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1) \(A=3\sqrt{\dfrac{1}{3}}-\dfrac{5}{2}\sqrt{12}-\sqrt{48}\)
\(=3\cdot\dfrac{\sqrt{1}}{\sqrt{3}}-\dfrac{5\sqrt{12}}{2}-\sqrt{4^2\cdot3}\)
\(=\dfrac{3\cdot1}{\sqrt{3}}-\dfrac{5\cdot2\sqrt{3}}{2}-4\sqrt{3}\)
\(=\sqrt{3}-5\sqrt{3}-4\sqrt{3}\)
\(=-8\sqrt{3}\)
2) \(A=\sqrt{12-4x}\) có nghĩa khi:
\(12-4x\ge0\)
\(\Leftrightarrow4x\le12\)
\(\Leftrightarrow x\le\dfrac{12}{4}\)
\(\Leftrightarrow x\le3\)
3) \(\dfrac{2x-2\sqrt{x}}{x-1}\)
\(=\dfrac{2\sqrt{x}\cdot\sqrt{x}-2\sqrt{x}}{\left(\sqrt{x}\right)^2-1^2}\)
\(=\dfrac{2\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{2\sqrt{\text{x}}}{\sqrt{x}+1}\)
1.\(\sqrt{27}+\sqrt{48}-\sqrt{108}-\sqrt{12}=3\sqrt{3}+4\sqrt{3}-6\sqrt{3}-2\sqrt{3}=-\sqrt{3}\)
2.\(P=\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}+1}\right).\dfrac{x-2\sqrt{x}+1}{2}\)
\(P=\left(\dfrac{\sqrt{x}+1-\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right).\dfrac{\left(\sqrt{x}-1\right)^2}{2}\)
\(P=\dfrac{2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\dfrac{\left(\sqrt{x}-1\right)^2}{2}\)
\(P=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)
\(a,\sqrt{27}+\sqrt{48}-\sqrt{108}-\sqrt{12}\\ =3\sqrt{3}+4\sqrt{3}-6\sqrt{3}-2\sqrt{3}\\ =-\sqrt{3}\)
\(b,P=\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}+1}\right).\dfrac{x-2\sqrt{x}+1}{2}\\ =\dfrac{\left(\sqrt{x}+1\right)-\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}.\dfrac{\left(\sqrt{x}-1\right)^2}{2}\\ =\dfrac{\sqrt{x}+1-\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}.\dfrac{\left(\sqrt{x}-1\right)^2}{2}\\ =\dfrac{2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}.\dfrac{\left(\sqrt{x}-1\right)^2}{2}\\ =\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)
a) \(\dfrac{\sqrt{7}+5}{\sqrt{7}-5}+\dfrac{\sqrt{7}-5}{\sqrt{7}+5}\)
\(=\dfrac{\left(\sqrt{7}+5\right)^2}{\left(\sqrt{7}-5\right)\left(\sqrt{7}+5\right)}+\dfrac{\left(\sqrt{7}-5\right)^2}{\left(\sqrt{7}+5\right)\left(\sqrt{7}-5\right)}\)
\(=\dfrac{\left(\sqrt{7}+5\right)^2+\left(\sqrt{7}-5\right)^2}{\left(\sqrt{7}-5\right)\left(\sqrt{7}+5\right)}\)
\(=\dfrac{\left(7+10\sqrt{7}+25\right)+\left(7-10\sqrt{7}+25\right)}{7-25}\)
\(=\dfrac{14+50}{7-25}\)
\(=\dfrac{64}{-18}\)
\(=\dfrac{-32}{9}\)
b) \(\sqrt{12}+\sqrt{48}-\sqrt{\left(\sqrt{75}-\sqrt{108}\right)^2}\)
\(=\sqrt{12}+\sqrt{48}-\left|\sqrt{75}-\sqrt{108}\right|\)
\(=\sqrt{12}+\sqrt{48}-\left(\sqrt{108}-\sqrt{75}\right)\) ( Vì \(\sqrt{75}< \sqrt{108}\) )
\(=\sqrt{12}+\sqrt{48}-\sqrt{108}+\sqrt{75}\)
\(=2\sqrt{3}+4\sqrt{3}-6\sqrt{3}+5\sqrt{3}\)
\(=5\sqrt{3}\)
a)\(\dfrac{\sqrt{7}+5+\sqrt{7}-5}{\sqrt{7}-5}=\dfrac{2\sqrt{7}}{\sqrt{7}-5}=\dfrac{-7-5\sqrt{7}}{9}\approx-2,25\)
Giải từ từ lần lượt các biểu thức trong dấu căn nhé:
\(\sqrt{13+\sqrt{48}}=\sqrt{\left(2\sqrt{3}\right)^2+2.2\sqrt{3}+1}=\sqrt{\left(2\sqrt{3}+1\right)^2}=2\sqrt{3}+1\)
\(\sqrt{5-2\sqrt{3}-1}=\sqrt{4-2\sqrt{3}}=\sqrt{\left(\sqrt{3}-1\right)^2}=\sqrt{3}-1\)
\(\sqrt{3+\sqrt{3}-1}=\sqrt{2+\sqrt{3}}\)
\(B=\frac{2\sqrt{2+\sqrt{3}}}{\sqrt{2}\left(\sqrt{3}-1\right)}=\frac{\sqrt{2}.\sqrt{2+\sqrt{3}}}{\sqrt{3}-1}\)
\(B=\frac{\sqrt{2}.\sqrt{2+\sqrt{3}}}{\sqrt{3}-1}=\frac{\sqrt{4+2\sqrt{3}}}{\sqrt{3}-1}=\frac{\sqrt{\left(\sqrt{3}+1\right)^2}}{\sqrt{3}-1}\)
\(B=\frac{\sqrt{3}+1}{\sqrt{3}-1}=\frac{\left(\sqrt{3}+1\right)^2}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}=\frac{3+2\sqrt{3}+1}{3-1}=\frac{4+2\sqrt{3}}{2}=2+\sqrt{3}\)
\(B=\frac{2\sqrt{3+\sqrt{5-\sqrt{1+4\sqrt{3}+12}}}}{\sqrt{6}-\sqrt{2}}\)
\(=\frac{2\sqrt{3+\sqrt{5-\sqrt{1+4\sqrt{3}+\left(2\sqrt{3}\right)^2}}}}{\sqrt{6}-\sqrt{2}}\)
\(=\frac{2\sqrt{3+\sqrt{5-\sqrt{\left(1+2\sqrt{3}\right)^2}}}}{\sqrt{6}-\sqrt{2}}\)
\(=\frac{2\sqrt{3+\sqrt{5-1-2\sqrt{3}}}}{\sqrt{6}-\sqrt{2}}=\frac{2\sqrt{3+\sqrt{4-2\sqrt{3}}}}{\sqrt{6}-\sqrt{2}}\)
\(=\frac{2\sqrt{3+\sqrt{1-2\sqrt{3}+\sqrt{3}^2}}}{\sqrt{6}-\sqrt{2}}=\frac{2\sqrt{3+\sqrt{\left(1-\sqrt{3}\right)^2}}}{\sqrt{6}-\sqrt{2}}\)
\(=\frac{2\sqrt{3+\sqrt{3}-1}}{\sqrt{6}-\sqrt{2}}=\frac{2\sqrt{2+\sqrt{3}}}{\sqrt{6}-\sqrt{2}}=\frac{2\sqrt{2+\sqrt{3}}\left(\sqrt{6}+\sqrt{2}\right)}{6-2}\)
\(\frac{\sqrt{2+\sqrt{3}}\left(\sqrt{6}+\sqrt{2}\right)}{2}\)
Bài 2 :
a) \(ĐKXĐ:\hept{\begin{cases}x;y>0\\x\ne y\end{cases}}\)
b) \(A=\left(\sqrt{x}+\frac{y-\sqrt{xy}}{\sqrt{x}-\sqrt{y}}\right):\frac{x\sqrt{xy}+y\sqrt{xy}}{\sqrt{xy}\left(y-x\right)}\)
\(\Leftrightarrow A=\frac{x-\sqrt{xy}+y-\sqrt{xy}}{\sqrt{x}-\sqrt{y}}:\frac{x+y}{y-x}\)
\(\Leftrightarrow A=\frac{\left(\sqrt{x}-\sqrt{y}\right)^2}{\sqrt{x}-\sqrt{y}}\cdot\frac{y-x}{x+y}\)
\(\Leftrightarrow A=\frac{\left(\sqrt{x}-\sqrt{y}\right)\left(y-x\right)}{x+y}\)
c) Thay \(x=4+2\sqrt{3},y=4-2\sqrt{3}\)vào A, ta được :
\(A=\frac{\left(\sqrt{4+2\sqrt{3}}-\sqrt{4-2\sqrt{3}}\right)\left(4-2\sqrt{3}-4-2\sqrt{3}\right)}{4+2\sqrt{3}+4-2\sqrt{3}}\)
\(\Leftrightarrow A=\frac{\left(\sqrt{\left(1+\sqrt{3}\right)^2}-\sqrt{\left(1-\sqrt{3}\right)^2}\right).\left(-4\sqrt{3}\right)}{8}\)
\(\Leftrightarrow A=\frac{\left(1+\sqrt{3}-\sqrt{3}+1\right).\left(-4\sqrt{3}\right)}{8}=\frac{-8\sqrt{3}}{8}=-\sqrt{3}\)
Vậy ....
Bài 1:
\(\frac{2\sqrt{8}-\sqrt{12}}{\sqrt{18}-\sqrt{48}}-\frac{\sqrt{5}+\sqrt{27}}{\sqrt{30}-\sqrt{2}}=\frac{2\sqrt{2\cdot4}-\sqrt{3\cdot4}}{\sqrt{2\cdot9}-\sqrt{16\cdot3}}-\frac{\sqrt{5}+\sqrt{9\cdot3}}{\sqrt{30}-\sqrt{2}}\)
\(=\frac{4\sqrt{2}-2\sqrt{3}}{3\sqrt{2}-4\sqrt{3}}-\frac{\sqrt{5}+3\sqrt{3}}{\sqrt{30}-\sqrt{2}}=\frac{\left(4\sqrt{2}-2\sqrt{3}\right)\left(\sqrt{30}-\sqrt{2}\right)-\left(\sqrt{5}+3\sqrt{3}\right)\left(3\sqrt{2}-4\sqrt{3}\right)}{\left(3\sqrt{2}-4\sqrt{3}\right)\left(\sqrt{30}-\sqrt{2}\right)}\)
\(=\frac{4\sqrt{60}-8-2\sqrt{90}+2\sqrt{6}-3\sqrt{10}+4\sqrt{15}-9\sqrt{6}+36}{3\sqrt{60}-6-4\sqrt{90}+4\sqrt{6}}\)
\(=\frac{8\sqrt{15}-8-6\sqrt{10}+2\sqrt{6}-3\sqrt{10}+4\sqrt{15}-9\sqrt{6}+36}{6\sqrt{15}-6-12\sqrt{10}+4\sqrt{6}}\)
\(=\frac{12\sqrt{15}-2\sqrt{10}-7\sqrt{6}+28}{6\sqrt{15}-12\sqrt{10}+4\sqrt{6}-6}\)
a) \(5\sqrt{48}-4\sqrt{27}-2\sqrt{57}+\sqrt{108}\)
\(=20\sqrt{3}-12\sqrt{3}-2\sqrt{57}+6\sqrt{3}\)
\(=\left(20-12+6\right)\sqrt{3}-2\sqrt{57}\)
\(=14\sqrt{3}-2\sqrt{57}\)
b) \(2\sqrt{24}-2\sqrt{54}+3\sqrt{6}-\sqrt{150}\)
\(=4\sqrt{6}-6\sqrt{6}+3\sqrt{6}-5\sqrt{6}\)
\(=\left(4-6+3-5\right)\sqrt{6}\)
\(=-4\sqrt{6}\)
a) \(2\sqrt{3}-\sqrt{27}+\sqrt{75}\)
\(=2\sqrt{3}-3\sqrt{3}+5\sqrt{3}\)
\(=\sqrt{3}\left(2-3+5\right)\)
\(=4\sqrt{3}\)
b)\(\sqrt{\left(1-\sqrt{3}\right)^2}+\sqrt{4+2\sqrt{3}}\)
\(=\sqrt{\left(1-\sqrt{3}\right)^2}+\sqrt{\left(1+\sqrt{3}\right)^2}\)
\(=\left(1-\sqrt{3}\right)+\left(1+\sqrt{3}\right)\)
\(=2\)
\(A=2\sqrt{5}-\sqrt{45}+2\sqrt{20}=2\sqrt{5}-\sqrt{3^2.5}+2\sqrt{2^2.5}=2\sqrt{5}-3\sqrt{5}+4\sqrt{5}=3\sqrt{5}\)
\(B=\left(\sqrt{18}-\frac{1}{2}\cdot\sqrt{32}+12\sqrt{2}\right):\sqrt{2}=\left(3\sqrt{2}-\frac{1}{2}\cdot4\sqrt{2}+12\sqrt{2}\right):\sqrt{2}\)
\(=13\sqrt{2}:\sqrt{2}=13\)
\(C=\left(\sqrt{12}+2\sqrt{27}-3\sqrt{3}\right)\cdot\sqrt{3}=\left(2\sqrt{3}+6\sqrt{3}-3\sqrt{3}\right)\cdot\sqrt{3}=5\sqrt{3}\cdot\sqrt{3}=15\)
\(D=\sqrt{20}-\sqrt{45}+3\sqrt{18}+\sqrt{72}=2\sqrt{5}-3\sqrt{5}+9\sqrt{2}+6\sqrt{2}=-\sqrt{5}+15\sqrt{2}\)
Ta có: \(A=2\sqrt{12}-\sqrt{48}+3\sqrt{27}-\sqrt{108}\)
\(=2\cdot2\sqrt{3}-4\sqrt{3}+3\cdot3\sqrt{3}-6\sqrt{3}\)
\(=4\sqrt{3}-4\sqrt{3}+9\sqrt{3}-6\sqrt{3}\)
\(=3\sqrt{3}\)
A=4\(\sqrt{3}\)-4\(\sqrt{3}\)+9\(\sqrt{3}\)-6\(\sqrt{3}\)
A= 3\(\sqrt{3}\)