Giải pt : A. √3 x – √2 x = √2 +√3 B. 3√5 x + √20 x – √5 = 0 C. √2 phần 3 x – √6 =0 30/07/2021 Bởi Maya Giải pt : A. √3 x – √2 x = √2 +√3 B. 3√5 x + √20 x – √5 = 0 C. √2 phần 3 x – √6 =0
a/ \(\sqrt 3x-\sqrt 2x=\sqrt 2+\sqrt 3\\↔x(\sqrt 3-\sqrt 2)=\sqrt 2+\sqrt 3\\↔x=\dfrac{\sqrt 2+\sqrt 3}{\sqrt 3-\sqrt 2}\)\\↔x=5+2\sqrt 6\) Vậy \(S=\{5+2\sqrt 6\}\) b/ \(3\sqrt 5 x+\sqrt {20} x-\sqrt 5=0\\↔x(3\sqrt 5+\sqrt{20})=\sqrt 5\\↔x=\dfrac{\sqrt 5}{3\sqrt 5+\sqrt{20}}\\↔x=0,2\) Vậy \(S=\{0,2\}\) c/ \(\sqrt{\dfrac{2}{3}}x-\sqrt 6=0\\↔\sqrt{\dfrac{2}{3}}x=\sqrt 6\\↔x=3\) Vậy \(S=\{3\}\) Bình luận
a/ \(\sqrt 3x-\sqrt 2x=\sqrt 2+\sqrt 3\\↔x(\sqrt 3-\sqrt 2)=\sqrt 2+\sqrt 3\\↔x=\dfrac{\sqrt 2+\sqrt 3}{\sqrt 3-\sqrt 2}\)\\↔x=5+2\sqrt 6\)
Vậy \(S=\{5+2\sqrt 6\}\)
b/ \(3\sqrt 5 x+\sqrt {20} x-\sqrt 5=0\\↔x(3\sqrt 5+\sqrt{20})=\sqrt 5\\↔x=\dfrac{\sqrt 5}{3\sqrt 5+\sqrt{20}}\\↔x=0,2\)
Vậy \(S=\{0,2\}\)
c/ \(\sqrt{\dfrac{2}{3}}x-\sqrt 6=0\\↔\sqrt{\dfrac{2}{3}}x=\sqrt 6\\↔x=3\)
Vậy \(S=\{3\}\)