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a: \(=2\cdot\dfrac{4}{3}\sqrt{3}-3\cdot\dfrac{1}{9}\sqrt{3}-6\cdot\dfrac{2}{15}\sqrt{3}\)
\(=\dfrac{8}{3}\sqrt{3}-\dfrac{1}{3}\sqrt{3}-\dfrac{4}{5}\sqrt{3}=\dfrac{23}{15}\sqrt{3}\)
b: \(=\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\left(2+\sqrt{3}\right)^2}\)
\(=2-\sqrt{3}+2+\sqrt{3}=4\)
c: \(=6\sqrt{3}-4\sqrt{3}+\dfrac{3}{5}\cdot5\sqrt{3}=2\sqrt{3}+3\sqrt{3}=5\sqrt{3}\)
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
\(a,2\sqrt{\dfrac{27}{4}}-\sqrt{\dfrac{48}{9}}-\dfrac{2}{5}.\sqrt{\dfrac{75}{16}}\)
\(\Leftrightarrow2.\dfrac{\sqrt{27}}{2}-\sqrt{\dfrac{48}{3}}-\dfrac{2}{5}.\dfrac{\sqrt{75}}{4}\)
\(\Leftrightarrow\sqrt{27}-\dfrac{4\sqrt{3}}{3}-\dfrac{1}{5}.\dfrac{5\sqrt{3}}{2}\)
\(\Leftrightarrow3\sqrt{3}-\dfrac{4\sqrt{3}}{3}-\dfrac{\sqrt{3}}{2}\)
\(\Leftrightarrow\dfrac{7\sqrt{3}}{6}\)
\(b,\left(1+\dfrac{5-\sqrt{5}}{1-\sqrt{5}}\right).\left(\dfrac{5+\sqrt{5}}{1+\sqrt{5}}+1\right)\)
\(\Leftrightarrow\)\(\left[1+\dfrac{\left(5-\sqrt{5}\right)\left(1+\sqrt{5}\right)}{-4}\right].\left[\dfrac{\left(5+\sqrt{5}\right).\left(1-\sqrt{5}\right)}{-4}+1\right]\)
\(\Leftrightarrow\)\(\left(1+\dfrac{5+5\sqrt{5}-\sqrt{5}-5}{-4}\right).\left(\dfrac{5-5\sqrt{5}+\sqrt{5}-5}{-4}+1\right)\)
\(\Leftrightarrow\)\(\left(1+\dfrac{4\sqrt{5}}{-4}\right)\left(\dfrac{-4\sqrt{5}}{-4}+1\right)\)
\(\Leftrightarrow\left(1-\sqrt{5}\right)\left(\sqrt{5}+1\right)\)
\(\Leftrightarrow\left(1-\sqrt{5}\right).\left(1+\sqrt{5}\right)\)
<=> 1-5
=-4
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: \(=\dfrac{\sqrt{2}+\sqrt{3}+2}{\left(\sqrt{2}+\sqrt{3}+2\right)+\sqrt{2}\left(\sqrt{2}+\sqrt{3}+2\right)}=\dfrac{1}{1+\sqrt{2}}=\sqrt{2}-1\)
B1 :
a) \(\sqrt{1,2.270}=\sqrt{0,4.3.90.3}=3\sqrt{36}=3.6=18\)
\(\sqrt{55.77.35}=\sqrt{5.11.7.11.7.5}=\sqrt{25.49.212}=\sqrt{25}.\sqrt{49}.\sqrt{121}=5.7.11=385\)
b) \(\left(\sqrt{3}-\sqrt{2}\right)^2=3-2.\sqrt{3}.\sqrt{2}+2=5-2\sqrt{6}\)
\(\left(3\sqrt{2}-1\right)\left(3\sqrt{2}+1\right)=3\sqrt{2}.3\sqrt{2}+3\sqrt{2}-3\sqrt{2}-1=18-1\)
\(\left(\sqrt{6}+2\right)\left(\sqrt{3}-2\right)=\sqrt{6}.\sqrt{3}-2\sqrt{6}+2\sqrt{3}-4=\sqrt{18}-2\sqrt{6}+2\sqrt{3}-4\)\(=3\sqrt{2}-2\sqrt{6}+2\sqrt{3}-4\)
\(c,\left(\sqrt{\dfrac{3}{2}}-\sqrt{\dfrac{2}{3}}\right)=\dfrac{\sqrt{3}}{\sqrt{2}}-\dfrac{\sqrt{2}}{\sqrt{3}}=\dfrac{3-2}{\sqrt{2}\sqrt{3}}\) = \(\dfrac{1}{\sqrt{6}}\)
\(\left(\sqrt{\dfrac{8}{3}}-\sqrt{24}+\sqrt{\dfrac{50}{3}}\right).\sqrt{6}=\sqrt{\dfrac{8}{3}}.\sqrt{6}-\sqrt{24}.\sqrt{6}+\sqrt{\dfrac{50}{3}}.\sqrt{6}\) = \(\dfrac{\sqrt{8}.\sqrt{6}}{\sqrt{3}}-\sqrt{144}+\dfrac{\sqrt{50}.\sqrt{6}}{\sqrt{3}}=\dfrac{\sqrt{48}}{\sqrt{3}}-12+\dfrac{\sqrt{300}}{\sqrt{3}}=\sqrt{\dfrac{48}{3}}-12+\sqrt{\dfrac{300}{3}}=4-12+10=2\)
B2 :
a) \(\sqrt{\dfrac{1}{8}}.\sqrt{2}.\sqrt{125}.\sqrt{\dfrac{1}{5}}=\sqrt{\dfrac{1}{8}.2.125.\dfrac{1}{5}}=\sqrt{\dfrac{25}{4}}=\dfrac{5}{2}\)
\(\sqrt{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}=\sqrt{2+\sqrt{2}-\sqrt{2}-1}=1\)
b) \(\sqrt{\left(\sqrt{2}-3\right)^2}.\sqrt{11+6\sqrt{2}}=\left|\sqrt{2}-3\right|.\sqrt{2+6\sqrt{2}+9}=\left(\sqrt{2}-3\right).\sqrt{\left(\sqrt{2}+3\right)^2}=\left(\sqrt{2}-3\right)\)\(\left(\sqrt{2}+3\right)=2+3\sqrt{2}-3\sqrt{2}-9=-7\)
\(\sqrt{\left(\sqrt{3}-3\right)^2}.\sqrt{\dfrac{1}{3-\sqrt{3}}}=\left|\sqrt{3}-3\right|.\dfrac{1}{3-\sqrt{3}}=-\left(3-\sqrt{3}\right).\left(\dfrac{1}{3-\sqrt{3}}\right)=-1\)
b: \(=\left(\sqrt{ab}+\dfrac{2\sqrt{ab}}{a}-\sqrt{\dfrac{a^2+1}{ab}}\right)\cdot\sqrt{ab}\)
\(=ab+\dfrac{2ab}{a}-\sqrt{a^2+1}=ab+2b-\sqrt{a^2+1}\)
c: \(=2\sqrt{6b}-6\sqrt{18}+10\sqrt{12}-\sqrt{48}\)
\(=2\sqrt{6b}-18\sqrt{2}+20\sqrt{3}-4\sqrt{3}\)
\(=2\sqrt{6n}-18\sqrt{2}+16\sqrt{3}\)
d: \(=\dfrac{\sqrt{3}\left(\sqrt{5}-\sqrt{2}\right)}{\sqrt{7}\left(\sqrt{5}-\sqrt{2}\right)}=\dfrac{\sqrt{21}}{7}\)
a) \(\sqrt{2-\sqrt{3}}\left(\sqrt{6}+\sqrt{2}\right)\)
\(=\sqrt{2-\sqrt{3}}\sqrt{\left(\sqrt{6}+\sqrt{2}\right)^2}\)
\(=\sqrt{\left(2-\sqrt{3}\right)\left(\sqrt{6}+\sqrt{2}\right)^2}\)
\(=\sqrt{\left(2-\sqrt{3}\right)\left(6+2\sqrt{12}+2\right)}\)
\(=\sqrt{\left(2-\sqrt{3}\right)\left(6+4\sqrt{3}+2\right)}\)
\(=\sqrt{\left(2-\sqrt{3}\right)\left(8+4\sqrt{3}\right)}\)
\(=\sqrt{\left(2-\sqrt{3}\right)\cdot4\left(2+\sqrt{3}\right)}\)
\(=\sqrt{\left(4-3\right)\cdot4}\)
\(=\sqrt{1\cdot4}\)
\(=\sqrt{4}\)
\(=2\)
b) \(\left(\sqrt{2}+1\right)^3-\left(\sqrt{2}-1\right)^3\)
\(=2\sqrt{2}+6+3\sqrt{2}+1-\left(2\sqrt{2}-6+3\sqrt{2}-1\right)\)
\(=2\sqrt{2}+6+3\sqrt{2}+1-\left(5\sqrt{2}-7\right)\)
\(=2\sqrt{2}+6+3\sqrt{2}+1-5\sqrt{2}+7\)
\(=0+14\)
\(=14\)
c) \(\dfrac{2\sqrt{8}-\sqrt{12}}{\sqrt{18}-\sqrt{48}}-\dfrac{\sqrt{5}+\sqrt{27}}{\sqrt{30}+\sqrt{162}}\)
dài quá ==' cả d, e, f nữa ==' có j rảnh lm cho nhé :D
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
\(1a.2\sqrt{40\sqrt{12}}-2\sqrt{\sqrt{75}}-3\sqrt{548}=2\sqrt{16.5\sqrt{3}}-2\sqrt{\sqrt{75}}-6\sqrt{137}=8\sqrt{\sqrt{75}}-2\sqrt{\sqrt{75}}-6\sqrt{137}=6\sqrt{\sqrt{75}}-6\sqrt{137}\) \(b.\left(\sqrt{12}+\sqrt{75}+\sqrt{27}\right):\sqrt{15}=\left(2\sqrt{3}+5\sqrt{3}+3\sqrt{3}\right).\dfrac{1}{\sqrt{15}}=10\sqrt{3}.\dfrac{1}{\sqrt{3}.\sqrt{5}}=2\sqrt{5}\) \(d.\left(12\sqrt{50}-8\sqrt{200}+7\sqrt{450}\right):\sqrt{10}=\left(60\sqrt{2}-80\sqrt{2}+105\sqrt{2}\right).\dfrac{1}{\sqrt{10}}=85\sqrt{2}.\dfrac{1}{\sqrt{2}.\sqrt{5}}=17\sqrt{5}\) \(e.\left(\sqrt{\dfrac{1}{7}}-\sqrt{\dfrac{16}{7}}+\sqrt{\dfrac{9}{7}}\right):\sqrt{7}=\left(\sqrt{\dfrac{1}{7}}-4\sqrt{\dfrac{1}{7}}+3\sqrt{\dfrac{1}{7}}\right).\dfrac{1}{\sqrt{7}}=0\) \(2a.A=\sqrt{3+\sqrt{5+2\sqrt{3}}}.\sqrt{3-\sqrt{5+2\sqrt{3}}}=\sqrt{9-5-2\sqrt{3}}=\sqrt{3-2\sqrt{3}+1}=\sqrt{3}-1\) \(b.B=\sqrt{4+\sqrt{8}}.\sqrt{2+\sqrt{2+\sqrt{2}}}.\sqrt{2-\sqrt{2+\sqrt{2}}}=\sqrt{2}.\sqrt{2+\sqrt{2}}.\sqrt{2-\sqrt{2}}=\sqrt{2}.\sqrt{4-2}=2\)
A: \(A=\sqrt{9}-3\sqrt{\dfrac{50}{9}}+3\sqrt{8}-\sqrt[3]{27}\)
\(=3-3\cdot\dfrac{5\sqrt{2}}{3}+6\sqrt{2}-3\)
\(=-5\sqrt{2}+6\sqrt{2}=\sqrt{2}\)
b: \(B=\sqrt{\left(2-\sqrt{3}\right)^2}+\dfrac{2}{\sqrt{3}-1}-6\cdot\sqrt{\dfrac{16}{3}}\)
\(=\left|2-\sqrt{3}\right|+\dfrac{2\left(\sqrt{3}+1\right)}{3-1}-6\cdot\dfrac{4}{\sqrt{3}}\)
\(=2-\sqrt{3}+\sqrt{3}+1-4\sqrt{3}\)
\(=3-4\sqrt{3}\)
\(A=\sqrt{9}-3\sqrt{\dfrac{50}{9}}+3\sqrt{8}-\sqrt[3]{27}\\ =3-3\cdot\dfrac{1}{3}\sqrt{25\cdot2}+3\sqrt{4\cdot2}-3\\ =3-1\cdot5\sqrt{2}+3\cdot2\sqrt{2}-3\\ =3-5\sqrt{2}+6\sqrt{2}-3\\ =\sqrt{2}\)
\(B=\sqrt{\left(2-\sqrt{3}\right)^2}+\dfrac{2}{\sqrt{3}-1}-6\sqrt{\dfrac{16}{3}}\\ =\left|2-\sqrt{3}\right|+\dfrac{2\left(\sqrt{3}+1\right)}{3-1}-6\cdot\dfrac{4\sqrt{3}}{3}\\ =2-\sqrt{3}+\sqrt{3}+1-8\sqrt{3}\\ =3-8\sqrt{3}\)