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Bài rút gọn
\(\sqrt{\left(x-1\right)^2}-x=\left|x-1\right|-x\)
\(=\left(x-1\right)-x=x-1-x=-1\left(x>1\right)\)
Bài gpt:
\(\sqrt{x^2-3x+2}+\sqrt{x^2-4x+3}=0\)
Đk:\(-1\le x\le3\)
\(pt\Leftrightarrow\sqrt{\left(x-1\right)\left(x-2\right)}+\sqrt{\left(x-1\right)\left(x-3\right)}=0\)
\(\Leftrightarrow\sqrt{x-1}\left(\sqrt{x-2}+\sqrt{x-3}\right)=0\)
Dễ thấy:\(\sqrt{x-2}+\sqrt{x-3}=0\) vô nghiệm
Nên \(\sqrt{x-1}=0\Rightarrow x-1=0\Rightarrow x=1\)
a) \(=\frac{7-4\sqrt{3}+7+4\sqrt{3}}{\left(7+4\sqrt{3}\right)\left(7-4\sqrt{3}\right)}=\frac{14}{49-48}=14\)
b) \(=\frac{15\left(\sqrt{6}-1\right)}{\left(\sqrt{6}+1\right)\left(\sqrt{6}-1\right)}-\frac{5\sqrt{6}}{5}+\frac{4\sqrt{3}-12\sqrt{2}}{\sqrt{6}\left(\sqrt{3}-\sqrt{2}\right)}\)
B=\(\frac{6-6\sqrt{3}}{1-\sqrt{3}}+\frac{3\sqrt{3}+3}{\sqrt{3}+1}=\frac{6\left(1-\sqrt{3}\right)}{1-\sqrt{3}}+\frac{3\left(\sqrt{3}+1\right)}{\sqrt{3}+1}=6+3=9\)
C=\(\frac{3+\sqrt{3}}{\sqrt{3}}+\frac{\sqrt{6}-\sqrt{3}}{1-\sqrt{2}}=\frac{3\left(1+\sqrt{3}\right)}{\sqrt{3}}+\frac{\sqrt{3}\left(\sqrt{2}-1\right)}{1-\sqrt{2}}=\sqrt{3}+1-\sqrt{3}=1\)
D=\(\frac{\sqrt{10}-\sqrt{2}}{\sqrt{5}-1}+\frac{2-\sqrt{2}}{\sqrt{2}-1}=\frac{\sqrt{2}\left(\sqrt{5}-1\right)}{\sqrt{5}-1}+\frac{\sqrt{2}\left(\sqrt{2}-1\right)}{\sqrt{2}-1}=\sqrt{2}+\sqrt{2}=2\sqrt{2}\)
E=\(\frac{\sqrt{15}-\sqrt{12}}{\sqrt{5}-2}+\frac{1}{2-\sqrt{3}}=\frac{\sqrt{3}\left(\sqrt{5}-2\right)}{\sqrt{5}-2}+\frac{1}{2-\sqrt{3}}=\sqrt{3}+\frac{1}{2-\sqrt{3}}=\frac{2\sqrt{3}-1}{2-\sqrt{3}}\)
a,
\(2\sqrt{3x}-\sqrt{48x}+\sqrt{108x}+\sqrt{3x}\\ =3\sqrt{3x}-\sqrt{4^2\cdot3x}+\sqrt{6^2\cdot3x}\\ =3\sqrt{3x}-4\sqrt{3x}+6\sqrt{3x}=5\sqrt{3x}\)
b,
\(2\sqrt{25xy}+\sqrt{5}\cdot\sqrt{45x^3y^3}-3y\sqrt{16x^3y}\\ =2\sqrt{5^2xy}+\sqrt{5\cdot45}\cdot\sqrt{\left(xy\right)^2\cdot xy}-3y\sqrt{\left(4x\right)^2\cdot xy}\\ =2\cdot5\sqrt{xy}+\sqrt{225}\cdot xy\sqrt{xy}-3y\cdot4x\sqrt{xy}\\ =10\sqrt{xy}+15xy\sqrt{xy}-12xy\sqrt{xy}=\sqrt{xy}\left(3xy+10\right)\)
c,
\(\frac{2}{\sqrt{3}-1}+\frac{3}{\sqrt{3}-2}+\frac{12}{3-\sqrt{13}}\\ =\frac{2\left(\sqrt{3}+1\right)}{3-1}+\frac{3\left(\sqrt{3}+2\right)}{3-4}+\frac{12\left(3+\sqrt{13}\right)}{9-13}\\ =\frac{2\left(\sqrt{3}+1\right)}{2}+\frac{3\left(\sqrt{3}+2\right)}{-1}+\frac{12\left(3+\sqrt{13}\right)}{-4}\\ =\sqrt{3}+1-3\sqrt{3}-6-9-3\sqrt{13}\\ =-14-2\sqrt{3}-3\sqrt{13}\)
d,
\(\frac{1}{\sqrt{3}-\sqrt{2}}-\frac{2}{\sqrt{3}+\sqrt{5}}-\frac{3}{\sqrt{5}-\sqrt{2}}+\frac{4}{\sqrt{7}+\sqrt{3}}\\ =\frac{\sqrt{3}+\sqrt{2}}{3-2}-\frac{2\left(\sqrt{5}-\sqrt{3}\right)}{5-3}-\frac{3\left(\sqrt{5}+\sqrt{2}\right)}{5-2}+\frac{4\left(\sqrt{7}-\sqrt{3}\right)}{7-3}\\ =\sqrt{3}+\sqrt{2}-\sqrt{5}+\sqrt{3}+\sqrt{5}+\sqrt{2}+\sqrt{7}-\sqrt{3}=\sqrt{7}+\sqrt{3}\)
Chúc bạn học tốt nha
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