Trục căn thức ở mẫu biểu thức 4 3 x + 2 y với x,y ≥ 0; a ≠ 4 9 y ta được:
A. 3 x - 2 y 9 x - 4 y
B. 12 x - 8 y 3 x + 2 y
C. 12 x + 8 y 9 x + 4 y
D. 12 x - 8 y 9 x - 4 y
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a, \(\dfrac{2}{\sqrt{5}-1}=\dfrac{2\left(\sqrt{5}+1\right)}{\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)}=\dfrac{2\left(\sqrt{5}+1\right)}{5-1}=\dfrac{\sqrt{5}+1}{2}\)
b, \(\left\{{}\begin{matrix}x-y=4\\2x+3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-\dfrac{3}{2}-y=4\\x=-\dfrac{3}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{11}{2}\\y=-\dfrac{3}{2}\end{matrix}\right.\)
a,\(\dfrac{2}{\left(\sqrt{5}-1\right)}=\dfrac{2\left(\sqrt{5}+1\right)}{\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)}=\dfrac{2\sqrt{5} +2}{5-1}=\dfrac{2\left(\sqrt{5}+1\right)}{4}=\dfrac{\sqrt{5}+1}{2}\)
b,\(\left\{{}\begin{matrix}x-y=4\\2x+3=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-y=4\\x=\dfrac{-3}{2}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{-11}{2}\\x=\dfrac{-3}{2}\end{matrix}\right.\)
Vậy hệ phương trình có nghiệm duy nhất là (x;y)=\(\left(\dfrac{-3}{2};\dfrac{-11}{2}\right)\)
-Chúc bạn học tốt-
a) \(\dfrac{14}{2\sqrt{3}-\sqrt{5}}\)
\(=\dfrac{14\left(2\sqrt{3}+\sqrt{5}\right)}{\left(2\sqrt{3}-\sqrt{5}\right)\left(2\sqrt{3}+\sqrt{5}\right)}\)
\(=\dfrac{14\left(2\sqrt{3}+\sqrt{5}\right)}{\left(2\sqrt{3}\right)^2-\sqrt{5^2}}=\dfrac{14\left(2\sqrt{3}+\sqrt{5}\right)}{12-5}\)
\(=\dfrac{14\left(2\sqrt{3}+\sqrt{5}\right)}{7}=2\left(2\sqrt{3}+\sqrt{5}\right)\)
\(=4\sqrt{3}+2\sqrt{5}\)
b) \(\dfrac{x^2-y}{x-\sqrt{y}}=\dfrac{\left(x-\sqrt{y}\right)\left(x+\sqrt{y}\right)}{x-\sqrt{y}}=x+\sqrt{y}\)
a, \(\frac{a}{\sqrt{a}}=\sqrt{a}\)
b, \(\frac{a}{\sqrt{ab}}=\frac{\sqrt{a}}{\sqrt{b}}=\frac{\sqrt{ab}}{b}\)
c, \(\frac{x}{\sqrt{3x^3}}=\frac{x}{x\sqrt{3x}}=\frac{1}{\sqrt{3x}}=\frac{\sqrt{3x}}{3x}\)
d, \(\frac{4y^2}{\sqrt{2y^5}}=\frac{4y^2}{y^2\sqrt{2y}}=\frac{4}{\sqrt{2y}}=\frac{4\sqrt{2y}}{2y}=\frac{2\sqrt{2y}}{y}\)
a)\(\dfrac{a}{\sqrt{a}}=\dfrac{a\sqrt{a}}{a}=\sqrt{a}\) b) \(\dfrac{a}{\sqrt{ab}}=\dfrac{a\sqrt{ab}}{\left(\sqrt{ab}\right)^2}=\dfrac{a\sqrt{ab}}{ab}=\dfrac{\sqrt{ab}}{b}\) c) \(\dfrac{x}{\sqrt{3x^3}}=\dfrac{x\sqrt{3x}}{\sqrt{3x^3.\sqrt{3x}}}=\dfrac{x\sqrt{3x}}{\left(\sqrt{3x^2}\right)^2}=\dfrac{x\sqrt{3x}}{\left(3x^2\right)^2}=\dfrac{x\sqrt{3x}}{3x^2}=\dfrac{\sqrt{3x}}{3x}\)