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Câu hỏi của ʚĭɞ Thị Quyên ʚĭɞ - Toán lớp 9 | Học trực tuyến
Từ \(\left(x+\sqrt{x^2+3}\right)\left(y+\sqrt{y^2+3}\right)=3\)
\(\Leftrightarrow\left(x+\sqrt{x^2+3}\right)\left(y+\sqrt{y^2+3}\right)\left(\sqrt{x^2+3}-x\right)=3\left(\sqrt{x^2+3}-x\right)\)
\(\Leftrightarrow y+\sqrt{y^2+3}=\sqrt{x^2+3}-x\)
Tương tự \(x+\sqrt{x^2+3}=\sqrt{y^2+3}-y\)
Cộng theo vế ta có: \(2\left(x+y\right)=0\)
\(\Leftrightarrow E=0\)
\(\left(x+\sqrt{x^2+3}\right)\left(x-\sqrt{x^2+3}\right)\left(y+\sqrt{y^2+3}\right)=3\left(x-\sqrt{x^2-3}\right)\)
\(\Leftrightarrow\left(x^2-x^2+3\right)\left(y+\sqrt{y^2+3}\right)=3\left(x-\sqrt{x^2+3}\right)\)
\(\Leftrightarrow y+\sqrt{y^2+3}=x-\sqrt{x^2+3}\) (1)
Tương tự \(x+\sqrt{x^2+3}=y-\sqrt{y^2+3}\) (2)
Từ (1) và (2)\(\Rightarrow x+y=0\)
\(Sửa:\left(x+\sqrt{x^2+3}\right)\left(y+\sqrt{y^2+3}\right)=3\\ \Leftrightarrow\left(x+\sqrt{x^2+3}\right)\left(x-\sqrt{x^2+3}\right)\left(y+\sqrt{y^2+3}\right)=3\left(x-\sqrt{x^2+3}\right)\\ \Leftrightarrow\left(x^2-x^2-3\right)\left(y+\sqrt{y^2+3}\right)=3\left(x-\sqrt{x^2+3}\right)\\ \Leftrightarrow-3\left(y+\sqrt{y^2+3}\right)=-3\left(\sqrt{x^2+3}-x\right)\\ \Leftrightarrow y+\sqrt{y^2+3}=\sqrt{x^2+3}-x\)
Cmtt: \(x+\sqrt{x^2+3}=\sqrt{y^2+3}-y\)
Cộng vế theo vế:
\(\Leftrightarrow x+\sqrt{x^2+3}+y+\sqrt{y^2+3}=\sqrt{x^2+3}-x+\sqrt{y^2+3}-y\\ \Leftrightarrow x+y=-x-y\\ \Leftrightarrow2\left(x+y\right)=0\Leftrightarrow x+y=0\)
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a. \(\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)=x-3\sqrt{x} +2\sqrt{x}-6=x-\sqrt{x}-6\)
b. \(\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)=x-y\)
c. \(\left(\sqrt{\dfrac{25}{3}}-\sqrt{\dfrac{49}{3}}+\sqrt{3}\right).\sqrt{3}\)
\(=\left(\dfrac{5}{\sqrt{3}}-\dfrac{7}{\sqrt{3}}+\sqrt{3}\right).\sqrt{3}=\dfrac{5}{3}-\dfrac{7}{3}+9=\dfrac{25}{3}\)
d. \(\left(1+\sqrt{3}-\sqrt{5}\right)\left(1+\sqrt{3}+\sqrt{5}\right)\)
\(=\left(1+\sqrt{3}\right)^2-5=1+2\sqrt{3}+3-5=2\sqrt{3}-1\)
Có \(x^3=3+2\sqrt{2}-3\sqrt[3]{\left(3+2\sqrt{2}\right)\left(3-2\sqrt{2}\right)}\left(\sqrt[3]{3+2\sqrt{2}}-\sqrt[3]{3-2\sqrt{2}}\right)-\left(3-2\sqrt{2}\right)\)
\(\Leftrightarrow x^3=4\sqrt{2}-3x\) \(\Leftrightarrow x^3+3x=4\sqrt{2}\) (1)
Có \(y^3=17+12\sqrt{2}-3\sqrt[3]{\left(17+12\sqrt{2}\right)\left(17-12\sqrt{2}\right)}\left(\sqrt[3]{17+12\sqrt{2}}-\sqrt[3]{17-12\sqrt{2}}\right)-\left(17-12\sqrt{2}\right)\)
\(\Leftrightarrow y^3=24\sqrt{2}-3y\) \(\Leftrightarrow y^3+3y=24\sqrt{2}\) (2)
Từ (1) (2)\(\Rightarrow x^3+3x-y^3-3y=-20\sqrt{2}\)
Có \(M=\left(x-y\right)^3+3\left(x-y\right)\left(xy+1\right)=\left(x-y\right)\left[\left(x-y\right)^2+3\left(xy+1\right)\right]\)
\(=\left(x-y\right)\left(x^2+xy+y^2+3\right)=x^3-y^3+3\left(x-y\right)=-20\sqrt{2}\)
Vậy \(M=-20\sqrt{2}\)
theo bài ra
\(x=\sqrt[3]{3+2\sqrt{2}}-\sqrt[3]{3-2\sqrt{2}}\)
\(=>x^3=\left(\sqrt[3]{3+2\sqrt{2}}-\sqrt[3]{3-2\sqrt{2}}\right)^3\)
\(x^3=4\sqrt{2}-3\left[\left(\sqrt[3]{3+2\sqrt{2}}\right)\left(\sqrt[3]{3-2\sqrt{2}}\right)\right]\left[\sqrt[3]{3+2\sqrt{2}}-\sqrt[3]{3-2\sqrt{2}}\right]\)
\(x^3=4\sqrt{2}-3\left[\sqrt[3]{\left(3+2\sqrt{2}\right)\left(3-2\sqrt{2}\right)}\right].x\)
\(x^3=4\sqrt{2}-3.\left[\sqrt[3]{9-\left(2\sqrt{2}\right)^2}\right]x\)
\(x^3=4\sqrt{2}-3.1x\)
\(x^3=4\sqrt{2}-3x\)
\(< =>x^3+3x-4\sqrt{2}=0\)
rồi làm y tương tự rồi thế vào M là ra
Ta có: \(\left(x+\sqrt{x^2+3}\right)\left(\sqrt{x^2+3}-x\right)=3\)
\(\left(y+\sqrt{y^2+3}\right)\left(\sqrt{y^2+3}-y\right)=3\)
Kết hợp với giả thiết ta có:
\(\sqrt{x^2+3}-x=y+\sqrt{y^2+3}\)
\(\sqrt{y^2+3}-y=x+\sqrt{x^2+3}\)
Cộng theo vế ta được: \(-\left(x+y\right)=x+y\)
\(\Rightarrow\)\(E=x+y=0\)
\(\left(x+\sqrt{x^2+3}\right)\left(y+\sqrt{y^2+3}\right)=3\)
\(\Leftrightarrow\left(x+\sqrt{x^2+3}\right)\left(x-\sqrt{x^2+3}\right)\left(y+\sqrt{y^2+3}\right)=3\left(x-\sqrt{x^2+3}\right)\)
\(\Leftrightarrow\left(x^2-x^2-3\right)\left(y+\sqrt{y^2+3}\right)=3\left(x-\sqrt{x^2+3}\right)\)
\(\Leftrightarrow-3\left(y+\sqrt{y^2+3}\right)=3\left(x-\sqrt{x^2+3}\right)\)
\(\Leftrightarrow-y-\sqrt{y^2+3}=x-\sqrt{x^2+3}\)(*)
Tương tự, nhân mỗi vế vs \(y-\sqrt{y^2+3}\), ta được:
\(-x-\sqrt{x^2+3}=y-\sqrt{y^2+3}\)(**)
Cộng (*) và (**) suy ra :
\(-y-x-\sqrt{y^2+3}-\sqrt{x^2+3}=x+y-\sqrt{x^2+3}-\sqrt{y^2+3}\)
\(\Leftrightarrow-y-x=x+y\Leftrightarrow2\left(x+y\right)=0\Leftrightarrow x+y=0\)
Vậy \(E=0.\)