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a,
\(=\sqrt{\left(\sqrt{5}\right)^2-2\sqrt{5}+1}=\sqrt{\left(\sqrt{5}-1\right)^2}=\left|\sqrt{5}-1\right|=\sqrt{5}-1\)
b, \(\sqrt{8-\sqrt{60}}=\sqrt{8-\sqrt{4.15}}=\sqrt{8-2\sqrt{15}}\)
\(=\sqrt{8-2\sqrt{3}\sqrt{5}}=\sqrt{\left(\sqrt{5}\right)^2-2\sqrt{3}\sqrt{5}+\left(\sqrt{3}\right)^2}\)
\(=\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}=\left|\sqrt{5}-\sqrt{3}\right|=\sqrt{5}-\sqrt{3}\)
2 câu cuối tự làm nhé
\(\frac{\sqrt{3}+\sqrt{7}}{\sqrt{3}-\sqrt{7}}+\frac{\sqrt{3}-\sqrt{7}}{\sqrt{3}+\sqrt{7}}\)
\(=\frac{\left(\sqrt{3}+\sqrt{7}\right)\left(\sqrt{3}+\sqrt{7}\right)+\left(\sqrt{3}-\sqrt{7}\right)\left(\sqrt{3}-\sqrt{7}\right)}{\left(\sqrt{3}-\sqrt{7}\right)\left(\sqrt{3}+\sqrt{7}\right)}\)
\(=\frac{\left(\sqrt{3}+\sqrt{7}\right)^2+\left(\sqrt{3}-\sqrt{7}\right)^2}{3-7}\)
\(=\frac{3+2\sqrt{3}.\sqrt{7}+7+3-2\sqrt{3}.\sqrt{7}+7}{-4}\)
\(=\frac{3+7+3+7}{-4}\)
\(=\frac{20}{-4}=-5\)
:) trình bày các bước đi bạn :)) ai lại làm thế :v Bấm casio à :)
\(H=\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}\)
\(H^2=4+\sqrt{7}+4-\sqrt{7}+2\sqrt{\left(4+\sqrt{7}\right)\left(4-\sqrt{7}\right)}\)
\(=8-2\sqrt{16-7}=8-6=2\)
\(\Rightarrow H=\sqrt{2}\Rightarrow\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}-2=0\)
Vậy .....................
\(A=\frac{\sqrt{2}-\sqrt{1}}{\left(\sqrt{2}-\sqrt{1}\right)\left(\sqrt{2}+\sqrt{1}\right)}+.......+\frac{\sqrt{n}-\sqrt{n-1}}{\left(\sqrt{n}-\sqrt{n-1}\right)\left(\sqrt{n}+\sqrt{n}-1\right)}\)
\(=\frac{\sqrt{2}-\sqrt{1}}{2-1}+........+\frac{\sqrt{n}-\sqrt{n-1}}{n-\left(n-1\right)}\)
\(=\sqrt{2}-\sqrt{1}+...........+\sqrt{n}-\sqrt{n-1}\)
\(=\sqrt{n}-\sqrt{1}=\sqrt{n}-1\)
bài B tương tự
a, \(M=\sqrt{x^2-4x+4}-\sqrt{x^2+4x+4}\) (ĐK : \(\forall x\in R\))
\(=\sqrt{\left(x-2\right)^2}-\sqrt{\left(x+2\right)^2}\)
* Nếu x\(\ge2\Rightarrow M=x-2-x-2=-4\)
*Nếu x<2 => M=2-x-x-2=-2x
b,Để M=2\(\ne-4\)
=>M=-2x
=>-2x=-4
=>x=2
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P=\(\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}\)
\(=\sqrt{x-1+2\sqrt{x-1}+1}+\sqrt{x-1-2\sqrt{x-1}+1}\)
\(=\sqrt{\left(\sqrt{x-1}+1\right)^2}+\sqrt{\left(\sqrt{x-1}-1\right)^2}\)
* Nếu \(x\ge2\Rightarrow P=\sqrt{x-1}+1+\sqrt{x-1}-1=2\sqrt{x-1}\)
* Nếu x<2 =>P=\(\sqrt{x-1}+1+1-\sqrt{x-1}=2\)
VẬY.......
Tk nha!
a, \(\sqrt{11-2\sqrt{10}}=\sqrt{\left(\sqrt{10}\right)^2-2\sqrt{10}+1}=\sqrt{\left(\sqrt{10}+1\right)^2}\)
\(=\left|\sqrt{10}+1\right|=\sqrt{10}+1\)
b, \(\sqrt{27-10\sqrt{2}}=\sqrt{5^2-10\sqrt{2}+\left(\sqrt{2}\right)^2}=\sqrt{\left(5-\sqrt{2}\right)^2}\)
\(=\left|5-\sqrt{2}\right|=5-\sqrt{2}\)
c, \(\sqrt{4+2\sqrt{3}}=\sqrt{\left(\sqrt{3}\right)^2+2\sqrt{3}+1}=\sqrt{\left(\sqrt{3}+1\right)^2}\)
\(=\left|\sqrt{3}+1\right|=\sqrt{3}+1\)
làm nốt 2 câu cuối nhé, cách làm y trên
d/\(\sqrt{9+4\sqrt{5}}\)
= \(\sqrt{2^2+4\sqrt{5}+\left(\sqrt{5}\right)^2}\)
=\(\sqrt{\left(2+\sqrt{5}\right)^2}\)
= \(\left|2+\sqrt{5}\right|\)
= \(2+\sqrt{5}\)
e/ \(\sqrt{21+4\sqrt{5}}\)
= \(\sqrt{20+4\sqrt{5}+1}\)
=\(\sqrt{\left(2\sqrt{5}\right)^2+2.2\sqrt{5}+1^2}\)
=\(\sqrt{\left(2\sqrt{5}+1\right)^2}\)
= \(\left|2\sqrt{5}+1\right|\)
= \(2\sqrt{5}+1\)