khỏi chứng minh
căn có cả dấu = nằm bên trong , mà còn dấu - của 4 , = - 4 ???

viết lại câu hỏi nhé !

\(x=\sqrt{7+4\sqrt{3}}\)

\(x=\sqrt{4+4\sqrt{3}+3}\)

\(x=\sqrt{\left(2+\sqrt{3}\right)^2}\)

\(x=2+\sqrt{3}\)

Ta có ;

A=\(\sqrt{3}-\sqrt{x^2-4x+4}\)

A=\(\sqrt{3}-\sqrt{\left(x-2\right)^2}\)

Thay x vào A ta có 

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

A=\(\sqrt{3}-\sqrt{\left(\sqrt{3}\right)^2}\)

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

A=0

Vậy A=0 <=>x=\(\sqrt{7+4\sqrt{3}}\)

\(2\sqrt{48}-2\sqrt{2}\sqrt{14}-8\dfrac{\sqrt{24}}{\sqrt{8}}+\sqrt{63}\)

\(=2\sqrt{16.3}-2\sqrt{4.7}-8\dfrac{\sqrt{8.3}}{\sqrt{8}}+\sqrt{9.7}\)

\(=2.4\sqrt{3}-2.2\sqrt{7}-8\sqrt{3}+3\sqrt{7}\)

\(=8\sqrt{3}-4\sqrt{7}-8\sqrt{3}+3\sqrt{7}\)

\(=-\sqrt{7}\)

đáp số = - 3 \(\sqrt{ }\)3

48 = 2 ^4. 3
14 = 7. 2
63 = 3 ². 7

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

\(=\left(-\sqrt{a}+\sqrt{2}\right)\left(\sqrt{2}+\sqrt{a}\right)\)

\(=2-a\)

lấy\(\sqrt{ }\) a và \(\sqrt{ }\)2 ra ngoài , trong ngoặc sẽ là :

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

`a)`\(D=\dfrac{a+1}{\sqrt{a}}+\dfrac{a\sqrt{a}-1}{a-\sqrt{a}}+\dfrac{a^2-a\sqrt{a}+\sqrt{a}-1}{\sqrt{a}-a\sqrt{a}}\);\(a\ge0\)

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

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

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

\(D=\dfrac{2a+\sqrt{a}+2}{\sqrt{a}}-\dfrac{a-\sqrt{a}+1}{\sqrt{a}}\)

\(D=\dfrac{2a+\sqrt{a}+2-a+\sqrt{a}-1}{\sqrt{a}}\)

\(D=\dfrac{a+2\sqrt{a}+1}{\sqrt{a}}\)

\(D=\dfrac{\left(\sqrt{a}+1\right)^2}{\sqrt{a}}\)

\(\dfrac{1}{5}\sqrt{25.2}-2\sqrt{16.6}-\sqrt{2}+\sqrt{\dfrac{144}{6}}=\sqrt{2}-8\sqrt{6}-\sqrt{2}+2\sqrt{6}=-6\sqrt{6}\)

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