= 1 nhé

19999 : 999 - 19999 : 999 + 1

= (19999 : 999 - 19999 : 999) + 1

= 1

Dễ mà!!

Chúc bn học tốt!!

Không hiểu 

a: Ta có: \(\sqrt{8+2\sqrt{15}}-\sqrt{6+2\sqrt{5}}\)

\(=\sqrt{5}+\sqrt{3}-\sqrt{5}-1\)

\(=\sqrt{3}-1\)

b: Ta có: \(\sqrt{17-2\sqrt{72}}+\sqrt{19+2\sqrt{18}}\)

\(=3-2\sqrt{2}+3\sqrt{2}+1\)

\(=4+\sqrt{2}\)

c: Ta có: \(\sqrt{12-2\sqrt{32}}+\sqrt{9+4\sqrt{2}}\)

\(=2\sqrt{2}-2+2\sqrt{2}+1\)

\(=4\sqrt{2}-1\)

a)

\(\sqrt{8+2\sqrt{15}}-\sqrt{6+2\sqrt{5}}\\ =\sqrt{5+2\sqrt{5}\cdot\sqrt{3}+3}-\sqrt{5+2\sqrt{5}\cdot\sqrt{1}+1}\\ =\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{5}+\sqrt{1}\right)^2}\\ =\sqrt{5}+\sqrt{3}-\sqrt{5}-\sqrt{1}\\ =\sqrt{3}-\sqrt{1}\)

b)

\(\sqrt{17-2\sqrt{72}}+\sqrt{19+2\sqrt{18}}\\ =\sqrt{9-2\sqrt{9}\cdot\sqrt{8}+8}+\sqrt{18+2\sqrt{18}\cdot\sqrt{1}+1}\\ =\sqrt{\left(3-2\sqrt{2}\right)^2}+\sqrt{\left(3\sqrt{2}+1\right)^2}\\ =3-2\sqrt{2}+3\sqrt{2}+1\\ =4+\sqrt{2}\)

c)

\(\sqrt{12-2\sqrt{32}}+\sqrt{9+4\sqrt{2}}\\ =\sqrt{8-2\sqrt{8}\cdot\sqrt{4}+4}+\sqrt{8+2\sqrt{8}\cdot\sqrt{1}+1}\\ =\sqrt{\left(2\sqrt{2}-2\right)^2}+\sqrt{\left(2\sqrt{2}+1\right)^2}\\ =2\sqrt{2}-2+2\sqrt{2}+1\\ =4\sqrt{2}-1\)

2016996

tích nhé tram pham

tinh cung kho so to nam

Ta có: \(A=\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-1}-\dfrac{\sqrt{a}-1}{\sqrt{a}+1}+4\sqrt{a}\right)\left(\sqrt{a}-\dfrac{1}{\sqrt{a}}\right)\)

\(=\dfrac{a+2\sqrt{a}+1-a+2\sqrt{a}-1+4\sqrt{a}\left(a-1\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\cdot\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\sqrt{a}}\)

\(=\dfrac{4\sqrt{a}\left(1+a-1\right)}{\sqrt{a}}\)

\(=4a\)

Để \(\sqrt{A}>A\) thì \(\sqrt{4a}>4a\)

\(\Leftrightarrow2\sqrt{a}-4a>0\)

\(\Leftrightarrow2\sqrt{a}\left(1-2\sqrt{a}\right)>0\)

\(\Leftrightarrow2\sqrt{a}< 1\)

\(\Leftrightarrow a< \dfrac{1}{4}\)

Kết hợp ĐKXĐ, ta được: \(0< a< \dfrac{1}{4}\)

\(A=\left(\frac{\left(\sqrt{a}+1\right)^2-\left(\sqrt{a}-1\right)^2}{a-1}+4\sqrt{a}\right)\left(\frac{a+1}{\sqrt{a}}\right)\)

\(A=\left(\frac{4\sqrt{a}}{a-1}+\frac{4\sqrt{a}\left(a-1\right)}{a-1}\right)\left(\frac{a+1}{\sqrt{a}}\right)\)

\(A=\frac{4a\sqrt{a}}{a-1}.\frac{a+1}{\sqrt{a}}=\frac{4a\left(a+1\right)}{a-1}\)

....... Tới đây được chưa bạn? 

\(A=\sqrt{4+\sqrt{7}}-\sqrt{4+\sqrt{7}}\Leftrightarrow\sqrt{2}A=\sqrt{8+2\sqrt{7}}-\sqrt{8+2\sqrt{7}}\)

\(\Leftrightarrow\sqrt{2}A=\sqrt{\sqrt{7}^2+2\sqrt{7}+1}-\sqrt{\sqrt{7}^2+2\sqrt{7}+1}\)

\(\Leftrightarrow\sqrt{2}A=\sqrt{7}+1-\sqrt{7}-1=0\)

\(\Leftrightarrow A=0\)

=0

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