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\(1a.2\sqrt{40\sqrt{12}}-2\sqrt{\sqrt{75}}-3\sqrt{548}=2\sqrt{16.5\sqrt{3}}-2\sqrt{\sqrt{75}}-6\sqrt{137}=8\sqrt{\sqrt{75}}-2\sqrt{\sqrt{75}}-6\sqrt{137}=6\sqrt{\sqrt{75}}-6\sqrt{137}\) \(b.\left(\sqrt{12}+\sqrt{75}+\sqrt{27}\right):\sqrt{15}=\left(2\sqrt{3}+5\sqrt{3}+3\sqrt{3}\right).\dfrac{1}{\sqrt{15}}=10\sqrt{3}.\dfrac{1}{\sqrt{3}.\sqrt{5}}=2\sqrt{5}\) \(d.\left(12\sqrt{50}-8\sqrt{200}+7\sqrt{450}\right):\sqrt{10}=\left(60\sqrt{2}-80\sqrt{2}+105\sqrt{2}\right).\dfrac{1}{\sqrt{10}}=85\sqrt{2}.\dfrac{1}{\sqrt{2}.\sqrt{5}}=17\sqrt{5}\) \(e.\left(\sqrt{\dfrac{1}{7}}-\sqrt{\dfrac{16}{7}}+\sqrt{\dfrac{9}{7}}\right):\sqrt{7}=\left(\sqrt{\dfrac{1}{7}}-4\sqrt{\dfrac{1}{7}}+3\sqrt{\dfrac{1}{7}}\right).\dfrac{1}{\sqrt{7}}=0\) \(2a.A=\sqrt{3+\sqrt{5+2\sqrt{3}}}.\sqrt{3-\sqrt{5+2\sqrt{3}}}=\sqrt{9-5-2\sqrt{3}}=\sqrt{3-2\sqrt{3}+1}=\sqrt{3}-1\) \(b.B=\sqrt{4+\sqrt{8}}.\sqrt{2+\sqrt{2+\sqrt{2}}}.\sqrt{2-\sqrt{2+\sqrt{2}}}=\sqrt{2}.\sqrt{2+\sqrt{2}}.\sqrt{2-\sqrt{2}}=\sqrt{2}.\sqrt{4-2}=2\)
Bài 1:
a/ \(\sqrt{\dfrac{2x^2-4x+2}{6}}=1\) .
\(\Leftrightarrow\dfrac{2\left(x^2-2x+1\right)}{6}=1\)
\(\Leftrightarrow\dfrac{\left(x-1\right)^2}{3}=1\)
\(\Leftrightarrow\left(x-1\right)^2=3\) \(\Rightarrow\left[{}\begin{matrix}x-1=\sqrt{3}\\x-1=-\sqrt{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{3}+1\\x=-\sqrt{3}+1\end{matrix}\right.\)
vậy tập nghiệm của phương trình S=\(\left\{1-\sqrt{3};\sqrt{3}+1\right\}\)
b/ ta có: \(\dfrac{6}{x-4}=\sqrt{2}\Leftrightarrow\sqrt{2}\left(x-4\right)=6\)
\(\Leftrightarrow x\sqrt{2}-4\sqrt{2}=6\)
\(\Leftrightarrow x\sqrt{2}=6+4\sqrt{2}\)
\(\Leftrightarrow x=\dfrac{6+4\sqrt{2}}{2}=4+3\sqrt{2}\)
vậy \(x=4+3\sqrt{2}\) là nghiệm của phương trình
c/ \(\sqrt{\dfrac{20}{2x^2-8x+8}}=\sqrt{5}\)
\(\Leftrightarrow\left(\sqrt{\dfrac{20}{2x^2-8x+8}}\right)^2=\left(\sqrt{5}\right)^2\)
\(\Leftrightarrow\dfrac{20}{2\left(x^2-4x+4\right)}=5\)
\(\Leftrightarrow\dfrac{10}{\left(x-2\right)^2}=\dfrac{10}{2}\)
\(\Rightarrow\left(x-2\right)^2=2\) \(\Leftrightarrow\left[{}\begin{matrix}x-2=\sqrt{2}\\x-2=-\sqrt{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2+\sqrt{2}\\x=2-\sqrt{2}\end{matrix}\right.\)
vậy tập nghiệm của phương trình \(S=\left\{2+\sqrt{2};2-\sqrt{2}\right\}\)
Bài 2:
a/ đặt A= \(\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}}\)
\(\Leftrightarrow A^2=3+\sqrt{5}+3-\sqrt{5}-2\sqrt{\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)}\)
\(\Leftrightarrow A^2=6-2\sqrt{9-5}\)
\(\Leftrightarrow A^2=6-2\sqrt{4}=6-4=2\)
\(\Rightarrow A=\sqrt{2}\)
\(\Rightarrow\)\(\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}}\) = \(\sqrt{2}\)
\(\Rightarrow\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}}-\sqrt{2}=\sqrt{2}-\sqrt{2}=0\)
b/ \(\left(\sqrt{12}+\sqrt{75}+\sqrt{27}\right):\sqrt{15}\)
\(=\dfrac{\sqrt{12}}{\sqrt{15}}+\dfrac{\sqrt{75}}{\sqrt{15}}+\dfrac{\sqrt{27}}{\sqrt{15}}=\sqrt{\dfrac{12}{15}}+\sqrt{\dfrac{75}{15}}+\sqrt{\dfrac{27}{15}}\)
\(=\dfrac{2\sqrt{5}}{5}+\sqrt{5}+\dfrac{3\sqrt{5}}{5}=\left(\dfrac{2\sqrt{5}}{5}+\dfrac{3\sqrt{5}}{5}\right)+\sqrt{5}\)
\(=\sqrt{5}+\sqrt{5}=2\sqrt{5}\)
c/ \(\left(12\sqrt{20}-8\sqrt{200}+7\sqrt{450}\right):\sqrt{10}\)
\(=\left(24\sqrt{5}-80\sqrt{2}+105\sqrt{2}\right):\sqrt{10}\)
\(=\left(24\sqrt{5}+25\sqrt{2}\right):\sqrt{10}=\dfrac{24\sqrt{5}}{\sqrt{10}}+\dfrac{25\sqrt{2}}{\sqrt{10}}\)
\(=12\sqrt{2}+5\sqrt{5}\)
Bài 1:
a: \(\sqrt{125}-2\sqrt{20}-3\sqrt{80}+4\sqrt{45}\)
\(=5\sqrt{5}-4\sqrt{5}-12\sqrt{5}+12\sqrt{5}=\sqrt{5}\)
b: \(\sqrt{\left(1-2\sqrt{7}\right)^2}+\sqrt{8+2\sqrt{7}}\)
\(=2\sqrt{7}-1+\sqrt{7}+1=3\sqrt{7}\)
c:\(\dfrac{1}{1-\sqrt{3}}-\dfrac{1}{1+\sqrt{3}}\)
\(=\dfrac{1+\sqrt{3}-1+\sqrt{3}}{-2}=-\dfrac{2\sqrt{3}}{2}=-\sqrt{3}\)
\(1.A=\dfrac{1}{2}\sqrt{48}-2\sqrt{75}-\dfrac{\sqrt{33}}{\sqrt{11}}+5\sqrt{1\dfrac{1}{3}}=\dfrac{1}{2}\sqrt{\dfrac{1}{3}.144}-2\sqrt{\dfrac{1}{3}.225}-\sqrt{\dfrac{1}{3}.9}+5\sqrt{\dfrac{4}{3}}=6\sqrt{\dfrac{1}{3}}-30\sqrt{\dfrac{1}{3}}-3\sqrt{\dfrac{1}{3}}+10\sqrt{\dfrac{1}{3}}=-17\sqrt{\dfrac{1}{3}}\) \(2.B=\left(2\sqrt{27}-3\sqrt{48}+3\sqrt{75}-\sqrt{192}\right)\left(1-\sqrt{3}\right)=\left(6\sqrt{3}-12\sqrt{3}+15\sqrt{3}-8\sqrt{3}\right)\left(1-\sqrt{3}\right)=\sqrt{3}\left(1-\sqrt{3}\right)=\sqrt{3}-3\) \(3.C=\left(2\sqrt{7}-2\sqrt{6}\right).\sqrt{6}-\sqrt{168}=2\sqrt{42}-12-2\sqrt{42}=-12\) \(4.D=\left(\sqrt{28}-2\sqrt{8}+\sqrt{7}\right).\sqrt{7}+4\sqrt{14}=\left(3\sqrt{7}-4\sqrt{2}\right)\sqrt{7}=21-4\sqrt{14}+4\sqrt{14}=21\)
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
1 , \(\left(\sqrt{12}-2\sqrt{75}\right).\sqrt{3}=\sqrt{12.3}-\sqrt{300.3}=6-30=-24\)
2 , \(\sqrt{3}.\left(\sqrt{12}.\sqrt{27}-\sqrt{3}\right)=\sqrt{12.27.3}-\sqrt{3.3}=18\sqrt{3}-3\)
3 , \(\left(7\sqrt{48}+3\sqrt{27}-\sqrt{12}\right):\sqrt{3}=\left(28\sqrt{3}+9\sqrt{3}-2\sqrt{3}\right):\sqrt{3}=35\)
4 , bạn làm tương tự nhé
5 , bạn làm tương tự nhé
6 , bạn làm tương tự nhé
a)\(\left(\sqrt{12}+\sqrt{75}+\sqrt{27}\right):\sqrt{15}\)
\(=\left(2\sqrt{3}+5\sqrt{3}+3\sqrt{3}\right):\sqrt{15}\)
\(=10\sqrt{3}:\sqrt{15}=\sqrt{300}:\sqrt{15}=\sqrt{20}=2\sqrt{5}\)
b) \(\frac{12\sqrt{50}-8\sqrt{200}+7\sqrt{450}}{\sqrt{10}}\)
\(=\frac{60\sqrt{2}-80\sqrt{2}+105\sqrt{2}}{\sqrt{10}}\)
\(=\frac{85\sqrt{2}}{10}=\frac{17\sqrt{2}}{2}\)
c)\(\frac{\sqrt{\frac{1}{7}}-\sqrt{\frac{16}{7}}+\sqrt{\frac{9}{7}}}{7}=\frac{\frac{1}{\sqrt{7}}-\frac{4}{\sqrt{7}}+\frac{3}{\sqrt{7}}}{7}=\frac{0}{7}=0\)
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) \(\sqrt{12}+2\sqrt{27}+3\sqrt{75}-9\sqrt{48}=2\sqrt{3}+6\sqrt{3}+15\sqrt{3}-36\sqrt{3}\)
\(=\left(2+6+15-36\right)\sqrt{3}=-13\sqrt{3}\)
b) \(2\sqrt{3}\left(\sqrt{27}+2\sqrt{48}-\sqrt{75}\right)=6\left(3+8-5\right)=36\)
a)\(\sqrt{12}+2\sqrt{27}+3\sqrt{75}-9\sqrt{48}\)
\(=\sqrt{4\cdot3}+2\sqrt{9\cdot3}+3\sqrt{25\cdot3}-9\sqrt{16\cdot3}\)
\(=2\sqrt{3}+6\sqrt{3}+15\sqrt{3}-36\sqrt{3}\)
\(=-13\sqrt{3}\)
b)\(2\sqrt{3}\left(\sqrt{27}+2\sqrt{48}-\sqrt{75}\right)\)
\(=2\sqrt{3}\left(\sqrt{9\cdot3}+2\sqrt{16\cdot3}-\sqrt{25\cdot3}\right)\)
\(=2\sqrt{3}\left(3\sqrt{3}+8\sqrt{3}-5\sqrt{3}\right)\)
\(2\sqrt{3}\cdot6\sqrt{3}=12\cdot3=36\)
a: Ta có: \(\sqrt{75}-2\sqrt{27}+\sqrt{48}\)
\(=5\sqrt{3}-2\cdot3\sqrt{3}+4\sqrt{3}\)
\(=3\sqrt{3}\)
c: Ta có: \(\sqrt{8+2\sqrt{7}}-\sqrt{11-4\sqrt{7}}\)
\(=\sqrt{7}+1-\sqrt{7}+2\)
=3