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\(A=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2+4\sqrt{ab}}{\sqrt{a}+\sqrt{b}}=\dfrac{a-2\sqrt{ab}+b+4\sqrt{ab}}{\sqrt{a}+\sqrt{b}}=\dfrac{a+2\sqrt{ab}+b}{\sqrt{a}+\sqrt{b}}=\dfrac{\left(\sqrt{a}+\sqrt{b}\right)^2}{\sqrt{a}+\sqrt{b}}=\sqrt{a}+\sqrt{b}\)
Cách 1 :\(A=\sqrt{6-2\sqrt{5}}-\sqrt{6+2\sqrt{5}}\)
\(=\sqrt{\sqrt{5}^2-2\sqrt{5}+\sqrt{1}^2}-\sqrt{\sqrt{5}^2+2\sqrt{5}+\sqrt{1}^2}\)
\(=\sqrt{\left(\sqrt{5}-\sqrt{1}\right)^2}-\sqrt{\left(\sqrt{5}+\sqrt{1}\right)^2}\)
\(=|\sqrt{5}-\sqrt{1}|-|\sqrt{5}+\sqrt{1}|=\sqrt{5}-\sqrt{1}-\sqrt{5}-\sqrt{1}=-2\)
Cách 2 \(A=\sqrt{6-2\sqrt{5}}-\sqrt{6+2\sqrt{5}}\)
\(< =>A^2=6-2\sqrt{5}-6-2\sqrt{5}+2\sqrt{36-20}\)
\(< =>A^2=8-2\sqrt{5}-2\sqrt{5}=8-2\left(2\sqrt{5}\right)=8-4\sqrt{5}\)
<=>...
\(B=\frac{\sqrt{3-2\sqrt{2}}}{\sqrt{17-12\sqrt{2}}}-\frac{\sqrt{3+2\sqrt{2}}}{\sqrt{17+12\sqrt{2}}}\)
\(=\frac{\sqrt{2}-\sqrt{1}}{\sqrt{17-12\sqrt{2}}}-\frac{\sqrt{2}+\sqrt{1}}{\sqrt{17+12\sqrt{2}}}\)
\(=\frac{\left(\sqrt{2}-\sqrt{1}\right)\sqrt{17+12\sqrt{2}}-\left(\sqrt{2}+1\right)\sqrt{17-12\sqrt{2}}}{\sqrt{17^2-\left(12\sqrt{2}\right)^2}}\)
tự làm tiếp đi , mình lười viết
a) Ta có: \(VP=\left(3+\sqrt{6}\right)^2\)
\(=3^2+2\cdot3\cdot\sqrt{6}+\left(\sqrt{6}\right)^2\)
\(=9+6\sqrt{6}+6\)
\(=15+6\sqrt{6}\)≠VP
=> Sai đề rồi bạn
a) \(\sqrt{5+\sqrt{21}}-\sqrt{6-\sqrt{35}}\) = \(\dfrac{\sqrt{10+2\sqrt{21}}}{\sqrt{2}}-\dfrac{\sqrt{12-2\sqrt{35}}}{\sqrt{2}}\)
= \(\dfrac{\sqrt{\left(\sqrt{7}+\sqrt{3}\right)^2}}{\sqrt{2}}-\dfrac{\sqrt{\left(\sqrt{7}-\sqrt{5}\right)^2}}{\sqrt{2}}\)
= \(\dfrac{\sqrt{7}+\sqrt{3}}{\sqrt{2}}-\dfrac{\sqrt{7}-\sqrt{5}}{\sqrt{2}}\) = \(\dfrac{\sqrt{7}+\sqrt{3}-\left(\sqrt{7}-\sqrt{5}\right)}{\sqrt{2}}\)
= \(\dfrac{\sqrt{7}+\sqrt{3}-\sqrt{7}+\sqrt{5}}{\sqrt{2}}=\dfrac{\sqrt{5}+\sqrt{3}}{\sqrt{2}}\)
câu b) hình như đề sai
\(\sqrt{9-6\sqrt{6}+6}+\sqrt{9+6\sqrt{6}+6}\\ =\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(3+\sqrt{6}\right)^2}=6\)
sửa lại lúc nhìm nhầm
\(\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{24-12\sqrt{6}+3}\\ =3-\sqrt{6}+2\sqrt{6}-3=\sqrt{6}\)