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\(\sqrt{x^2-9}-3\sqrt{x-3}=0\)
\(ĐKXĐ:x\ge3\)
\(pt\Leftrightarrow\sqrt{\left(x+3\right)\left(x-3\right)}-3\sqrt{x-3}=0\)
\(\Leftrightarrow\sqrt{x+3}.\sqrt{x-3}-3.\sqrt{x-3}=0\)
\(\Leftrightarrow\sqrt{x-3}\left(\sqrt{x+3}-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x-3}=0\\\sqrt{x+3}-3=0\end{cases}}\)
\(TH1:\sqrt{x-3}=0\Leftrightarrow x-3=0\Leftrightarrow x=3\left(tm\right)\)
\(TH2:\sqrt{x+3}-3=0\Leftrightarrow\sqrt{x+3}=3\)
\(\Leftrightarrow\sqrt{x+3}=\sqrt{9}\Leftrightarrow x+3=9\Leftrightarrow x=9\left(tm\right)\)
Vậy pt có 2 nghiệm là 3 và 9
ĐK: \(x\ge0;x\ne9\)
\(A=\frac{\sqrt{x}}{\sqrt{x}+3}+\frac{2\sqrt{x}}{\sqrt{x}-3}+\frac{3x+9}{x-9}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}-3\right)+2\sqrt{x}\left(\sqrt{x}-3\right)+3x+9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{x-3\sqrt{x}+2x-6\sqrt{x}+3x+9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{-9x+9}{x-9}\)
\(x=\sqrt[3]{9+4\sqrt{5}}+\sqrt[3]{9-4\sqrt{5}}\)
\(\Leftrightarrow x^3=18+3x\)
Tương tự co:
\(y^3=6+3y\)
\(\Rightarrow P=18+3x+6+3y-3\left(x+y\right)+2019=2043\)
a)\(=4\sqrt{6}-3\sqrt{6}+1-\sqrt{6}\)
\(=1\)
b)ĐK: \(x>0,x\ne9\)
\(=\left(\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)-x-9}{x-9}\right):\dfrac{3\sqrt{x}+1-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-3\right)}\)
\(=\dfrac{3\sqrt{x}-9}{\sqrt{x}+3}.\dfrac{\sqrt{x}}{2\sqrt{x}+4}\)
\(=\dfrac{3\sqrt{x}\left(\sqrt{x}-3\right)}{2\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)}\)
b: \(\Leftrightarrow\sqrt{x-3}\left(\sqrt{x+3}-3\right)=0\)
=>x-3=0 hoặc x+3=9
=>x=3 hoặc x=6
c: \(\Leftrightarrow\sqrt{x^3-6x^2+9}=2x-6=\sqrt{4x^2-24x+36}\)
\(\Leftrightarrow x^3-6x^2+9-4x^2+24x-36=0\)
=>\(x^3-10x^2+24x-27=0\)
hay \(x\in\left\{7.18\right\}\)
\(\sqrt{3+9+3+3}\times\sqrt{2}=\sqrt{15}\times\sqrt{2}=\sqrt{15\times2}=\sqrt{30}\)
\(\sqrt{9+9}=\sqrt{18}\)
CÂU ĐẦU = 6
CÂU SAU = \(3\sqrt{2}\)