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\(x^3=9+4\sqrt{5}+9-4\sqrt{5}+3\cdot x\cdot1\)
=>x^3-3x-18=0
=>x=3
\(y^3=3+2\sqrt{2}+3-2\sqrt{2}+3y\)
=>y^3-3y-6=0
=>y=2,36
\(P=\left(x+y\right)^3-3xy\left(x+y\right)-3\left(x+y\right)+1993\)
\(=\left(3+2.36\right)^3-3\cdot3\cdot2.26\left(3+2.26\right)-3\left(3+2.36\right)+1993\)
=2023,922256
Đặt \(a=\sqrt[3]{9+4\sqrt{5}},b=\sqrt[3]{9-4\sqrt{5}}\)
\(\Rightarrow\hept{\begin{cases}a^3+b^3=18\\ab=1\end{cases};a+b=x}\)
Ta có: \(x=a+b\Leftrightarrow x^3=\left(a+b\right)^3=a^3+b^3+3ab\left(a+b\right)\)\(\Rightarrow x^3=18+3x\Leftrightarrow x^3-3x=18\)(1)
Tương tự: Đặt \(c=\sqrt[3]{3+2\sqrt{2}},d=\sqrt[3]{3-2\sqrt{2}}\)
\(\Rightarrow\hept{\begin{cases}c^3+d^3=6\\cd=1\end{cases};c+d=y}\)
Ta có: \(y=c+d\Leftrightarrow y^3=\left(c+d\right)^3=c^3+d^3+3cd\left(c+d\right)\)\(\Rightarrow y^3=6+3y\)
\(\Leftrightarrow y^3-3y=6\)(2)
Từ (1) và (2) suy ra \(A=x^3-3x+y^3-3y+2020=18+6+2020=2048\)
Ta có: \(x=\sqrt[3]{9+4\sqrt{5}}+\sqrt[3]{9-4\sqrt{5}}\Leftrightarrow x^3=9+4\sqrt{5}+9-4\sqrt{5}+\sqrt[3]{\left(9-4\sqrt{5}\right)\left(9+4\sqrt{5}\right)}x\Leftrightarrow x^3=18+3x\) làm tương tự ⇒ y3 = 9+ 3x
Thay x=..., y=... vào A ta có:
\(A=18+3x+9+3y-3x-3y+2020\)
A= 2047
\(x=\sqrt[3]{9+4\sqrt{5}}+\sqrt[3]{9-4\sqrt{5}}\)
\(\Leftrightarrow x^3=18+3x\)
Tương tự co:
\(y^3=6+3y\)
\(\Rightarrow P=18+3x+6+3y-3\left(x+y\right)+2019=2043\)
\(a,\dfrac{x+2\sqrt{x}-3}{\sqrt{x}-1}\)
\(\Leftrightarrow\dfrac{x+3\sqrt{x}-\sqrt{x}-3}{\sqrt{x}-1}\)
\(\Leftrightarrow\dfrac{\sqrt{x}.\left(\sqrt{x}+3\right)-\left(\sqrt{x}+3\right)}{\sqrt{x}-1}\)
\(\Leftrightarrow\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\)
\(\Rightarrow\sqrt{x}+3\)
\(b,\dfrac{4y+3\sqrt{y}-7}{4\sqrt{y}+7}\)
\(\Leftrightarrow\dfrac{4y+7\sqrt{y}-4\sqrt{y}-7}{4\sqrt{y}+7}\)
\(\Leftrightarrow\dfrac{\sqrt{y}.\left(4\sqrt{y}\right)-\left(4\sqrt{y}+7\right)}{4\sqrt{y}+7}\)
\(\Leftrightarrow\dfrac{\left(4\sqrt{y}+7\right).\left(\sqrt{y}-1\right)}{4\sqrt{y}+7}\)
\(\Rightarrow\sqrt{y}-1\)
\(c,\dfrac{x\sqrt{y}-y\sqrt{x}}{\sqrt{x}-\sqrt{y}}\)
\(\Leftrightarrow\dfrac{\sqrt{xy}.\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{x}-\sqrt{y}}\)
\(\Rightarrow\sqrt{xy}\)
\(d,\dfrac{x-3\sqrt{x}-4}{x-\sqrt{x}-12}\)
\(\Leftrightarrow\dfrac{x+\sqrt{x}-4\sqrt{x}-4}{x+3\sqrt{x}-4\sqrt{x}-12}\)
\(\Leftrightarrow\dfrac{\sqrt{x}.\left(\sqrt{x}+1\right)-4\left(\sqrt{x}+1\right)}{\sqrt{x}.\left(x+3\right)-4\left(\sqrt{x}+3\right)}\)
\(\Leftrightarrow\dfrac{\left(\sqrt{x}+1\right).\left(\sqrt{x}-4\right)}{\left(\sqrt{x}+3\right).\left(\sqrt{x}-4\right)}\)
\(\Leftrightarrow\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\)
\(\Rightarrow\dfrac{x-2\sqrt{x}-3}{x-9}\)
\(e,\dfrac{1+\sqrt{x}+\sqrt{y}+\sqrt{xy}}{1+\sqrt{4}}\)
\(\Leftrightarrow\dfrac{1+\sqrt{x}+\sqrt{y}+\sqrt{xy}}{1+2}\)
\(\Rightarrow\dfrac{1+\sqrt{x}+\sqrt{y}+\sqrt{xy}}{3}\)