Thu gọn biểu thức sau:
\(B=5\left(\sqrt{2+\sqrt{3}}+\sqrt{3-\sqrt{5}}-\sqrt{\frac{5}{2}}\right)^2-\left(\sqrt{2-\sqrt{3}}+\sqrt{3+\sqrt{5}}-\sqrt{\frac{3}{2}}\right)^2\)
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Rút gọn biểu thức
A= \(\frac{\sqrt{2}\left(3+\sqrt{5}\right)}{2\sqrt{2}+\sqrt{3+\sqrt{5}}}+\frac{\sqrt{2}\left(3-\sqrt{5}\right)}{2\sqrt{2}-\sqrt{3-\sqrt{5}}}\)
Rút gọn các biểu thức sau:a) left(sqrt{8}-3sqrt{2}+sqrt{10}right)left(sqrt{2}-3sqrt{0,4}right) b) left(sqrt{28}-2sqrt{14}+sqrt{7}right).sqrt{7}+7sqrt{8}c) 2sqrt{left(sqrt{2}-3right)^2}+sqrt{2left(-3right)^2}-5sqrt{left(-1right)^4} d) left(frac{sqrt{14}-sqrt{7}}{1-sqrt{2}}+frac{sqrt{15}-sqrt{5}}{1-sqrt{3}}right):frac{1}{sqrt{7}-sqrt{5}}
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Rút gọn các biểu thức sau:
a) \(\left(\sqrt{8}-3\sqrt{2}+\sqrt{10}\right)\left(\sqrt{2}-3\sqrt{0,4}\right)\) b) \(\left(\sqrt{28}-2\sqrt{14}+\sqrt{7}\right).\sqrt{7}+7\sqrt{8}\)
c) \(2\sqrt{\left(\sqrt{2}-3\right)^2}+\sqrt{2\left(-3\right)^2}-5\sqrt{\left(-1\right)^4}\) d) \(\left(\frac{\sqrt{14}-\sqrt{7}}{1-\sqrt{2}}+\frac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}\right):\frac{1}{\sqrt{7}-\sqrt{5}}\)
Rút gọn biểu thức:
Afrac{1}{sqrt{1}-sqrt{2}}-frac{1}{sqrt{2}-sqrt{3}}+frac{1}{sqrt{3}-sqrt{4}}-frac{1}{sqrt{4}-sqrt{5}}
Bleft(frac{5-sqrt{5}}{sqrt{5}}-2right)left(frac{4}{1+sqrt{5}}+4right)
Cleft(frac{3+2sqrt{3}}{sqrt{3}+2}+frac{2+sqrt{2}}{sqrt{2}+1}right):left(1:frac{1}{sqrt{2}+sqrt{3}}right)
D2sqrt{50}-frac{1}{sqrt{2}-1}+4sqrt{frac{9}{2}}-sqrt{3-2sqrt{2}}
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Rút gọn biểu thức:
\(A=\frac{1}{\sqrt{1}-\sqrt{2}}-\frac{1}{\sqrt{2}-\sqrt{3}}+\frac{1}{\sqrt{3}-\sqrt{4}}-\frac{1}{\sqrt{4}-\sqrt{5}}\)
\(B=\left(\frac{5-\sqrt{5}}{\sqrt{5}}-2\right)\left(\frac{4}{1+\sqrt{5}}+4\right)\)
\(C=\left(\frac{3+2\sqrt{3}}{\sqrt{3}+2}+\frac{2+\sqrt{2}}{\sqrt{2}+1}\right):\left(1:\frac{1}{\sqrt{2}+\sqrt{3}}\right)\)
\(D=2\sqrt{50}-\frac{1}{\sqrt{2}-1}+4\sqrt{\frac{9}{2}}-\sqrt{3-2\sqrt{2}}\)
Bài 1: Rút gọn các biểu thức sau
a, \(\frac{\left(\sqrt{3}-\sqrt{5}\right)^2+4\sqrt{15}}{\sqrt{3}+\sqrt{5}}-\sqrt{5}\)
b,\(\frac{\left(\sqrt{2}+1\right)^2-4\sqrt{2}}{\sqrt{2}-1}.\left(\sqrt{2}+1\right)\)
a) \(\frac{\left(\sqrt{3}-\sqrt{5}\right)^2+4\sqrt{15}}{\sqrt{3}+\sqrt{5}}-\sqrt{5}\)
= \(\frac{3-2\sqrt{15}+5+4\sqrt{15}}{\sqrt{3}+\sqrt{5}}-\sqrt{5}\)
=\(\frac{8+2\sqrt{15}-\sqrt{5}\left(\sqrt{3}+\sqrt{5}\right)}{\sqrt{3}+\sqrt{5}}\)
= \(\frac{8+2\sqrt{15}-\sqrt{15}-\sqrt{25}}{\sqrt{3}+\sqrt{5}}\)
= \(\frac{3+\sqrt{15}}{\sqrt{3}+\sqrt{5}}\)
= \(\frac{\sqrt{3}\left(\sqrt{3}+\sqrt{5}\right)}{\sqrt{3}+\sqrt{5}}\)
= \(\sqrt{3}\)
b) \(\frac{\left(\sqrt{2}+1\right)^2-4\sqrt{2}}{\sqrt{2}-1}.\left(\sqrt{2}+1\right)\)
= \(\frac{2+2\sqrt{2}+1-4\sqrt{2}}{\sqrt{2}-1}.\left(\sqrt{2}+1\right)\)
= \(\frac{\left(\sqrt{2}-1\right)^2.\left(\sqrt{2}+1\right)}{\sqrt{2}-1}\)
= \(\left(\sqrt{2}-1\right).\left(\sqrt{2}+1\right)\)
= 2 - 1
= 1
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Rút gọn các biểu thức sau:
a) \(0,2\sqrt{\left(-10\right)^2.3}+2\sqrt{\left(\sqrt{3}-\sqrt{5}\right)^2}\) b) \(\left(15\sqrt{50}+5\sqrt{200}-3\sqrt{450}\right):\sqrt{10}\)
c) \(\left(\frac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}-\frac{\sqrt{216}}{3}\right):\sqrt{6}\) d) \(\frac{\sqrt{5+2\sqrt{6}}+\sqrt{8-2\sqrt{15}}}{\sqrt{7+2\sqrt{10}}}\)
Rút gọn biểu thức:
a)\(\frac{2}{\sqrt{5}-\sqrt{3}}+\frac{3}{\sqrt{6}+\sqrt{3}}\)
b)\(\frac{3+2\sqrt{3}}{\sqrt{3}}+\frac{2+\sqrt{2}}{\sqrt{2}+1}-\left(2+\sqrt{3}\right)\)
c)\(\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}}-\sqrt{2}\)
d)\(\left(1+tan^2a\right)\left(1-sin^2a\right)+\left(1+cotan^2a\right)\left(1-cos^2a\right)\)
Thu gọn biểu thức:
\(B=21\cdot\left(\sqrt{2+\sqrt{3}}+\sqrt{3-\sqrt{5}}\right)^2-6\left(\sqrt{2-\sqrt{3}}+\sqrt{3+\sqrt{5}}\right)^2-15\sqrt{15}\)
\(B=\dfrac{21}{2}\left(\sqrt{4+2\sqrt{3}}+\sqrt{6-2\sqrt{5}}\right)^2-3\left(\sqrt{4-2\sqrt{3}}+\sqrt{6+2\sqrt{5}}\right)^2-15\sqrt{15}\)
\(=\dfrac{21}{2}\left(\sqrt{3}+1+\sqrt{5}-1\right)^2-3\left(\sqrt{3}-1+\sqrt{5}+1\right)^2-15\sqrt{15}\)
\(=\dfrac{21}{2}\left(\sqrt{3}+\sqrt{5}\right)^2-3\left(\sqrt{3}+\sqrt{5}\right)^2-15\sqrt{15}\)
\(=\dfrac{15}{2}\left(8+2\sqrt{15}\right)-15\sqrt{15}\)
\(=60+15\sqrt{15}-15\sqrt{15}=60\)
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* rút gọn biểu thức:
a, sqrt{frac{3}{20}}+sqrt{frac{1}{60}}-2sqrt{frac{1}{15}}
b, left(frac{1}{sqrt{5}-sqrt{3}}+frac{1}{sqrt{5}+sqrt{3}}right).sqrt{5}
c, left(5sqrt{3}+3sqrt{5}right):sqrt{15}
d, left(2+sqrt{5}right)^2-left(2+sqrt{5}right)^2
e, frac{1}{3}sqrt{48}+3sqrt{75}-sqrt{27}-10sqrt{1frac{1}{3}}
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* rút gọn biểu thức:
a, \(\sqrt{\frac{3}{20}}+\sqrt{\frac{1}{60}}-2\sqrt{\frac{1}{15}}\)
b, \(\left(\frac{1}{\sqrt{5}-\sqrt{3}}+\frac{1}{\sqrt{5}+\sqrt{3}}\right).\sqrt{5}\)
c, \(\left(5\sqrt{3}+3\sqrt{5}\right):\sqrt{15}\)
d, \(\left(2+\sqrt{5}\right)^2-\left(2+\sqrt{5}\right)^2\)
e, \(\frac{1}{3}\sqrt{48}+3\sqrt{75}-\sqrt{27}-10\sqrt{1\frac{1}{3}}\)
a, = \(\frac{\sqrt{15}}{10}\) + \(\frac{\sqrt{15}}{30}\) - \(\frac{2\sqrt{15}}{15}\)
= \(\sqrt{15}\left(\frac{1}{10}+\frac{1}{30}-\frac{2}{15}\right)\)
= \(\sqrt{15}.0\)
= 0
b, = \(\left(\frac{\sqrt{5}+\sqrt{3}}{5-3}+\frac{\sqrt{5}-\sqrt{3}}{5-3}\right).\sqrt{5}\)
= \(\frac{\sqrt{5}+\sqrt{3}+\sqrt{5}-\sqrt{3}}{2}.\sqrt{5}\)
= \(\frac{2\sqrt{5}}{2}.\sqrt{5}\)
= \(\sqrt{5}.\sqrt{5}\)
= 5
c, = \(\frac{5\sqrt{3}}{\sqrt{15}}+\frac{3\sqrt{5}}{\sqrt{15}}\)
= \(\sqrt{5}+\sqrt{3}\)
d, Mình sửa lại đề bài cho bạn : \(\left(2+\sqrt{5}\right)^2-\left(2-\sqrt{5}\right)^2\)
= \(\left(2+\sqrt{5}-2+\sqrt{5}\right)\left(2+\sqrt{5}+2-\sqrt{5}\right)\)
= \(2\sqrt{5}.4\)
= \(8\sqrt{5}\)
e, = \(\frac{4\sqrt{3}}{3}+15\sqrt{3}-3\sqrt{3}-\frac{20\sqrt{3}}{3}\)
= \(\sqrt{3}.\left(\frac{4}{3}+15-3-\frac{20}{3}\right)\)
= \(\sqrt{3}.\frac{20}{3}\)
= \(\frac{20\sqrt{3}}{3}\)
a, 320+160−2115
b, (15−3+15+3).5
c, (53+35):15
d, (2+5)2−(2+5)2
e, 1348+375−27−10113
a) Rút gọn biểu thức:\(\left(\frac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\frac{\sqrt{5}-5}{1-\sqrt{5}}\right):\frac{1}{\sqrt{2}-\sqrt{5}}\)
b) Tìm giá trị nhỏ nhất của biểu thức B=\(x^2-x\sqrt{3}+1\)
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 \(x=\dfrac{\sqrt{3}}{2}\)
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a)
\(\left(\dfrac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\dfrac{\sqrt{5}-5}{1-\sqrt{5}}\right):\dfrac{1}{\sqrt{2}-\sqrt{5}}\\ =\left(-\dfrac{\sqrt{2}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}-\dfrac{\sqrt{5}\left(1-\sqrt{5}\right)}{1-\sqrt{5}}\right).\left(\sqrt{2}-\sqrt{5}\right)\\ =\left(-\sqrt{2}-\sqrt{5}\right).\left(\sqrt{2}-\sqrt{5}\right)\\ =-\left(\sqrt{2}+\sqrt{5}\right)\left(\sqrt{2}-\sqrt{5}\right)\\ =-\left(\sqrt{2}^2-\sqrt{5}^2\right)\\ =-\left(2-5\right)\\ =-\left(-3\right)\\ =3\)
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