rút gọn các biểu thức sau a, 9 căn5 + 3căn20 - 7căn45 b, 2căn6 + căn40 trên căn3 + căn5
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a: \(\dfrac{5+2\sqrt{5}}{\sqrt{5}+\sqrt{2}}=\dfrac{\left(5+2\sqrt{5}\right)\left(\sqrt{5}-\sqrt{2}\right)}{3}=\dfrac{5\sqrt{5}-5\sqrt{2}+10-2\sqrt{10}}{3}\)
b: \(\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}=\sqrt{\left(2-\sqrt{3}\right)^2}=2-\sqrt{3}\)
\(x^2-\left(\sqrt{3}+\sqrt{5}\right).x+\sqrt{3}.\sqrt{5}=0\)
\(\Leftrightarrow x^2-\sqrt{3}.x-\sqrt{5}.x+\sqrt{3}.\sqrt{5}=0\)
\(\Leftrightarrow x^2-\sqrt{3}.x-\sqrt{5}.x+\sqrt{3}.\sqrt{5}=0\)
\(\Leftrightarrow x\left(x-\sqrt{3}\right)-\sqrt{5}\left(x-\sqrt{3}\right)=0\)
\(\Leftrightarrow\left(x-\sqrt{5}\right)\left(x-\sqrt{3}\right)=0\)
\(\Leftrightarrow\int^{x-\sqrt{5}=0}_{x-\sqrt{3}=0}\Leftrightarrow\int^{x=\sqrt{5}}_{x=\sqrt{3}}\)
Vậy x \(\in\left\{\sqrt{3};\sqrt{5}\right\}\)
\(\dfrac{1}{1-\sqrt{5}}+\dfrac{1}{\sqrt{5}-1}=\dfrac{-1+1}{\sqrt{5}-1}=\dfrac{0}{\sqrt{5}-1}=0\)
\(\dfrac{1}{1-\sqrt{5}}+\dfrac{1}{\sqrt{5}-1}=\dfrac{1}{1-\sqrt{5}}-\dfrac{1}{1-\sqrt{5}}=0\)
\(<=>x^2-\sqrt{3}x-\sqrt{5}x+\sqrt{15}=0<=>x\left(x-\sqrt{3}\right)-\sqrt{5}\left(x-\sqrt{3}\right)=0<=>\left(x-\sqrt{3}\right)\left(x-\sqrt{5}\right)=0\)
<=>Tự làm
\(\sqrt{3}-\frac{5}{2}>\sqrt{3}-4\text{ vì }-\frac{5}{2}>-4\)
\(\Rightarrow2.\left(\sqrt{3}-\frac{5}{2}\right)>\sqrt{3}-4\)
\(\Rightarrow2.\sqrt{3}-5>\sqrt{3}-4\)
\(a,9\sqrt{5}+3\sqrt{20}-7\sqrt{45}=9\sqrt{5}+6\sqrt{5}-21\sqrt{5}=-6\sqrt{5}\\ b,\dfrac{2\sqrt{6}+\sqrt{40}}{\sqrt{3}+\sqrt{5}}=\dfrac{2\sqrt{6}+2\sqrt{10}}{\sqrt{3}+\sqrt{5}}\\ =\dfrac{2\sqrt{2}\left(\sqrt{3}+\sqrt{5}\right)\left(\sqrt{5}-\sqrt{3}\right)}{\left(\sqrt{3}+\sqrt{5}\right)\left(\sqrt{5}-\sqrt{3}\right)}=\dfrac{2\sqrt{2}\left(5-3\right)}{5-3}=2\sqrt{2}\)