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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
a) \(\left(7\sqrt{48}+3\sqrt{27}-2\sqrt{12}\right):\sqrt{3}=7\sqrt{16}+3\sqrt{9}-2\sqrt{4}\)
\(=7.4+3.3-2.2=28+9-4=33\)
b) \(\left(\sqrt{125}+\sqrt{245}-\sqrt{5}\right):\sqrt{5}=\sqrt{25}+\sqrt{49}-1\)
\(=5+7-1=11\)
c) \(\left(\sqrt{\dfrac{1}{7}}-\sqrt{\dfrac{16}{7}}+\sqrt{7}\right):\sqrt{7}=\left(\dfrac{\sqrt{1}}{\sqrt{7}}-\dfrac{\sqrt{16}}{\sqrt{7}}+\sqrt{7}\right):\sqrt{7}\)
\(=\left(\dfrac{1}{\sqrt{7}}-\dfrac{4}{\sqrt{7}}+\sqrt{7}\right):\sqrt{7}=\dfrac{1}{\sqrt{7}.\sqrt{7}}-\dfrac{4}{\sqrt{7}.\sqrt{7}}+1\)
\(=\dfrac{1}{7}-\dfrac{4}{7}+1=\dfrac{1}{7}-\dfrac{4}{7}+\dfrac{7}{7}\Leftrightarrow\dfrac{1-4+7}{7}=\dfrac{4}{7}\)
bạn ghi rõ tại sao từ cái đề mà có ngay phép tính thứ hai cho mình với
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)\(2\sqrt{3}\left(\sqrt{27}+2\sqrt{48}-\sqrt{75}\right)\)
\(=2\sqrt{3}\left(\sqrt{9\cdot3}+2\sqrt{16\cdot3}-\sqrt{25\cdot3}\right)\)
\(=2\sqrt{3}\left(3\sqrt{3}+8\sqrt{3}-5\sqrt{3}\right)\)
\(2\sqrt{3}\cdot6\sqrt{3}=12\cdot3=36\)
\(a,\sqrt{12}+2\sqrt{27}+3\sqrt{75}-9\sqrt{48}=2\sqrt{3}+6\sqrt{3}+15\sqrt{3}-36\sqrt{3}\)
\(=-13\sqrt{3}\)
\(b,2\sqrt{3}.\left(\sqrt{27}+2\sqrt{48}-\sqrt{75}\right)=2\sqrt{3}\left(3\sqrt{3}+8\sqrt{3}-5\sqrt{3}\right)\)
\(=2\sqrt{3}.6\sqrt{3}=36\)
\(c,\left(2\sqrt{2}-\sqrt{3}\right)^2=8-4\sqrt{6}+3\)
\(=11-4\sqrt{6}\)
\(d,\left(1+\sqrt{3}-\sqrt{2}\right)\left(1+\sqrt{3}+\sqrt{2}\right)=1+2\sqrt{3}+3-2\)
\(=2+2\sqrt{3}\)
\(\sqrt{3\cdot27}-\sqrt{\dfrac{144}{36}}\)=\(\sqrt{81}-\sqrt{4}\)=9-2=7
\(\dfrac{2\cdot3+3\cdot6}{4}\)=6
\(\sqrt{7}-\sqrt{7-2\cdot\sqrt{7}+1}\)=\(\sqrt{7}-\left(\sqrt{7}-1\right)\)=1
\(\dfrac{\sqrt{3-2\cdot\sqrt{3}+1}}{\sqrt{2}\cdot\left(\sqrt{3}-1\right)}\)=\(\dfrac{\sqrt{3}-1}{\sqrt{2}\cdot\left(\sqrt{3}-1\right)}\)=\(\dfrac{1}{\sqrt{2}}\)
\(\dfrac{\sqrt{5}\cdot\left(\sqrt{5}+3\right)}{\sqrt{5}}\)+\(\dfrac{\sqrt{3}\cdot\left(1+\sqrt{3}\right)}{\sqrt{3}+1}\)-(\(\sqrt{5}+3\))
=(\(\sqrt{5}+3\))+\(\sqrt{3}\)-(\(\sqrt{5}+3\))=\(\sqrt{3}\)
\(\sqrt{3}\cdot\sqrt{9}+5\cdot\sqrt{4}\cdot3-2\sqrt{3}\)
=\(\sqrt{3}\cdot\left(3+10-2\right)\)
=\(11\sqrt{3}\)
a)\(3\sqrt{2}-\sqrt{8}+\sqrt{50}-4\sqrt{32}=3\sqrt{2}-2\sqrt{2}+5\sqrt{2}-16\sqrt{2}=-10\sqrt{2}\)
b) \(5\sqrt{48}-4\sqrt{27}-2\sqrt{75}+\sqrt{108}=20\sqrt{3}-12\sqrt{3}-10\sqrt{3}+6\sqrt{3}=4\sqrt{3}\)
c)\(\sqrt{12}+2\sqrt{75}-3\sqrt{48}-\frac{2}{7}\sqrt{147}=2\sqrt{3}+10\sqrt{3}-12\sqrt{3}-2\sqrt{3}=-2\sqrt{3}\)
d) \(\sqrt{\left(3+\sqrt{5}\right)^2}-\sqrt{9-4\sqrt{5}}\)
\(=\left|3+\sqrt{5}\right|-\sqrt{\left(\sqrt{5}-2\right)^2}=3+\sqrt{5}-\left|\sqrt{5}-2\right|=3+\sqrt{5}-\sqrt{5}+2=5\)
e) \(\left(\frac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\frac{5}{\sqrt{5}}\right):\frac{\sqrt{5}+\sqrt{2}}{3}\)
\(=\left[\frac{\sqrt{2}\left(\sqrt{3}-1\right)}{1-\sqrt{3}}-\sqrt{5}\right]\cdot\frac{3}{\sqrt{5}+\sqrt{2}}\)
\(=-\left(\sqrt{2}+\sqrt{5}\right)\cdot\frac{3}{\sqrt{5}+\sqrt{2}}=-3\)
Nản k lm nữa ^^
Phần lớn bạn nên nhân từng cái nha
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