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Bài 1: Rút gọn biểu thức1) \(\sqrt{12}-\sqrt{27}+\sqrt{48}\)              2) \(\left(\sqrt{25}+\sqrt{20}-\sqrt{80}\right):\sqrt{5}\)3) \(2\sqrt{27}-\sqrt{\frac{16}{3}}-\sqrt{48}-\sqrt{8\frac{1}{3}}\)      4) \(\frac{1}{\sqrt{5}-\sqrt{3}}-\frac{1}{\sqrt{5}+\sqrt{3}}\)5) \(\left(\sqrt{125}-\sqrt{12}-2\sqrt{5}\right)\left(3\sqrt{5}-\sqrt{3}+\sqrt{27}\right)\) ...
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Bài 1: Rút gọn biểu thức

1) \(\sqrt{12}-\sqrt{27}+\sqrt{48}\)              2) \(\left(\sqrt{25}+\sqrt{20}-\sqrt{80}\right):\sqrt{5}\)

3) \(2\sqrt{27}-\sqrt{\frac{16}{3}}-\sqrt{48}-\sqrt{8\frac{1}{3}}\)      4) \(\frac{1}{\sqrt{5}-\sqrt{3}}-\frac{1}{\sqrt{5}+\sqrt{3}}\)

5) \(\left(\sqrt{125}-\sqrt{12}-2\sqrt{5}\right)\left(3\sqrt{5}-\sqrt{3}+\sqrt{27}\right)\)   6) \(\left(3\sqrt{20}-\sqrt{125}-15\sqrt{\frac{1}{5}}\right).\sqrt{5}\)

7) \(\left(6\sqrt{128}-\frac{3}{5}\sqrt{50}+7\sqrt{8}\right):3\sqrt{2}\)  8) \(\left(2\sqrt{48}-\frac{3}{2}\sqrt{\frac{4}{3}}+\sqrt{27}\right).2\sqrt{3}\)

9) \(\sqrt{\left(3-2\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{8}-4\right)^2}\)    10) \(\sqrt{\left(4-\sqrt{15}\right)^2}+\sqrt{\left(\sqrt{15}-3\right)^2}\)

11) \(\frac{\sqrt{10}-\sqrt{2}}{\sqrt{5}-1}+\frac{2-\sqrt{2}}{\sqrt{2}-1}\)      12) \(\left(1-\frac{5+\sqrt{5}}{1+\sqrt{5}}\right)\left(\frac{5-\sqrt{5}}{1-\sqrt{5}}-1\right)\)

13) \(\sqrt{15-6\sqrt{6}}\)    14) \(\sqrt{8-2\sqrt{15}}\)    15) \(\sqrt[3]{-2}.\sqrt[3]{32}+\sqrt{2}.\sqrt{32}\)

 

1
26 tháng 11 2017

Giúp mình :<

Rút gọn biểu thức: 1) \(\sqrt{12}+5\sqrt{3}-\sqrt{48}\) 2) \(5\sqrt{5}+\sqrt{20}-3\sqrt{45}\) 3) \(2\sqrt{32}+4\sqrt{8}-5\sqrt{18}\) 4) \(3\sqrt{12}-4\sqrt{27}+5\sqrt{48}\) 5) \(\sqrt{12}+\sqrt{75}-\sqrt{27}\) 6) \(2\sqrt{18}-7\sqrt{2}+\sqrt{162}\) 7) \(3\sqrt{20}-2\sqrt{45}+4\sqrt{5}\) 8) \(\left(\sqrt{2}+2\right)\sqrt{2}-2\sqrt{2}\) 9) \(\dfrac{1}{\sqrt{5}-1}-\dfrac{1}{\sqrt{5}+}\) 10) \(\dfrac{1}{\sqrt{5}-2}+\dfrac{1}{\sqrt{5}+2}\) 11)...
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Rút gọn biểu thức:

1) \(\sqrt{12}+5\sqrt{3}-\sqrt{48}\)

2) \(5\sqrt{5}+\sqrt{20}-3\sqrt{45}\)

3) \(2\sqrt{32}+4\sqrt{8}-5\sqrt{18}\)

4) \(3\sqrt{12}-4\sqrt{27}+5\sqrt{48}\)

5) \(\sqrt{12}+\sqrt{75}-\sqrt{27}\)

6) \(2\sqrt{18}-7\sqrt{2}+\sqrt{162}\)

7) \(3\sqrt{20}-2\sqrt{45}+4\sqrt{5}\)

8) \(\left(\sqrt{2}+2\right)\sqrt{2}-2\sqrt{2}\)

9) \(\dfrac{1}{\sqrt{5}-1}-\dfrac{1}{\sqrt{5}+}\)

10) \(\dfrac{1}{\sqrt{5}-2}+\dfrac{1}{\sqrt{5}+2}\)

11) \(\dfrac{2}{4-3\sqrt{2}}-\dfrac{2}{4+3\sqrt{2}}\)

12) \(\dfrac{2+\sqrt{2}}{1+\sqrt{2}}\)

13) \(\left(\sqrt{28}-2\sqrt{14}+\sqrt{7}\right)\sqrt{7}+7\sqrt{8}\)

14) \(\left(\sqrt{14}-3\sqrt{2}\right)^2+6\sqrt{28}\)

15) \(\left(\sqrt{6}-\sqrt{5}\right)^2-\sqrt{120}\)

16) \(\left(2\sqrt{3}-3\sqrt{2}\right)^2+2\sqrt{6}+3\sqrt{24}\)

17) \(\sqrt{\left(1-\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{2}+3\right)^2}\)

18) \(\sqrt{\left(\sqrt{3}-2\right)^2}+\sqrt{\left(\sqrt{3}-1\right)^2}\)

19) \(\sqrt{\left(\sqrt{5}-3\right)^2}+\sqrt{\left(\sqrt{5}-2\right)^2}\)

20) \(\left(\sqrt{19}-3\right)\left(\sqrt{19}+3\right)\)

4
3 tháng 1 2019

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


4 tháng 1 2019
https://i.imgur.com/pmexRQv.jpg

1: \(=\sqrt{36}=6\)

2: \(=\sqrt{\left(15-9\right)\left(15+9\right)}=\sqrt{24\cdot6}=12\)

3: \(=3\sqrt{5}-1-3\sqrt{5}-1=-2\)

4: \(=3\sqrt{2}+\sqrt{3}-3\sqrt{2}+\sqrt{3}=2\sqrt{3}\)

5: \(=\left(2+\sqrt{5}\right)\left(\sqrt{5}-2\right)=5-4=1\)

12 tháng 7 2018

@Phùng Khánh Linh Cậu ơi giúp tớ với.

12 tháng 7 2018

A = \(\sqrt{2}\left(\sqrt{8}-\sqrt{32}-2\sqrt{18}\right)=\sqrt{16}-\sqrt{64}-2\sqrt{36}=4-8-2\cdot6=-4-12=-16\)

--

\(B=\sqrt{2}-\sqrt{3-\sqrt{5}}=\dfrac{2-\sqrt{6-2\sqrt{5}}}{\sqrt{2}}=\dfrac{2-\sqrt{\left(\sqrt{5}-1\right)^2}}{\sqrt{2}}=\dfrac{2-\sqrt{5}+1}{\sqrt{2}}=\dfrac{3-\sqrt{5}}{\sqrt{2}}\)

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\(C=\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}=\dfrac{\sqrt{8-2\sqrt{7}}}{\sqrt{2}}-\dfrac{\sqrt{8+2\sqrt{7}}}{\sqrt{2}}=\dfrac{\sqrt{\left(\sqrt{7}-1\right)^2}-\sqrt{\left(\sqrt{7}+1\right)^2}}{\sqrt{2}}=\dfrac{\sqrt{7}-1-\sqrt{7}-1}{\sqrt{2}}=-\dfrac{2}{\sqrt{2}}=-\sqrt{2}\)

còn lại lúc nx mk lm nốt nhé, h bận

20 tháng 6 2018

a) \(3\sqrt{8}-4\sqrt{18}+5\sqrt{32}-\sqrt{50}=3\sqrt{4.2}-4\sqrt{9.2}+5\sqrt{16.2}-\sqrt{25.2}=6\sqrt{2}-12\sqrt{2}+20\sqrt{2}-5\sqrt{2}=9\sqrt{2}\)b) \(\left(15\sqrt{50}+5\sqrt{200}-3\sqrt{450}\right):10=\left(15\sqrt{50}+5\sqrt{50.4}-3\sqrt{50.9}\right):10=\left(15\sqrt{50}+10\sqrt{50}-9\sqrt{50}\right):10=\dfrac{16\sqrt{50}}{10}=\dfrac{16\sqrt{25.2}}{10}=\dfrac{80\sqrt{2}}{10}=8\sqrt{2}\) c) \(2\sqrt{28}+2\sqrt{63}-3\sqrt{175}+\sqrt{112}=2\sqrt{7.4}+2\sqrt{7.9}-3\sqrt{7.25}+\sqrt{7.16}=4\sqrt{7}+6\sqrt{7}-15\sqrt{7}+4\sqrt{7}=-\sqrt{7}\)

d)
\(\left(\sqrt{14}-3\sqrt{2}\right)^2+6\sqrt{28}=14-2.3\sqrt{2.14}+18+6\sqrt{28}=32-6\sqrt{28}+6\sqrt{28}=32\)