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3: \(\sqrt{12-3\sqrt{7}}-\sqrt{12-3\sqrt{7}}=0\)
4: \(\sqrt{7-2\sqrt{10}}-\sqrt{7+2\sqrt{10}}\)
\(=\sqrt{5}-\sqrt{2}-\sqrt{5}-\sqrt{2}\)
\(=-2\sqrt{2}\)
6: \(3\sqrt{3}+4\sqrt{12}-5\sqrt{27}\)
\(=3\sqrt{3}+8\sqrt{3}-15\sqrt{3}\)
\(=-4\sqrt{3}\)
\(a,\dfrac{-3}{5}.\sqrt{\left(-0.5\right)^2}\\ =\dfrac{-3}{5}.0,5\\ =\dfrac{-3}{5}.\dfrac{1}{2}\\ =-\dfrac{3}{10}\)
Câu (b) nhìn hơi lạ lạ á :v
\(c,\sqrt{\left(1-\sqrt{7}\right)^2}+\sqrt{7}\\ =\sqrt{7}-1+\sqrt{7}\\ =2\sqrt{7}-1\)
\(d,\sqrt{11+6\sqrt{2}}-\sqrt{11-6\sqrt{2}}\\ =\sqrt{\left(3+\sqrt{2}\right)^2}-\sqrt{\left(3-\sqrt{2}\right)^2}\\ =3+\sqrt{2}-\left(3-\sqrt{2}\right)\\ =3+\sqrt{2}-3+\sqrt{2}\\ =2\sqrt{2}\)
\(a.\dfrac{3\sqrt{2}-2\sqrt{3}}{\sqrt{3}-\sqrt{2}}-\dfrac{3}{3-\sqrt{6}}=\dfrac{\sqrt{6}\left(\sqrt{3}-\sqrt{2}\right)}{\sqrt{3}-\sqrt{2}}-\dfrac{\sqrt{3}.\sqrt{3}}{\sqrt{3}\left(\sqrt{3}-\sqrt{2}\right)}=\sqrt{6}-\dfrac{\sqrt{3}}{\sqrt{3}-\sqrt{2}}=\dfrac{3\sqrt{2}-3\sqrt{3}}{\sqrt{3}-\sqrt{2}}=\dfrac{-3\left(\sqrt{3}-\sqrt{2}\right)}{\sqrt{3}-\sqrt{2}}=-3\) \(b.\left(2\sqrt{2}-\sqrt{3}\right)^2-2\sqrt{3}\left(\sqrt{3}-2\sqrt{2}\right)=\left(2\sqrt{2}-\sqrt{3}\right)\left(2\sqrt{2}+\sqrt{3}\right)=8-3=5\) \(c.\left(\dfrac{1}{3-\sqrt{5}}-\dfrac{1}{3+\sqrt{5}}\right):\dfrac{5-\sqrt{5}}{\sqrt{5}-1}=\dfrac{3+\sqrt{5}-3+\sqrt{5}}{9-5}:\sqrt{5}=\dfrac{2\sqrt{5}}{4}.\dfrac{1}{\sqrt{5}}=\dfrac{\sqrt{5}}{2}.\dfrac{1}{\sqrt{5}}=\dfrac{1}{2}\) \(d.\left(3-\dfrac{a-2\sqrt{a}}{\sqrt{a}-2}\right)\left(3+\dfrac{\sqrt{ab}-3\sqrt{a}}{\sqrt{b}-3}\right)=\left(3-\sqrt{a}\right)\left(3+\sqrt{a}\right)=9-a\)
1.
a) \(\sqrt{3-2\sqrt{2}}+\sqrt{6-4\sqrt{2}}+\sqrt{9-4\sqrt{2}}=\sqrt{2-2\sqrt{2}+1}+\sqrt{4-2.2.\sqrt{2}+2}+\sqrt{8-2.2\sqrt{2}.1+1}=\sqrt{\left(\sqrt{2}\right)^2-2.\sqrt{2}.1+1^2}+\sqrt{2^2-2.2.\sqrt{2}+\left(\sqrt{2}\right)^2}+\sqrt{\left(2\sqrt{2}\right)^2-2.2\sqrt{2}.1+1^2}=\sqrt{\left(\sqrt{2}-1\right)^2}+\sqrt{\left(2-\sqrt{2}\right)^2}+\sqrt{\left(2\sqrt{2}-1\right)^2}=\left|\sqrt{2}-1\right|+\left|2-\sqrt{2}\right|+\left|2\sqrt{2}-1\right|=\sqrt{2}-1+2-\sqrt{2}+2\sqrt{2}-1=2\sqrt{2}\)
b) \(\sqrt{\left(4+\sqrt{10}\right)^2}-\sqrt{\left(4-\sqrt{10}\right)^2}=\left|4+\sqrt{10}\right|-\left|4-\sqrt{10}\right|=4+\sqrt{10}-4+\sqrt{10}=2\sqrt{10}\)
c) \(\dfrac{1}{\sqrt{2013}-\sqrt{2014}}-\dfrac{1}{\sqrt{2014}-\sqrt{2015}}=\dfrac{\sqrt{2013}+\sqrt{2014}}{\left(\sqrt{2013}-\sqrt{2014}\right)\left(\sqrt{2013}+\sqrt{2014}\right)}-\dfrac{\sqrt{2014}+\sqrt{2015}}{\left(\sqrt{2014}-\sqrt{2015}\right)\left(\sqrt{2014}+\sqrt{2015}\right)}=\dfrac{\sqrt{2013}+\sqrt{2014}}{2013-2014}-\dfrac{\sqrt{2014}+\sqrt{2015}}{2014-2015}=-\left(\sqrt{2013}+\sqrt{2014}\right)+\sqrt{2014}+\sqrt{2015}=-\sqrt{2013}-\sqrt{2014}+\sqrt{2014}+\sqrt{2015}=\sqrt{2015}-\sqrt{2013}\)
2.
a) \(x^2-2\sqrt{5}x+5=0\Leftrightarrow x^2-2.x.\sqrt{5}+\left(\sqrt{5}\right)^2=0\Leftrightarrow\left(x-\sqrt{5}\right)^2=0\Leftrightarrow x-\sqrt{5}=0\Leftrightarrow x=\sqrt{5}\)Vậy S={\(\sqrt{5}\)}
b) ĐK:x\(\ge-3\)
\(\sqrt{x+3}=1\Leftrightarrow\left(\sqrt{x+3}\right)^2=1^2\Leftrightarrow x+3=1\Leftrightarrow x=-2\left(tm\right)\)
Vậy S={-2}
3.
a) \(A=\dfrac{x-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}}+\dfrac{2\left(x-1\right)}{\sqrt{x}-1}=\dfrac{\sqrt{x}\left(x\sqrt{x}-1\right)}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}+\dfrac{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}-1}=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-\left(2\sqrt{x}+1\right)+2\left(\sqrt{x}+1\right)=\sqrt{x}\left(\sqrt{x}-1\right)-2\sqrt{x}-1+2\sqrt{x}+2=x-\sqrt{x}+1\)
b) Ta có \(A=x-\sqrt{x}+1=x-2\sqrt{x}.\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}=\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)
Ta có \(\left(\sqrt{x}-\dfrac{1}{2}\right)^2\ge0\Leftrightarrow\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\Leftrightarrow A\ge\dfrac{3}{4}\)
Dấu bằng xảy ra khi x=\(\dfrac{1}{4}\)
Vậy GTNN của A=\(\dfrac{3}{4}\)
\(1a.2\sqrt{40\sqrt{12}}-2\sqrt{\sqrt{75}}-3\sqrt{548}=2\sqrt{16.5\sqrt{3}}-2\sqrt{\sqrt{75}}-6\sqrt{137}=8\sqrt{\sqrt{75}}-2\sqrt{\sqrt{75}}-6\sqrt{137}=6\sqrt{\sqrt{75}}-6\sqrt{137}\) \(b.\left(\sqrt{12}+\sqrt{75}+\sqrt{27}\right):\sqrt{15}=\left(2\sqrt{3}+5\sqrt{3}+3\sqrt{3}\right).\dfrac{1}{\sqrt{15}}=10\sqrt{3}.\dfrac{1}{\sqrt{3}.\sqrt{5}}=2\sqrt{5}\) \(d.\left(12\sqrt{50}-8\sqrt{200}+7\sqrt{450}\right):\sqrt{10}=\left(60\sqrt{2}-80\sqrt{2}+105\sqrt{2}\right).\dfrac{1}{\sqrt{10}}=85\sqrt{2}.\dfrac{1}{\sqrt{2}.\sqrt{5}}=17\sqrt{5}\) \(e.\left(\sqrt{\dfrac{1}{7}}-\sqrt{\dfrac{16}{7}}+\sqrt{\dfrac{9}{7}}\right):\sqrt{7}=\left(\sqrt{\dfrac{1}{7}}-4\sqrt{\dfrac{1}{7}}+3\sqrt{\dfrac{1}{7}}\right).\dfrac{1}{\sqrt{7}}=0\) \(2a.A=\sqrt{3+\sqrt{5+2\sqrt{3}}}.\sqrt{3-\sqrt{5+2\sqrt{3}}}=\sqrt{9-5-2\sqrt{3}}=\sqrt{3-2\sqrt{3}+1}=\sqrt{3}-1\) \(b.B=\sqrt{4+\sqrt{8}}.\sqrt{2+\sqrt{2+\sqrt{2}}}.\sqrt{2-\sqrt{2+\sqrt{2}}}=\sqrt{2}.\sqrt{2+\sqrt{2}}.\sqrt{2-\sqrt{2}}=\sqrt{2}.\sqrt{4-2}=2\)
1) \(\sqrt{12}\)+\(5\sqrt{3}-\sqrt{48}\)
= \(2\sqrt{3}+5\sqrt{3}-4\sqrt{3}\)
= (2+5-4).\(\sqrt{3}\)
= \(3\sqrt{3}\)
2)\(5\sqrt{5}+\sqrt{20}-3\sqrt{45}\)
= \(5\sqrt{5}+2\sqrt{5}-3.3\sqrt{5}\)
= \(5\sqrt{5}+2\sqrt{5}-9\sqrt{5}\)
= \(\left(5+2-9\right).\sqrt{5}\)
= -2\(\sqrt{2}\)
3)\(3\sqrt{32}+4\sqrt{8}-5\sqrt{18}\)
= \(3.4\sqrt{2}+4.2\sqrt{2}-5.3\sqrt{2}
\)
= 12\(\sqrt{2}\) \(+8\sqrt{2}\) \(-15\sqrt{2}\)
= \(\left(12+8-15\right).\sqrt{2}\)
= \(5\sqrt{2}\)
4)\(3\sqrt{12}-4\sqrt{27}+5\sqrt{48}\)
= \(3.2\sqrt{3}-4.3\sqrt{3}+5.4\sqrt{3}\)
= \(6\sqrt{3}-12\sqrt{3}+20\sqrt{3}\)
= \(\left(6-12+20\right).\sqrt{3}\)
= \(14\sqrt{3}\)
5)\(\sqrt{12}+\sqrt{75}-\sqrt{27}\)
= \(2\sqrt{3}+5\sqrt{3}-3\sqrt{3}\)
= \(\left(2+5-3\right).\sqrt{3}\)
= \(4\sqrt{3}\)
6) \(2\sqrt{18}-7\sqrt{2}+\sqrt{162}\)
= \(2.3\sqrt{2}-7\sqrt{2}+9\sqrt{2}\)
= 6\(\sqrt{2}-7\sqrt{2}+9\sqrt{2}\)
= \(\left(6-7+9\right).\sqrt{2}\)
= 8\(\sqrt{2}\)
7)\(3\sqrt{20}-2\sqrt{45}+4\sqrt{5}\)
= \(3.2\sqrt{5}-2.3\sqrt{5}+4\sqrt{5}\)
= \(6\sqrt{5}-6\sqrt{5}+4\sqrt{5}\)
= \(4\sqrt{5}\)
8)\(\left(\sqrt{2}+2\right).\sqrt{2}-2\sqrt{2}\)
= \(\left(\sqrt{2}\right)^2+2\sqrt{2}-2\sqrt{2}\)
= 2
Bài 50:
\(\dfrac{5}{\sqrt{10}}=\dfrac{5\sqrt{10}}{10}=\dfrac{\sqrt{10}}{2}\)
\(\dfrac{5}{2\sqrt{5}}=\dfrac{\sqrt{5}}{2}\)
\(\dfrac{1}{3\sqrt{20}}=\dfrac{1}{6\sqrt{5}}=\dfrac{\sqrt{5}}{30}\)
\(\dfrac{2\sqrt{2}+2}{5\sqrt{2}}=\dfrac{\sqrt{2}\left(2+\sqrt{2}\right)}{5\sqrt{2}}=\dfrac{2+\sqrt{2}}{5}\)
a: \(=\dfrac{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}-\sqrt{ab}=\sqrt{ab}-\sqrt{ab}=0\)
b: \(=\dfrac{\left(\sqrt{x}-2\sqrt{y}\right)^2}{\sqrt{x}-2\sqrt{y}}+\dfrac{\sqrt{y}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}\)
\(=\sqrt{x}-2\sqrt{y}+\sqrt{y}=\sqrt{x}-\sqrt{y}\)
c: \(=\sqrt{x}+2-\dfrac{x-4}{\sqrt{x}-2}\)
\(=\sqrt{x}+2-\sqrt{x}-2=0\)
a) \(2\sqrt{50}-3\sqrt{32}-\sqrt{162}+5\sqrt{98}\)
=\(2.5\sqrt{2}-3.4\sqrt{2}-9\sqrt{2}+5.7\sqrt{2}\)
= \(10\sqrt{2}-12\sqrt{2}-9\sqrt{2}+35\sqrt{2}\)
= \(24\sqrt{2}\)
b) \(\sqrt{8+2\sqrt{7}}+\sqrt{11-4\sqrt{7}}\)
= \(\sqrt{7+2\sqrt{7}+1}+\sqrt{7-4\sqrt{7}+4}\)
= \(\sqrt{\left(\sqrt{7}+1\right)^2}+\sqrt{\left(\sqrt{7}-2\right)^2}\)
= \(\sqrt{7}+1+\sqrt{7}-2\)
= \(2\sqrt{7}-1\)
c) \(\dfrac{10}{\sqrt{5}}+\dfrac{8}{3+\sqrt{5}}-\dfrac{\sqrt{18}-3\sqrt{5}}{\sqrt{2}-\sqrt{5}}\)
= \(2\sqrt{5}+6-2\sqrt{5}-3\)
= 3