Tìm GTLN:
\(x^2\sqrt{9-x2}\)
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26 tháng 3 2022
Ta có: \(4\ge2\left(x^2+y^2\right)\ge\left(x+y\right)^2\)
\(\Rightarrow x+y\le2\)
Ta có: \(P=\sqrt{x\left(14x+10y\right)}+\sqrt{y\left(14y+10x\right)}\)
\(=\sqrt{\dfrac{24x\left(14x+10y\right)}{24}}+\sqrt{\dfrac{24y\left(14y+10x\right)}{24}}\le\dfrac{\dfrac{24x+14x+10y}{2}}{\sqrt{24}}+\dfrac{\dfrac{24y+14y+10x}{2}}{\sqrt{24}}\)
\(\Leftrightarrow P\le\dfrac{24\left(x+y\right)}{2\sqrt{6}}\le\dfrac{24.2}{2\sqrt{6}}=4\sqrt{6}\)
Dấu "=" xảy ra ⇔ x = y = 1
AT
12 tháng 7 2021
a) \(P=\dfrac{3x+3\sqrt{x}-9}{x+\sqrt{x}-2}+\dfrac{\sqrt{x}+3}{\sqrt{x}+2}-\dfrac{\sqrt{x}-2}{\sqrt{x}-1}\left(x\ge0,x\ne1\right)\)
\(=\dfrac{3x+3\sqrt{x}-9}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}+\dfrac{\sqrt{x}+3}{\sqrt{x}+2}-\dfrac{\sqrt{x}-2}{\sqrt{x}-1}\)
\(=\dfrac{3x+3\sqrt{x}-9+\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{3x+5\sqrt{x}-8}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}=\dfrac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+8\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}=\dfrac{3\sqrt{x}+8}{\sqrt{x}+2}\)
b) \(P=\dfrac{3\sqrt{x}+8}{\sqrt{x}+2}=\dfrac{3\sqrt{x}+6+2}{\sqrt{x}+2}=3+\dfrac{2}{\sqrt{x}+2}\)
Để \(P\in Z\Rightarrow2⋮\sqrt{x}+2\Rightarrow\sqrt{x}+2=2\left(\sqrt{x}+2\ge2\right)\)
\(\Rightarrow x=0\)
c) Ta có: \(\sqrt{x}\ge0\Rightarrow\sqrt{x}+2\ge2\Rightarrow\dfrac{2}{\sqrt{x}+2}\le1\Rightarrow3+\dfrac{2}{\sqrt{x}+2}\le4\)
\(\Rightarrow P_{max}=4\) khi \(x=0\)
NM
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0
31 tháng 8 2017
\(A=x^2\sqrt{9-x^2}=\frac{x^2}{\sqrt{3}}.\sqrt{3}.\sqrt{9-x^2}\)
\(\le\frac{x^2}{\sqrt{3}}.\frac{\left(3+9-x^2\right)}{2}\)
\(=\frac{1}{2\sqrt{3}}.\left(-x^4+12x^2\right)=\frac{1}{2\sqrt{3}}.\left(-x^4+12x^2-36+36\right)\)
\(=\frac{1}{2\sqrt{3}}.\left(36-\left(x^2-6\right)^2\right)\le\frac{36}{2\sqrt{3}}=6\sqrt{3}\)
T
3 tháng 9 2023
\(\dfrac{M}{N}=\left(\dfrac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{\sqrt{x}+2}{3-\sqrt{x}}\right):\left(\dfrac{\sqrt{x}+2}{\sqrt{x}-3}\right)\) (ĐKXĐ: \(x\ge0;x\ne4;x\ne9\))
\(=\left[\dfrac{2\sqrt{x}-9}{x-2\sqrt{x}-3\sqrt{x}+6}-\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}+\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right]\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+2}\)\(=\left[\dfrac{2\sqrt{x}-9}{\sqrt{x}\left(\sqrt{x}-2\right)-3\left(\sqrt{x}-2\right)}-\dfrac{x-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}+\dfrac{x-4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right]\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+2}\)
\(=\left[\dfrac{2\sqrt{x}-9-x+9+x-4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right]\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+2}\)
\(=\dfrac{2\sqrt{x}-4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+2}\)
\(=\dfrac{2\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)}\cdot\dfrac{1}{\sqrt{x}+2}\)
\(=\dfrac{2}{\sqrt{x}+2}\)
\(\Rightarrow P=\dfrac{M}{N}+1=\dfrac{2}{\sqrt{x}+2}+1\)
Ta thấy: \(\sqrt{x}\ge0\forall x\)
\(\Rightarrow\sqrt{x}+2\ge2\forall x\)
\(\Rightarrow\dfrac{2}{\sqrt{x}+2}\le1\forall x\)
\(\Rightarrow\dfrac{2}{\sqrt{x}+2}+1\le2\forall x\)
\(\Rightarrow Max_P=2\Leftrightarrow\dfrac{2}{\sqrt{x}+2}+1=2\)
\(\Leftrightarrow\dfrac{2}{\sqrt{x}+2}=1\)
\(\Leftrightarrow\sqrt{x}+2=2\)
\(\Leftrightarrow\sqrt{x}=0\)
\(\Leftrightarrow x=0\left(tm\right)\)
#Urushi☕
3 tháng 9 2023
Bạn tự rút gọn nha .
c) Ta có : \(P\text{=}\dfrac{M}{N}+1\text{=}\dfrac{2}{\sqrt{x}+2}+1\)
Để P có giá trị lớn nhất.
\(\Leftrightarrow\dfrac{2}{\sqrt{x}+2}cóGTLN\)
\(\Leftrightarrow\sqrt{x}+2cóGTNN\)
Mà : \(\sqrt{x}+2\ge2\)
\(\Rightarrow\) Để : \(\left(\sqrt{x}+2\right)_{min}\) \(\Leftrightarrow\sqrt{x}\text{=}0\Leftrightarrow x\text{=}0\)
Vậy............
\(x^2\sqrt{9-x^2}\)
Ta có:
\(9-x^2\le9\forall x\)
\(Minx^2\sqrt{9-x^2}=0\Leftrightarrow x=0\)