Tính:
a,\(\sqrt{19-6\sqrt{2}}\)
b,\(\sqrt{21+12\sqrt{3}}\)
c,\(\sqrt{57-40\sqrt{2}}\)
d,\(\sqrt{\left(5-2\sqrt{6}\right)\left(4-2\sqrt{3}\right)}\)
e,\(\sqrt{21+6\sqrt{6}}+\sqrt{21-6\sqrt{6}}\)
g,\(\sqrt{7-4\sqrt{3}}-\sqrt{7+4\sqrt{3}}\)
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a) \(\sqrt{19-6\sqrt{2}}=3\sqrt{2}-1\)
b) \(\sqrt{11-6\sqrt{2}}=3-\sqrt{2}\)
d) \(\sqrt{21+12\sqrt{3}}=2\sqrt{3}+3\)
e) \(\sqrt{57-40\sqrt{2}}=4\sqrt{2}-5\)
a: \(=\sqrt{5}+2+\sqrt{3}+1-\sqrt{5}-\sqrt{3}=3\)
b: \(=\left(-\sqrt{5}-2+\sqrt{5}-\sqrt{3}\right)\cdot\left(2\sqrt{3}+3\right)\)
\(=-\sqrt{3}\left(2+\sqrt{3}\right)\cdot\left(2+\sqrt{3}\right)\)
\(=-\sqrt{3}\left(7+4\sqrt{3}\right)=-7\sqrt{3}-12\)
c: \(=\dfrac{\sqrt{2}+\sqrt{3}+2}{\left(\sqrt{2}+\sqrt{3}+2\right)+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+2\right)}=\dfrac{1}{1+\sqrt{2}}=\sqrt{2}-1\)
Bài 2:
a)\(\dfrac{1}{3}\sqrt{x-2}-\dfrac{2}{3}\sqrt{9x-18}+6\sqrt{\dfrac{x-2}{81}}=-4\) (đk: \(x\ge2\))
\(\Leftrightarrow\dfrac{1}{3}\sqrt{x-2}-\dfrac{2}{3}\sqrt{9\left(x-2\right)}+\dfrac{6}{\sqrt{81}}\sqrt{x-2}=-4\)
\(\Leftrightarrow\dfrac{1}{3}\sqrt{x-2}-2\sqrt{x-2}+\dfrac{2}{3}\sqrt{x-2}=-4\)
\(\Leftrightarrow-\sqrt{x-2}=-4\) \(\Leftrightarrow x-2=16\)
\(\Leftrightarrow x=18\) (thỏa)
Vậy...
b)\(\sqrt{9x^2+12x+4}=4x\)(Đk:\(9x^2+12x+4\ge0\))
\(\Leftrightarrow\left\{{}\begin{matrix}4x\ge0\\9x^2+12x+4=16x^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\-7x^2+12x+4=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\-7x^2+14x-2x+4=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\\left(x-2\right)\left(-7x-2\right)=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\\left[{}\begin{matrix}x=2\\x=-\dfrac{2}{7}\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow x=2\) (tm đk)
Vậy...
c) \(\sqrt{x-2\sqrt{x-1}}=\sqrt{x-1}\) (đk: \(x\ge1\))
\(\Leftrightarrow x-2\sqrt{x-1}=x-1\)
\(\Leftrightarrow\sqrt{x-1}=\dfrac{1}{2}\) \(\Leftrightarrow x=\dfrac{5}{4}\) (tm)
Vậy...
Bài 1
a) Đặt VT = A
<=> \(2\sqrt{2}A=\left(8+2\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{8-2\sqrt{15}}\)
<=> \(2\sqrt{2}A=\left(\sqrt{5}+\sqrt{3}\right)^2.\sqrt{2}\left(\sqrt{5}-\sqrt{3}\right).\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\)
<=> \(2A=\left(\sqrt{5}+\sqrt{3}\right)^2.\left(\sqrt{5}-\sqrt{3}\right)^2\)
<=> 2A = \(\left(5-3\right)^2=4\)
<=> A = 2
b) Đặt VT = B
<=> \(2\sqrt{2}B=\left(10+2\sqrt{21}\right).\left(\sqrt{14}-\sqrt{6}\right)\sqrt{10-2\sqrt{21}}\)
<=> \(2\sqrt{2}B=\left(\sqrt{7}+\sqrt{3}\right)^2.\sqrt{2}\left(\sqrt{7}-\sqrt{3}\right).\sqrt{\left(\sqrt{7}-\sqrt{3}\right)^2}\)
<=> \(2B=\left(\sqrt{7}+\sqrt{3}\right)^2.\left(\sqrt{7}-\sqrt{3}\right)^2=\left(7-3\right)^2=16\)
<=> B = 8
Bài 2
Đặt VT = A
<=> A2 = \(\dfrac{\sqrt{5}+2+\sqrt{5}-2+2\sqrt{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)}}{2}\)
<=> A2 = \(\dfrac{2\sqrt{5}+2\sqrt{5-4}}{2}=\dfrac{2\sqrt{5}+2}{2}=\sqrt{5}+1\)
<=> \(A=\sqrt{\sqrt{5}+1}\)
a)\(\left(\sqrt{21}+7\right)\cdot\sqrt{10-2\sqrt{21}}\)
\(=\left(\sqrt{21}+7\right)\sqrt{\left(\sqrt{7}-\sqrt{3}\right)^2}\)
\(=\sqrt{7}\left(\sqrt{3}+\sqrt{7}\right)\left(\sqrt{7}-\sqrt{3}\right)\)
\(=\sqrt{7}\left(7-3\right)=4\sqrt{7}\)
b)\(\left(7+\sqrt{14}\right)\sqrt{9-2\sqrt{14}}\)
\(=\left(7+\sqrt{14}\right)\sqrt{\left(\sqrt{7}-\sqrt{2}\right)^2}\)
\(=\sqrt{7}\left(\sqrt{7}+\sqrt{2}\right)\left(\sqrt{7}-\sqrt{2}\right)\)
\(=\sqrt{7}\left(7-2\right)=5\sqrt{7}\)
b: \(=\sqrt{5}-1-\sqrt{5}-1=-2\)
c: \(=\dfrac{\left(2\sqrt{2}+\sqrt{3}-2\sqrt{2}+\sqrt{3}\right)}{2\sqrt{3}}=1\)
d: \(=\dfrac{\sqrt{6-2\sqrt{5}}-\sqrt{6+2\sqrt{5}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{5}-1-\sqrt{5}-1}{\sqrt{2}}=-\sqrt{2}\)
Ta có :
a)\(\left(2\sqrt{5}-\sqrt{7}\right)\left(2\sqrt{5}-\sqrt{7}\right)=\left(2\sqrt{5}\right)^2-\left(\sqrt{7}\right)^2=20-7=13\)
b)\(\left(5\sqrt{2}+2\sqrt{3}\right)\left(2\sqrt{3}-5\sqrt{2}\right)=\left(2\sqrt{3}\right)^2-\left(5\sqrt{2}\right)^2=12-50=-38\)
c)\(\sqrt{9+4\sqrt{5}}=\sqrt{2^2+2.2.\sqrt{5}+\left(\sqrt{5}\right)^2}=\sqrt{\left(2+\sqrt{5}\right)^2}=\left|2+\sqrt{5}\right|=2+\sqrt{5}\)
a: \(\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}\)
\(=\sqrt{9-2\cdot3\cdot\sqrt{6}+6}+\sqrt{24-2\cdot2\sqrt{6}\cdot3+9}\)
\(=\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(2\sqrt{6}-3\right)^2}\)
\(=3-\sqrt{6}+2\sqrt{6}-3=\sqrt{6}\)
b: \(\sqrt{\left(3+\sqrt{5}\right)^2}+\sqrt{14-6\sqrt{5}}\)
\(=\sqrt{\left(3+\sqrt{5}\right)^2}+\sqrt{\left(3-\sqrt{5}\right)^2}\)
\(=\left|3+\sqrt{5}\right|+\left|3-\sqrt{5}\right|\)
\(=3+\sqrt{5}+3-\sqrt{5}=6\)
c: \(\dfrac{3}{2\sqrt{3}+3}+\dfrac{3}{2\sqrt{3}-3}\)
\(=\dfrac{3\left(2\sqrt{3}-3\right)+3\left(2\sqrt{3}+3\right)}{12-9}\)
\(=2\sqrt{3}-3+2\sqrt{3}+3=4\sqrt{3}\)
d: \(\sqrt{\left(\sqrt{3}+4\right)\cdot\sqrt{19-8\sqrt{3}}+3}\)
\(=\sqrt{\left(4+\sqrt{3}\right)\cdot\sqrt{\left(4-\sqrt{3}\right)^2}+3}\)
\(=\sqrt{\left(4+\sqrt{3}\right)\cdot\left(4-\sqrt{3}\right)+3}\)
\(=\sqrt{16-3+3}=\sqrt{16}=4\)
e: \(\dfrac{9-2\sqrt{3}}{3\sqrt{6}-2\sqrt{2}}+\dfrac{3}{3+\sqrt{6}}\)
\(=\dfrac{\sqrt{3}\left(3\sqrt{3}-2\right)}{\sqrt{2}\left(3\sqrt{3}-2\right)}+\dfrac{3\left(3-\sqrt{6}\right)}{3}\)
\(=\dfrac{\sqrt{6}}{2}+3-\sqrt{6}=3-\dfrac{\sqrt{6}}{2}\)
\(a.\sqrt{19-6\sqrt{2}}=\sqrt{18-2.3\sqrt{2}+1}=3\sqrt{2}-1\)
\(b.\sqrt{21+12\sqrt{3}}=\sqrt{12+2.2\sqrt{3}.3+9}=2\sqrt{3}+3\)
\(c.\sqrt{57-40\sqrt{2}}=\sqrt{32-2.4\sqrt{2}.5+25}=4\sqrt{2}-5\)
\(d.\sqrt{\left(5-2\sqrt{6}\right)\left(4-2\sqrt{3}\right)}=\sqrt{3-2\sqrt{3}.\sqrt{2}+2}.\sqrt{3-2\sqrt{3}+1}=\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}-1\right)\) \(e.\sqrt{21+6\sqrt{6}}+\sqrt{21-6\sqrt{6}}=\sqrt{18+2.3\sqrt{2}.\sqrt{3}+3}+\sqrt{18-2.3\sqrt{2}.\sqrt{3}+3}=3\sqrt{2}+\sqrt{3}+3\sqrt{2}-\sqrt{3}=6\sqrt{2}\) \(g.\sqrt{7-4\sqrt{3}}-\sqrt{7+4\sqrt{3}}=\sqrt{4-2.2\sqrt{3}+3}-\sqrt{4+2.2\sqrt{3}+3}=2-\sqrt{3}-2-\sqrt{3}=-2\sqrt{3}\)
a)
=\(\sqrt{18-2.3\sqrt{2}.1+1}\)
\(=\sqrt{\left(3\sqrt{2}-1\right)^2}\)
\(=3\sqrt{2}-1\)
b)
=\(\sqrt{12+2.2\sqrt{3}.3+9}\)
=\(\sqrt{\left(2\sqrt{3}+3\right)^2}\)
=\(2\sqrt{3}+3\)
c)
=\(\sqrt{25-2.5.4\sqrt{2}+32}\)
=\(\sqrt{\left(5-4\sqrt{2}\right)^2}\)
=\(4\sqrt{2}-5\)
d)
\(=\sqrt{\left(3-2.\sqrt{3}.\sqrt{2}+2\right)\left(3-2\sqrt{3}+1\right)}\\ =\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2\left(\sqrt{3}-1\right)^2}\\ =\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}-1\right)\\ =3-\sqrt{3}-\sqrt{6}+\sqrt{2}\)
e)
\(=\sqrt{18+2.3\sqrt{2}.\sqrt{3}+3}+\sqrt{18-2.3\sqrt{2}.\sqrt{3}+3}\\ =\sqrt{\left(3\sqrt{2}+\sqrt{3}\right)^2}+\sqrt{\left(3\sqrt{2}-\sqrt{3}\right)^2}\\ =3\sqrt{2}+\sqrt{3}+3\sqrt{2}-\sqrt{3}\\ =6\sqrt{2}\)
g)
\(=\sqrt{4-2.2.\sqrt{3}+3}-\sqrt{4+2.2.\sqrt{3}+3}\\ =\sqrt{\left(2-\sqrt{3}\right)^2}-\sqrt{\left(2+\sqrt{3}\right)^2}\\ =2-\sqrt{3}-2-\sqrt{3}\\ =-2\sqrt{3}\)