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\(1.a.A=\left(1-\dfrac{\sqrt{x}}{1+\sqrt{x}}\right):\left(\dfrac{\sqrt{x}+3}{\sqrt{x}-2}+\dfrac{\sqrt{x}+2}{3-\sqrt{x}}+\dfrac{\sqrt{x}+2}{x-5\sqrt{x}+6}\right)=\dfrac{1}{\sqrt{x}+1}:\dfrac{x-9-x+4+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\dfrac{1}{\sqrt{x}+1}.\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}{\sqrt{x}-3}=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\left(x\ge0;x\ne4;x\ne9\right)\)
\(b.A< 0\Leftrightarrow\dfrac{\sqrt{x}-2}{\sqrt{x}+1}< 0\)
\(\Leftrightarrow\sqrt{x}-2< 0\)
\(\Leftrightarrow x< 4\)
Kết hợp với ĐKXĐ , ta có : \(0\le x< 4\)
KL............
\(2.\) Tương tự bài 1.
\(3a.A=\dfrac{1}{x-\sqrt{x}+1}=\dfrac{1}{x-2.\dfrac{1}{2}\sqrt{x}+\dfrac{1}{4}+\dfrac{3}{4}}=\dfrac{1}{\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}}\le\dfrac{4}{3}\)
\(\Rightarrow A_{Max}=\dfrac{4}{3}."="\Leftrightarrow x=\dfrac{1}{4}\)
Bài 2:
a: \(\sqrt{4-x^2}>=0\)
Dấu '=' xảy ra khi x=2 hoặc x=-2
b: \(\sqrt{x^2-x+3}=\sqrt{x^2-x+\dfrac{1}{4}+\dfrac{11}{4}}\)
\(=\sqrt{\left(x-\dfrac{1}{2}\right)^2+\dfrac{11}{4}}>=\dfrac{\sqrt{11}}{2}\)
Dấu '=' xảy ra khi x=1/2
c: \(x+\sqrt{x}+1>=1\)
=>1/(x+căn x+1)<=1
Dấu '=' xảy ra khi x=0
a) Để biểu thức P xác định thì \(\left\{{}\begin{matrix}x\ge0\\x\ne9\end{matrix}\right.\)
Vậy ĐKXĐ:x\(\ge0\),x\(\ne9\)
\(P=\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+3}{x-9}\right):\left(\dfrac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)=\left[\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-\dfrac{3x+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right]:\dfrac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}=\left[\dfrac{2x-6\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\dfrac{x+3\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-\dfrac{3x+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right]:\dfrac{\sqrt{x}+1}{\sqrt{x}-3}=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\dfrac{\sqrt{x}-3}{\sqrt{x}+1}=\dfrac{\left(-3\sqrt{x}-3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)\left(\sqrt{x}+1\right)}=\dfrac{-3\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)\left(\sqrt{x}+1\right)}=\dfrac{-3}{\sqrt{x}+3}\)
b) Ta có \(P< \dfrac{1}{2}\Leftrightarrow\dfrac{-3}{\sqrt{x}+3}< \dfrac{1}{2}\Leftrightarrow-6< \sqrt{x}+3\Leftrightarrow\sqrt{x}>-9\)
Vì \(\sqrt{x}\ge0\) và 0>-9
Vậy \(x\ge0\)
Kết hợp với ĐKXĐ, Vậy \(x\ge0\) và \(x\ne9\) thì P<\(\dfrac{1}{2}\)
2.
a/ Áp dụgn hệ quả bđt cô si,ta có :
\(A=xy+yz+zx\le\dfrac{\left(x+y+z\right)}{3}=\dfrac{a^2}{3}\)
Vậy GTLN A =a^2/3 khi x= y =z =a/3
b/Áp dụng BĐT Cô-Si dạng Engel,ta có :
\(B=\dfrac{x^2}{1}+\dfrac{y^2}{1}+\dfrac{z^2}{z}\ge\dfrac{\left(x+y+z\right)^2}{3}=\dfrac{a^2}{3}\)
Vậy GTNN của B = a^2/2 khi x=y=z =a/3
\(B=\dfrac{3x}{1-x}+\dfrac{4\left(1-x\right)}{x}+7\ge2\sqrt{\dfrac{3x}{1-x}.\dfrac{4\left(1-x\right)}{x}}+7=7+4\sqrt{3}=\left(2+\sqrt{3}\right)^2\)
Vậy min B = \(\left(2+\sqrt{3}\right)^2\) khi \(\dfrac{3x}{1-x}=\dfrac{4\left(1-x\right)}{x}\Leftrightarrow x=\left(\sqrt{3}-1\right)^2\)
Bài 1:
\(P=x\sqrt{3-x^2}=\sqrt{x^2}\cdot\sqrt{3-x^2}\)
\(=\sqrt{x^2\left(3-x^2\right)}\)\(\le\frac{x^2+3-x^2}{2}=\frac{3}{2}\)
Dấu = khi \(x=\sqrt{\frac{3}{2}}\)
Vậy MaxP=\(\frac{3}{2}\Leftrightarrow x=\sqrt{\frac{3}{2}}\)