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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
\(x=\sqrt[3]{9+4\sqrt{5}}+\sqrt[3]{9-4\sqrt{5}}\\ \Leftrightarrow x^3=9+4\sqrt{5}+9-4\sqrt{5}+3\sqrt[3]{\left(9-4\sqrt{5}\right)\left(9+4\sqrt{5}\right)}\left(\sqrt[3]{9+4\sqrt{5}}+\sqrt[3]{9-4\sqrt{5}}\right)\\ \Leftrightarrow x^3=18+3x\sqrt[3]{81-80}=18-3x\\ \Leftrightarrow x^3-3x=18\\ y=\sqrt[3]{3-2\sqrt{2}}+\sqrt[3]{3+2\sqrt{2}}\\ \Leftrightarrow y^3=6+3\sqrt[3]{\left(3-2\sqrt{2}\right)\left(3+2\sqrt{2}\right)}\left(\sqrt[3]{3-2\sqrt{2}}+\sqrt[3]{3+2\sqrt{2}}\right)\\ \Leftrightarrow y^3=6+3y\sqrt[3]{9-8}=6+3y\\ \Leftrightarrow y^3-3y=6\\ \Leftrightarrow P=x^3+y^3-3\left(x+y\right)+1993\\ P=x^3+y^3-3x-3y+1993=18+6+1993=2017\)
Áp dụng: \(\left(a+b\right)^3=a^3+3a^2b+3ab^2+b^3=a^3+b^3+3ab\left(a+b\right)\)
\(x=\sqrt[3]{9+4\sqrt{5}}+\sqrt[3]{9-4\sqrt{5}}\)
\(\Rightarrow x^3=9+4\sqrt{5}+9-4\sqrt{5}+3\sqrt[3]{\left(9+4\sqrt{5}\right)\left(9-4\sqrt{5}\right)}\left(\sqrt[3]{9+4\sqrt{5}}+\sqrt[3]{9-4\sqrt{5}}\right)\)
\(=18+3\sqrt[3]{81-80}.x=18+3x\)
\(y=\sqrt[3]{3-2\sqrt{2}}+\sqrt[3]{3+2\sqrt{2}}\)
\(\Rightarrow y^3=3-2\sqrt{2}+3+2\sqrt{2}+3\sqrt[3]{\left(3-2\sqrt{2}\right)\left(3+2\sqrt{2}\right)}\left(\sqrt[3]{3-2\sqrt{2}}+\sqrt[3]{3+2\sqrt{2}}\right)\)
\(=6+3\sqrt[3]{9-8}y=6+3y\)
\(P=x^3+y^3-3\left(x+y\right)+1993\)
\(=18+3x+6+3y-3x-3y+1993=2017\)
Bài 2:
\(x=\sqrt{4+2\sqrt{3}}=\sqrt{3}+1\)
Ta có: \(P=x^2-2x+2020\)
\(=4+2\sqrt{3}-2\left(\sqrt{3}-1\right)+2020\)
\(=4+2\sqrt{3}-2\sqrt{3}+2+2020\)
=2026
Bài 1:
\(A=-\dfrac{3}{4}\cdot\sqrt{9-4\sqrt{5}}\cdot\sqrt{\left(-8\right)^2\cdot\left(2+\sqrt{5}\right)^2}\)
\(=\dfrac{-3}{4}\cdot8\cdot\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)\)
=-6
a: Thay x=36 vào B, ta được:
\(B=\dfrac{36+2}{36+6+1}=\dfrac{38}{43}\)
b: Ta có: \(A=\dfrac{1}{\sqrt{x}-1}-\dfrac{x-\sqrt{x}+3}{x\sqrt{x}-1}\)
\(=\dfrac{x+\sqrt{x}+1-x+\sqrt{x}-3}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\dfrac{2}{x+\sqrt{x}+1}\)
a) \(ĐKXĐ:\hept{\begin{cases}x\ge0\\x\ne9\end{cases}}\)
\(M=\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}+1}{\sqrt{x}-3}+\frac{3-11\sqrt{x}}{9-x}\)
\(=\frac{2\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\frac{11\sqrt{x}-3}{x-9}\)
\(=\frac{2x-6\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\frac{x+4\sqrt{x}+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\frac{11\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{2x-6\sqrt{x}+x+4\sqrt{x}+3+11\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{3x+9\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\frac{3\sqrt{x}.\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\frac{3\sqrt{x}}{\sqrt{x}-3}\)
b) Ta có: \(x=\sqrt{\sqrt{3}-\sqrt{4-2\sqrt{3}}}=\sqrt{\sqrt{3}-\sqrt{3-2\sqrt{3}+1}}\)
\(=\sqrt{\sqrt{3}-\sqrt{\left(\sqrt{3}-1\right)^2}}=\sqrt{\sqrt{3}-\left|\sqrt{3}-1\right|}\)
\(=\sqrt{\sqrt{3}-\sqrt{3}+1}=\sqrt{1}=1\)( thỏa mãn ĐKXĐ )
Thay \(x=1\)vào M ta được:
\(M=\frac{3\sqrt{1}}{\sqrt{1}-3}=\frac{3}{1-3}=\frac{-3}{2}\)
c) \(M=\frac{3\sqrt{x}}{\sqrt{x}-3}=\frac{3\sqrt{x}-9+9}{\sqrt{x}-3}=\frac{3\left(\sqrt{x}-3\right)+9}{\sqrt{x}-3}=3+\frac{9}{\sqrt{x}-3}\)
Vì \(x\inℕ\)\(\Rightarrow\)Để M là số tự nhiên thì \(\frac{9}{\sqrt{x}-3}\inℕ\)
\(\Rightarrow9⋮\left(\sqrt{x}-3\right)\)\(\Rightarrow\sqrt{x}-3\inƯ\left(9\right)\)(1)
Vì \(x\ge0\)\(\Rightarrow\sqrt{x}\ge0\)\(\Rightarrow\sqrt{x}-3\ge-3\)(2)
Từ (1) và (2) \(\Rightarrow\sqrt{x}-3\in\left\{-3;-1;1;3;9\right\}\)
\(\Rightarrow\sqrt{x}\in\left\{0;2;4;6;12\right\}\)\(\Rightarrow x\in\left\{0;4;16;36;144\right\}\)( thỏa mãn ĐKXĐ )
Thử lại với \(x=4\)ta thấy M không là số tự nhiên
Vậy \(x\in\left\{0;16;36;144\right\}\)
x3 + y3 - 3(x +y) +2020 nha các cậu
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 \(A=x^3-3x+y^3-3y+2020=18+6+2020=2048\)