrt{1-a}\)
b: Để A=căn A thì A=1 hoặc A=0
=>A=1
=>1-a=1
=>a=0
c: Thay \(a=\dfrac{\sqrt{3}}{2+\sqrt{3}}=\sqrt{3}\left(2-\sqrt{3}\right)=2\sqrt{3}-3\) vào A, ta được:
\(A=\sqrt{1-2\sqrt{3}+3}=\sqrt{4-2\sqrt{3}}=\sqrt{3}-1\)
\(4x+\dfrac{1}{4x}-\dfrac{4\sqrt{x}+3}{x+1}+2017\)
\(=\left(4x+\dfrac{1}{4x}\right)-4+\dfrac{4x-4\sqrt{x}+1}{x+1}+2017\)
\(=\left(4x+\dfrac{1}{4x}\right)+\dfrac{\left(2\sqrt{x}-1\right)}{x+1}+2013\)
\(\ge2+0+2013=2015\)
Dấu = xảy ra khi \(x=\dfrac{1}{4}\)
a: \(P=\dfrac{9x+6\sqrt{x}+1-9x+6\sqrt{x}-1+4x}{9-x}:\dfrac{5\sqrt{x}-4\sqrt{x}-2}{\sqrt{x}\left(3-\sqrt{x}\right)}\)
\(=\dfrac{4x+12\sqrt{x}}{9-x}\cdot\dfrac{\sqrt{x}\left(3-\sqrt{x}\right)}{\sqrt{x}-2}\)
\(=\dfrac{4x}{\sqrt{x}-2}\)
b: Để P^2=40P thì P(P-40)=0
=>P=0(loại) hoặc P=40
=>4x=40 căn x-80
=>4x-40 căn x+80=0
=>x-10 căn x+20=0
=>căn x=5+căn 5 hoặc căn x=5-căn 5
=>x=30+10 căn 5 hoặc x=30-10 căn 5
Lời giải:
ĐKXĐ:......
a) Ta có:
\(\frac{3+\sqrt{x}}{3-\sqrt{x}}-\frac{3-\sqrt{x}}{3+\sqrt{x}}-\frac{4x}{x-9}=\frac{(3+\sqrt{x})^2-(3-\sqrt{x})^2}{(3-\sqrt{x})(3+\sqrt{x})}-\frac{4x}{x-9}\)
\(=\frac{9+x+6\sqrt{x}-(9+x-6\sqrt{x})}{9-x}-\frac{4x}{x-9}=\frac{-12\sqrt{x}}{x-9}-\frac{4x}{x-9}=\frac{-4\sqrt{x}(3+\sqrt{x})}{(\sqrt{x}-3)(\sqrt{x}+3)}=\frac{4\sqrt{x}}{3-\sqrt{x}}\)
Và:
\(\frac{5}{3-\sqrt{x}}-\frac{4\sqrt{x}+2}{3\sqrt{x}-x}=\frac{5\sqrt{x}}{3\sqrt{x}-x}-\frac{4\sqrt{x}+2}{3\sqrt{x}-x}=\frac{\sqrt{x}-2}{\sqrt{x}(3-\sqrt{x})}\)
Do đó:
\(C=\frac{4\sqrt{x}}{3-\sqrt{x}}: \frac{\sqrt{x}-2}{\sqrt{x}(3-\sqrt{x})}=\frac{4\sqrt{x}}{3-\sqrt{x}}.\frac{\sqrt{x}(3-\sqrt{x})}{\sqrt{x}-2}=\frac{4x}{\sqrt{x}-2}\)
b)
Nếu $C\leq 0$ thì \(|C|=-C\) (không thỏa mãn)
Nếu $C>0$ thì \(|C|=C>0>-C\) (thỏa mãn)
Vậy để \(|C|> -C\) thì \(C>0\Leftrightarrow \frac{4x}{\sqrt{x}-2}>0\Leftrightarrow \sqrt{x}-2>0\) (do \(x>0)\)
\(\Leftrightarrow x> 4\)
Kết hợp đkxđ suy ra điều kiện của $x$ là \(x>4; x\neq 9\)
c)
\(C^2=40C\Leftrightarrow C(C-40)=0\Leftrightarrow \left[\begin{matrix} C=0\\ C=40\end{matrix}\right.\)
Nếu $C=0$ thì \(\frac{4x}{\sqrt{x}-2}=0\Rightarrow x=0\) (không t/m ĐKXĐ)
Nếu \(C=40\Leftrightarrow \frac{4x}{\sqrt{x}-2}=40\Leftrightarrow x=10(\sqrt{x}-2)\)
\(\Rightarrow \sqrt{x}=5\pm \sqrt{5}\Rightarrow x=(5\pm \sqrt{5})^2\)
Na: cái này là giải pt bậc 2 đơn giản thôi bạn:
\(x=10(\sqrt{x}-2)\)
\(\Rightarrow x-10\sqrt{x}+20=0\)
\(\Rightarrow (\sqrt{x}-5)^2-5=0\Rightarrow (\sqrt{x}-5)^2=5\)
\(\Rightarrow \sqrt{x}-5=\pm \sqrt{5}\Rightarrow \sqrt{x}=5\pm \sqrt{5}\) đó bạn.
điều kiện xác định : \(x>0;x\ne9\)
a) ta có : \(C=\left(\dfrac{3+\sqrt{x}}{3-\sqrt{x}}-\dfrac{3-\sqrt{x}}{3+\sqrt{x}}-\dfrac{4x}{x-9}\right):\left(\dfrac{5}{3-\sqrt{x}}-\dfrac{4\sqrt{x}+2}{3\sqrt{x}-x}\right)\)
\(\Leftrightarrow C=\left(\dfrac{3+\sqrt{x}}{3-\sqrt{x}}-\dfrac{3-\sqrt{x}}{3+\sqrt{x}}+\dfrac{4x}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}\right):\left(\dfrac{5}{3-\sqrt{x}}-\dfrac{4\sqrt{x}+2}{\sqrt{x}\left(3-\sqrt{x}\right)}\right)\) \(\Leftrightarrow C=\left(\dfrac{3+\sqrt{x}}{3-\sqrt{x}}-\dfrac{3-\sqrt{x}}{3+\sqrt{x}}+\dfrac{4x}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}\right):\left(\dfrac{5\sqrt{x}-4\sqrt{x}-2}{\sqrt{x}\left(3-\sqrt{x}\right)}\right)\) \(\Leftrightarrow C=\left(\dfrac{\left(3+\sqrt{x}\right)^2-\left(3-\sqrt{x}\right)^2+4x}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}\right):\left(\dfrac{\sqrt{x}-2}{\sqrt{x}\left(3-\sqrt{x}\right)}\right)\) \(\Leftrightarrow C=\left(\dfrac{12\sqrt{x}+4x}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}\right)\left(\dfrac{\sqrt{x}\left(3-\sqrt{x}\right)}{\sqrt{x}-2}\right)\) \(\Leftrightarrow C=\left(\dfrac{4\sqrt{x}\left(\sqrt{x}+3\right)}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}\right)\left(\dfrac{\sqrt{x}\left(3-\sqrt{x}\right)}{\sqrt{x}-2}\right)=\dfrac{4x}{\sqrt{x}-2}\)b) để \(\left|C\right|>-C\) \(\Leftrightarrow C< 0\) \(\Leftrightarrow\dfrac{4x}{\sqrt{x}-2}< 0\) \(\Leftrightarrow\sqrt{x}-2< 0\Leftrightarrow\sqrt{x}< 2\Leftrightarrow0< x< 4\)
c) để \(C^2=40C\Leftrightarrow C^2-40C=0\Leftrightarrow C\left(C-40\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}C=0\\C=40\end{matrix}\right.\)
+) \(C=0\Leftrightarrow\dfrac{4x}{\sqrt{x}-2}=0\) \(\Leftrightarrow x=0\left(loại\right)\)
+) \(C=40\Leftrightarrow\dfrac{4x}{\sqrt{x}-2}=40\Leftrightarrow x=10\sqrt{x}-20\)
\(\Leftrightarrow x-10\sqrt{x}+20=0\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=5+3\sqrt{5}\left(N\right)\\\sqrt{x}=5-3\sqrt{5}\left(L\right)\end{matrix}\right.\)
ta có : \(\sqrt{x}=5+3\sqrt{5}\Leftrightarrow x=70+30\sqrt{5}\)
vậy ..............................................................................................................................
Sửa đề: \(A=\dfrac{5}{2\sqrt{x}+1}+\dfrac{3}{2\sqrt{x}-1}+\dfrac{12\sqrt{x}}{1-4x}\)
a: \(A=\dfrac{10\sqrt{x}-5+6\sqrt{x}+3-12\sqrt{x}}{4x-1}\)
\(=\dfrac{4\sqrt{x}-2}{4x-1}=\dfrac{2\left(2\sqrt{x}-1\right)}{4x-1}=\dfrac{2}{2\sqrt{x}+1}\)
b: Để A>1/3 thì A-1/3>0
\(\Leftrightarrow\dfrac{2}{2\sqrt{x}+1}-\dfrac{1}{3}>0\)
\(\Leftrightarrow6-2\sqrt{x}-1>0\)
\(\Leftrightarrow5-2\sqrt{x}>0\)
=>\(2\sqrt{x}< 5\)
=>x<25/4
Vậy: 0<x<25/4 và x<>1/4
\(P=\dfrac{y}{x}+\dfrac{x}{y}+\left(\dfrac{x}{3y}+3xy+\dfrac{1}{3}+\dfrac{1}{3}\right)+12\left(xy+\dfrac{1}{9}\right)-2\)
\(P\ge2\sqrt{\dfrac{xy}{xy}}+4\sqrt[4]{\dfrac{3x^2y}{27y}}+12.2\sqrt{\dfrac{xy}{9}}-2\)
\(P\ge4\sqrt{\dfrac{x}{3}}+8\sqrt{xy}=4\left(2\sqrt{xy}+\sqrt{\dfrac{x}{3}}\right)=4\)
\(P_{min}=4\) khi \(x=y=\dfrac{1}{3}\)