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a) \(\left(\dfrac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\dfrac{\sqrt{5}-5}{1-\sqrt{5}}\right):\dfrac{1}{\sqrt{2}-\sqrt{5}}\)
\(=\left[-\dfrac{\sqrt{2}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}-\dfrac{\sqrt{5}\left(1-\sqrt{5}\right)}{1-\sqrt{5}}\right]\cdot\left(\sqrt{2}-\sqrt{5}\right)\)
\(=\left(-\sqrt{2}-\sqrt{5}\right)\left(\sqrt{2}-\sqrt{5}\right)\)
\(=-\left(\sqrt{2}+\sqrt{5}\right)\left(\sqrt{2}-\sqrt{5}\right)\)
\(=-\left(2-5\right)\)
\(=-\left(-3\right)\)
\(=3\)
b) Ta có:
\(x^2-x\sqrt{3}+1\)
\(=x^2-2\cdot\dfrac{\sqrt{3}}{2}\cdot x+\left(\dfrac{\sqrt{3}}{2}\right)^2+\dfrac{1}{4}\)
\(=\left(x-\dfrac{\sqrt{3}}{2}\right)^2+\dfrac{1}{4}\)
Mà: \(\left(x-\dfrac{\sqrt{3}}{2}\right)^2\ge0\forall x\) nên
\(\left(x-\dfrac{\sqrt{3}}{2}\right)^2+\dfrac{1}{4}\ge\dfrac{1}{4}\forall x\)
Dấu "=" xảy ra:
\(\left(x-\dfrac{\sqrt{3}}{2}\right)^2+\dfrac{1}{4}=\dfrac{1}{4}\)
\(\Leftrightarrow x=\dfrac{\sqrt{3}}{2}\)
Vậy: GTNN của biểu thức là \(\dfrac{1}{4}\) tại \(x=\dfrac{\sqrt{3}}{2}\)
a)
\(\left(\dfrac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\dfrac{\sqrt{5}-5}{1-\sqrt{5}}\right):\dfrac{1}{\sqrt{2}-\sqrt{5}}\\ =\left(-\dfrac{\sqrt{2}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}-\dfrac{\sqrt{5}\left(1-\sqrt{5}\right)}{1-\sqrt{5}}\right).\left(\sqrt{2}-\sqrt{5}\right)\\ =\left(-\sqrt{2}-\sqrt{5}\right).\left(\sqrt{2}-\sqrt{5}\right)\\ =-\left(\sqrt{2}+\sqrt{5}\right)\left(\sqrt{2}-\sqrt{5}\right)\\ =-\left(\sqrt{2}^2-\sqrt{5}^2\right)\\ =-\left(2-5\right)\\ =-\left(-3\right)\\ =3\)
a ) \(A=\frac{1}{\sqrt{5}+\sqrt{3}}-\frac{1}{\sqrt{5}-\sqrt{3}}\)
\(=\frac{\left(\sqrt{5}-\sqrt{3}\right)-\left(\sqrt{5}+\sqrt{3}\right)}{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}\)
\(=\frac{\sqrt{5}-\sqrt{3}-\sqrt{5}-\sqrt{3}}{5-3}\)
\(=\frac{-2\sqrt{3}}{2}\)
\(=-\sqrt{3}\)
c ) \(\frac{1}{2+\sqrt{3}}+\frac{\sqrt{2}}{\sqrt{6}}-\frac{2}{3+\sqrt{3}}\)
\(=\frac{1}{2+\sqrt{3}}+\frac{1}{\sqrt{3}}-\frac{2}{\sqrt{3}\left(\sqrt{3}+1\right)}\)
\(=\frac{\sqrt{3}\left(\sqrt{3}+1\right)+\left(2+\sqrt{3}\right)\left(\sqrt{3}+1\right)-2\left(2+\sqrt{3}\right)}{\sqrt{3}\left(\sqrt{3}+1\right)\left(2+\sqrt{3}\right)}\)
\(=\frac{2\sqrt{3}+4}{\sqrt{3}\left(\sqrt{3}+1\right)\left(2+\sqrt{3}\right)}\)
\(=\frac{2\left(\sqrt{3}+2\right)}{\sqrt{3}\left(\sqrt{3}+1\right)\left(2+\sqrt{3}\right)}\)
\(=\frac{2.\sqrt{3}\left(\sqrt{3}-1\right)}{3\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}\)
\(=\frac{2\sqrt{3}\left(\sqrt{3}-1\right)}{3.\left(3-1\right)}\)
\(=\frac{\sqrt{3}\left(\sqrt{3}-1\right)}{3}\)
\(=\frac{3-\sqrt{3}}{3}\)
\(=1-\frac{\sqrt{3}}{3}\)
1:
\(A=\sqrt{x^2+\dfrac{2x^2}{3}}=\sqrt{\dfrac{5x^2}{3}}=\left|\sqrt{\dfrac{5}{3}}x\right|=-x\sqrt{\dfrac{5}{3}}\)
2: \(=\left(\dfrac{\sqrt{100}+\sqrt{40}}{\sqrt{5}+\sqrt{2}}+\sqrt{6}\right)\cdot\dfrac{2\sqrt{5}-\sqrt{6}}{2}\)
\(=\dfrac{\left(2\sqrt{5}+\sqrt{6}\right)\left(2\sqrt{5}-\sqrt{6}\right)}{2}\)
\(=\dfrac{20-6}{2}=7\)
a: \(=-6\sqrt{b}-\dfrac{1}{3}\cdot3\sqrt{3b}+\dfrac{1}{5}\cdot5\sqrt{6b}\)
\(=-6\sqrt{b}-\sqrt{3}\cdot\sqrt{b}+\sqrt{6}\cdot\sqrt{b}\)
\(=\sqrt{b}\left(-6-\sqrt{3}+\sqrt{6}\right)\)
c: \(=\sqrt{\left(5+2\sqrt{6}\right)^2}+\sqrt{\left(5-2\sqrt{6}\right)^2}\)
\(=5+2\sqrt{6}+5-2\sqrt{6}=10\)
d: \(A=\sqrt{\sqrt{5}-\sqrt{3-2\sqrt{5}+3}}\)
\(=\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{5}+1}=1\)
e: \(B=\sqrt{6+2\sqrt{5-2\sqrt{3}-1}}\)
\(=\sqrt{6+2\cdot\left(\sqrt{3}-1\right)}=\sqrt{4+2\sqrt{3}}=\sqrt{3}+1\)
1 a/ Trục căn thức ở mẫu
\(VT=\frac{-\sqrt{1}+\sqrt{2}}{2-1}+\frac{-\sqrt{2}+\sqrt{3}}{3-2}+...+\frac{-\sqrt{47}+\sqrt{48}}{48-47}\)\(=-\sqrt{1}+\sqrt{2}-\sqrt{2}+\sqrt{3}-....-\sqrt{47}+\sqrt{48}=\sqrt{48}-1>3=VP\)
b/
\(2\left(10+3\sqrt{11}\right)=11+2.\sqrt{11}.3+9=\left(\sqrt{11}+3\right)^2\)
\(VT=\left(\sqrt{11}-3\right)\sqrt{2}\sqrt{10+3\sqrt{11}}=\left(\sqrt{11}-3\right)\left(\sqrt{11}+3\right)=11-9=2=VP\)
2/
\(B=\left(5+\sqrt{21}\right)\left(\sqrt{7}-\sqrt{3}\right)\sqrt{2\left(5+\sqrt{3}.\sqrt{7}\right)}\)
\(2\left(5+\sqrt{21}\right)=7+2\sqrt{7}.\sqrt{3}+3=\left(\sqrt{7}+\sqrt{3}\right)^2\)
\(B=\left(5+\sqrt{21}\right)\left(\sqrt{7}-\sqrt{3}\right)\left(\sqrt{7}+\sqrt{3}\right)=\left(5+\sqrt{21}\right).4\)
\(=20+4\sqrt{21}\)
A chắc không rút gọn được.
a) Đặt \(A=\sqrt{5-2\sqrt{6}}+\sqrt{5+2\sqrt{6}}\)
\(A^2=5-2\sqrt{6}+2\sqrt{\left(5-2\sqrt{6}\right)\left(5+2\sqrt{6}\right)}+5+2\sqrt{6}\)
\(=10+2\sqrt{25-4.6}=10+2\sqrt{1}=10+2=12\)
\(\Rightarrow A=\sqrt{12}\)
b)\(\frac{\sqrt{10}-\sqrt{2}}{\sqrt{5}-1}+\frac{2-\sqrt{2}}{\sqrt{2}-1}=\frac{\sqrt{2}.\sqrt{5}-\sqrt{2}}{\sqrt{5}-1}+\frac{\sqrt{2}.\sqrt{2}-\sqrt{2}}{\sqrt{2}-1}\)
\(=\frac{\sqrt{2}\left(\sqrt{5}-1\right)}{\sqrt{5}-1}+\frac{\sqrt{2}\left(\sqrt{2}-1\right)}{\sqrt{2}-1}=\sqrt{2}+\sqrt{2}=2\sqrt{2}\)