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Đặt \(\sqrt[4]{5}=x\) thì \(x^4=5\). Ta có :
A = \(\frac{2}{\sqrt{4-3x+2x^2-x^3}}\)= \(\frac{2\left(x+1\right)}{\sqrt{\left(x+1\right)^2\left(4-3x+2x^2-x^3\right)}}\)= \(\frac{2\left(x+1\right)}{\sqrt{-x^5+5x+4}}\)
Ta thấy \(-x^5+5x\) = \(x\left(5-x^4\right)\)= \(0\)
nên A = \(\frac{2\left(x+1\right)}{\sqrt{4}}\)= \(x+1\)=\(\sqrt[4]{5}+1\)
a) \(2\sqrt{3}+\sqrt{\left(2-\sqrt{3}\right)^2}\)
\(=2\sqrt{3}+2-\sqrt{3}\)
\(=\left(2\sqrt{3}-\sqrt{3}\right)+2\)
\(=\sqrt{3}+2\)
b) \(\frac{5+\sqrt{5}}{5-\sqrt{5}}+\frac{5-\sqrt{5}}{5+\sqrt{5}}\)
\(=\frac{1+\sqrt{5}}{\sqrt{5}-1}+\frac{\sqrt{5}-1}{1+\sqrt{5}}\)
\(=\frac{\left(\sqrt{5}+1\right)^2}{\left(\sqrt{5}-1\right)\left(1+\sqrt{5}\right)}+\frac{\left(\sqrt{5}-1\right)^2}{\left(\sqrt{5}-1\right)\left(1+\sqrt{5}\right)}\)
\(=\frac{\left(\sqrt{5}+1\right)^2+\left(\sqrt{5}-1\right)^2}{\left(\sqrt{5}-1\right)\left(1+\sqrt{5}\right)}\)
\(=\frac{12}{4}=3\)
c) \(\frac{1}{7+4\sqrt{3}}+\frac{1}{7-4\sqrt{3}}\)
\(=\frac{7-4\sqrt{3}}{\left(7+4\sqrt{3}\right)\left(7-4\sqrt{3}\right)}+\frac{7+4\sqrt{3}}{\left(7+4\sqrt{3}\right)\left(7-4\sqrt{3}\right)}\)
\(=\frac{7-4\sqrt{3}+7+4\sqrt{3}}{\left(7+4\sqrt{3}\right)\left(7-4\sqrt{3}\right)}\)
\(=\frac{14}{1}=14\)
mọi ng ơi mk viết thiếu dấu ngoặc nha.thiếu ngoặc lownns nha. đóng ngoắc ở trước dấu chia
\(D=\sqrt{5}-\sqrt{13-4\sqrt{\left(\sqrt{5}-2\right)^2}}=\sqrt{5}-\sqrt{13-4\left(\sqrt{5}-2\right)}\)
\(=\sqrt{5}-\sqrt{21-4\sqrt{5}}=\sqrt{5}-\sqrt{\left(2\sqrt{5}-1\right)^2}\)
\(=\sqrt{5}-2\sqrt{5}+1=1-\sqrt{5}\)
\(B=10\sqrt{5}+\left|1-\sqrt{5}\right|-\frac{4\left(\sqrt{5}-1\right)}{\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)}\)
\(=10\sqrt{5}+\sqrt{5}-1-\sqrt{5}+1=10\sqrt{5}\)
\(C=\frac{2\left(\sqrt{3}-1\right)}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}+\frac{2+\sqrt{3}}{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}+\frac{12\left(3-\sqrt{3}\right)}{\left(3+\sqrt{3}\right)\left(3-\sqrt{3}\right)}\)
\(=\sqrt{3}-1+2+\sqrt{3}+2\left(3-\sqrt{3}\right)=7\)