Đáp án:1a)\(-6(1+\sqrt{5})\)
b)2
2a)\(\frac{4(2\sqrt{x}+1)}{(2-\sqrt{x})(2+\sqrt{x})}\)
Giải thích các bước giải:
Câu 1:a)\(A=\frac{4}{\sqrt{5}+3}-\sqrt{20}=\frac{4}{\sqrt{5}+3}-2\sqrt{5}=4-10-6\sqrt{5}=-6-6\sqrt{5}\)
=\(-6(1+\sqrt{5})\)
b)\((1+\sqrt{3})\sqrt{4-2\sqrt{5}}=(1+\sqrt{3})\sqrt{3-2\sqrt{3}+1}=(1+\sqrt{3})\sqrt{(\sqrt{3}-1)^{2}}\)
=\((1+\sqrt{3})|\sqrt{3}-1|=(1+\sqrt{3})(\sqrt{3}-1)=3-1=2\)
Câu 2:
a)đk:\(x\neq4\)
\(\frac{2+\sqrt{x}}{2-\sqrt{x}}-\frac{2-\sqrt{x}}{2+\sqrt{x}}-\frac{4}{x-4}\)
=\(\frac{2+\sqrt{x}}{2-\sqrt{x}}-\frac{2-\sqrt{x}}{2+\sqrt{x}}+\frac{4}{4-x}\)
=\(\frac{(2+\sqrt{x})(2+\sqrt{x})-(2-\sqrt{x})(2-\sqrt{x})+4}{(2-\sqrt{x})(2+\sqrt{x})}\)
=\(\frac{x+4\sqrt{x}+4-x+4\sqrt{x}-4+4}{(2-\sqrt{x})(2+\sqrt{x})}\)
=\(\frac{8\sqrt{x}+4}{(2-\sqrt{x})(2+\sqrt{x})}\)
=\(\frac{4(2\sqrt{x}+1)}{(2-\sqrt{x})(2+\sqrt{x})}\)
