\(a,\sqrt{9}-4\sqrt{5}-\sqrt{5}=3-3\sqrt{5}\)

\(b,\sqrt{3}-2\sqrt{2}-\sqrt{3}+2\sqrt{2}=0\)

a) \(\sqrt{9}-4\sqrt{5}-\sqrt{5}=3-5\sqrt{5}\)

b) \(\sqrt{3}-2\sqrt{2}-\sqrt{3}+2\sqrt{2}=0\)

c) \(\sqrt{11}-6\sqrt{2}+3+\sqrt{2}=\sqrt{11}-5\sqrt{2}+3\)

​\(a,\sqrt{9}-4\sqrt{5}-\sqrt{5}=\sqrt{3^2}-4\sqrt{5}-\sqrt{5}=3-5\sqrt{5}\)​

\(b,\sqrt{3}-2\sqrt{2}-\sqrt{3}+2\sqrt{2}=0\)

\(c,\sqrt{11}-6\sqrt{2}+3+\sqrt{2}=\sqrt{11}-5\sqrt{2}+3\)

\(=\left(1-\dfrac{\sqrt{5}\left(\sqrt{5}-1\right)}{\sqrt{5}-1}\right)\left(\dfrac{\sqrt{5}\left(\sqrt{5}+1\right)}{\sqrt{5}+1}+1\right)\)

=(1-căn 5)(1+căn 5)

=1-5

=-4

\(\left(1+\dfrac{5-\sqrt{5}}{1-\sqrt{5}}\right)\left(\dfrac{5+\sqrt{5}}{1+\sqrt{5}}+1\right)\)

\(=\left(1-\dfrac{\sqrt{5}\left(\sqrt{5}-1\right)}{\sqrt{5}-1}\right)\left(\dfrac{\sqrt{5}\left(\sqrt{5}+1\right)}{\sqrt{5}+1}+1\right)\)

\(=\left(1-\sqrt{5}\right)\left(\sqrt{5}+1\right)\)

\(=1^2-\left(\sqrt{5}\right)^2\)

\(=1-5\)

\(=-4\)

`(1+(5-\sqrt{5})/(1-\sqrt{5})).((5+\sqrt{5})/(1+\sqrt{5})+1)`

`=[1-(\sqrt{5}(\sqrt{5}-1))/(\sqrt{5}-1)].[(\sqrt{5}(\sqrt{5}+1))/(1+\sqrt{5})+1]`

`=(1-\sqrt{5})(1+\sqrt{5})`

`=1-5`

`=-4`

a: \(\dfrac{\sqrt{5}}{\sqrt{7}}=\dfrac{\sqrt{5\cdot7}}{7}=\dfrac{\sqrt{35}}{7}\)

b: \(\dfrac{2}{\sqrt{a}-1}=\dfrac{2\left(\sqrt{a}+1\right)}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}=\dfrac{2\sqrt{a}+2}{a-1}\)

\(\dfrac{1}{1-\sqrt{5}}+\dfrac{1}{\sqrt{5}-1}=\dfrac{-1+1}{\sqrt{5}-1}=\dfrac{0}{\sqrt{5}-1}=0\)

\(\dfrac{1}{1-\sqrt{5}}+\dfrac{1}{\sqrt{5}-1}=\dfrac{1}{1-\sqrt{5}}-\dfrac{1}{1-\sqrt{5}}=0\)

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: \(\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}=\sqrt{\left(2-\sqrt{3}\right)^2}=2-\sqrt{3}\)

Cho hàm số y=(1-√5)x-1
a, Hàm số đồng biến hay nghịch biến trên R?vì sao
Hàm số nghịch biến vi (1-√5<0
b,Tính y khi x=1+√5
y=(1-√5)(1+√5)-1
y = -5