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Bài 1:
a) \(\frac{2}{\sqrt{3}-1}-\frac{2}{\sqrt{3}+1}\)
\(=\frac{2\left(\sqrt{3}+1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}-\frac{2\left(\sqrt{3}-1\right)}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}\)
\(=\frac{2\left(\sqrt{3}+1\right)}{2}-\frac{2\left(\sqrt{3}-1\right)}{2}\)
\(=\sqrt{3}+1-\left(\sqrt{3}-1\right)=2\)
b) \(\frac{2}{5-\sqrt{3}}+\frac{3}{\sqrt{6}+\sqrt{3}}\)
\(=\frac{2\left(5+\sqrt{3}\right)}{\left(5-\sqrt{3}\right)\left(5+\sqrt{3}\right)}+\frac{3\left(\sqrt{6}-\sqrt{3}\right)}{\left(\sqrt{6}+\sqrt{3}\right)\left(\sqrt{6}-\sqrt{3}\right)}\)
\(=\frac{2\left(5+\sqrt{3}\right)}{2}+\frac{3\left(\sqrt{6}-\sqrt{3}\right)}{3}\)
\(=5+\sqrt{3}+\sqrt{6}-\sqrt{3}=5+\sqrt{6}\)
c) ĐK: \(a\ge0;a\ne1\)
\(\left(1+\frac{a+\sqrt{a}}{1+\sqrt{a}}\right).\left(1-\frac{a-\sqrt{a}}{\sqrt{a}-1}\right)+a\)
\(=\left(1+\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{1+\sqrt{a}}\right).\left(1-\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right)+a\)
\(=\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)+a\)
\(=1-a+a=1\)
mi tích tau tau tích mi xong tau trả lời nka
việt nam nói là làm
Đk: \(-1\le a\le1\)
\(P=\frac{\sqrt{1+\sqrt{1-a^2}}\left(\sqrt{\left(1+a\right)^3}-\sqrt{\left(1-a\right)^3}\right)}{2+\sqrt{1-a^2}}\)
\(P=\frac{\sqrt{1+\sqrt{1-a^2}}\left(\sqrt{1+a}-\sqrt{1-a}\right)\left(\sqrt{1+a}^2+\sqrt{1-a^2}+\sqrt{1-a}^2\right)}{2+\sqrt{1-a^2}}\)
\(P=\frac{\sqrt{1+\sqrt{1-a^2}}\left(\sqrt{1+a}-\sqrt{1-a}\right)\left(1+a+\sqrt{1-a^2}+1-a\right)}{2+\sqrt{1-a^2}}\)
\(P=\frac{\sqrt{2+2\sqrt{1-a^2}}\left(\sqrt{1+a}-\sqrt{1-a}\right).\left(2+\sqrt{1-a^2}\right)}{2\left(2+\sqrt{1+a^2}\right)}\)
\(P=\frac{\sqrt{1+a+2\sqrt{1-a}+1-a}\left(\sqrt{1+a}-\sqrt{1-a}\right)}{2}\)
\(P=\frac{\sqrt{\left(\sqrt{1-a}+\sqrt{1+a}\right)^2}\left(\sqrt{1+a}-\sqrt{1-a}\right)}{2}\)
\(P=\frac{\left(\sqrt{1+a}+\sqrt{1-a}\right)\left(\sqrt{1+a}-\sqrt{1-a}\right)}{2}=\frac{1+a-1+a}{2}=\frac{2a}{2}=a\)