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14 tháng 9 2019

a) \(\sqrt{2}.\sqrt{98}=\sqrt{196}=14\)

b) \(\sqrt{75}:\sqrt{3}=\sqrt{25}=5\)

c) \(\sqrt{\left(3-\sqrt{11}\right)^2}=3-\sqrt{11}\)

d) \(\left(2\sqrt{7}+4\sqrt{3}\right).\sqrt{3}-\sqrt{84}=2\sqrt{21}+12-2\sqrt{21}=12\)

Nhớ tick

NV
16 tháng 9 2019

a/ \(\sqrt{2.98}=\sqrt{2.2.49}=\sqrt{2^2.7^2}=2.7=14\)

b/ \(\frac{\sqrt{75}}{\sqrt{3}}=\frac{\sqrt{25}.\sqrt{3}}{\sqrt{3}}=\sqrt{25}=5\)

c/ \(\sqrt{\left(3-\sqrt{11}\right)^2}=\left|3-\sqrt{11}\right|=\sqrt{11}-3\)

d/ \(\left(2\sqrt{7}+4\sqrt{3}\right)\sqrt{3}-\sqrt{84}\)

\(=2\sqrt{21}+4.3-2\sqrt{21}\)

\(=12\)

7 tháng 6 2019

Thêm câu này hộ tớ nx nhé !
e) \(\left(\sqrt{8}-3\sqrt{2}+\sqrt{10}\right).\left(\sqrt{2}-3\sqrt{0.4}\right)\)

14 tháng 7 2019

\(a,\left(\frac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}-\frac{\sqrt{216}}{3}\right)\cdot\frac{1}{\sqrt{6}}\)

\(=\left(\frac{\sqrt{12}-\sqrt{6}}{2\left(\sqrt{2}-1\right)}-\frac{6\sqrt{6}}{3}\right)\cdot\frac{1}{\sqrt{6}}\)

\(=\left(\frac{\sqrt{6}\left(\sqrt{2}-1\right)}{2\left(\sqrt{2}-1\right)}-2\sqrt{6}\right)\cdot\frac{1}{\sqrt{6}}\)

\(=\left(\frac{\sqrt{6}}{2}-\frac{4\sqrt{6}}{2}\right)\cdot\frac{1}{\sqrt{6}}\)

\(=\frac{\sqrt{6}-4\sqrt{6}}{2}\cdot\frac{1}{\sqrt{6}}\)

\(=\frac{-3\sqrt{6}}{2}\cdot\frac{1}{\sqrt{6}}\)

\(=-\frac{3}{2}\)

21 tháng 6 2023

\(2\sqrt{2}\left(\sqrt{3}-2\right)+\left(1+2\sqrt{2}\right)^2-2\sqrt{6}=9\)

21 tháng 8 2018

giúp mình với ạ

21 tháng 8 2018

\(a,\dfrac{-3}{5}.\sqrt{\left(-0.5\right)^2}\\ =\dfrac{-3}{5}.0,5\\ =\dfrac{-3}{5}.\dfrac{1}{2}\\ =-\dfrac{3}{10}\)

Câu (b) nhìn hơi lạ lạ á :v

\(c,\sqrt{\left(1-\sqrt{7}\right)^2}+\sqrt{7}\\ =\sqrt{7}-1+\sqrt{7}\\ =2\sqrt{7}-1\)

\(d,\sqrt{11+6\sqrt{2}}-\sqrt{11-6\sqrt{2}}\\ =\sqrt{\left(3+\sqrt{2}\right)^2}-\sqrt{\left(3-\sqrt{2}\right)^2}\\ =3+\sqrt{2}-\left(3-\sqrt{2}\right)\\ =3+\sqrt{2}-3+\sqrt{2}\\ =2\sqrt{2}\)

25 tháng 10 2018

Bài 1

a) √81a - √36a - √144a = 9√a - 6√a - 12√a = -9√a

b) √75 - √48 - √300 = 5√3 - 4√3 - 10√3 = -9√3

Bài 2

a) √2x-3 = 7

⇒ 2x-3 = 49 ⇔ 2x = 52 ⇔ x =26

c) √16x - √9x = 2

⇔ 4√x - 3√x = 2 ⇔ √x = 2 ⇔ x = 4

Bài 3

a) √(2-√5)2 = l 2-√5 l = √5-2

b) (a - 3)2 + (a - 9)

= a2 - 6a + 9 + a - 9 = a2 - 5a

c) A=\(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+3}{x-9}:\left(\dfrac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)

=\(\left(\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-3x-3}{x-9}\right):\left(\dfrac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\right)\)

=\(\left(\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{x-9}\right):\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\right)\)

=\(\left(\dfrac{-3\sqrt{x}-3}{x-9}\right).\left(\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\right)\)

=\(\left(\dfrac{-3\left(\sqrt{x}+1\right)}{x-9}\right).\left(\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\right)\)

=\(\dfrac{-3\sqrt{x}+9}{x-9}\)

25 tháng 10 2018

mình cảm ơn bạn nhiều lắm

28 tháng 8 2018

a) ( \(\sqrt{12}\) + \(\sqrt{3}\) ) (\(\sqrt{27}\) - \(\sqrt{3}\) )

= (\(\sqrt{3}.\sqrt{4}\) + \(\sqrt{3}\) ) ( \(\sqrt{3}.\sqrt{9}\) - \(\sqrt{3}\) )

= \(\sqrt{3}\left(\sqrt{4}+1\right)\). \(\sqrt{3}\left(\sqrt{9}-1\right)\)

= \(\sqrt{3}.3.\sqrt{3}.2\)

= 3. 6 = 18

b) \(\left(5\sqrt{3}-2\sqrt{7}\right)\)\(\left(5\sqrt{3}+2\sqrt{7}\right)\)

=\(\left(5\sqrt{3}\right)^2\) \(-\left(2\sqrt{7}\right)^2\)

= 75 - 28 = 47

28 tháng 8 2018

hay :)) ok

1 tháng 8 2018

\(a.\sqrt{19-6\sqrt{2}}=\sqrt{18-2.3\sqrt{2}+1}=3\sqrt{2}-1\)

\(b.\sqrt{21+12\sqrt{3}}=\sqrt{12+2.2\sqrt{3}.3+9}=2\sqrt{3}+3\)

\(c.\sqrt{57-40\sqrt{2}}=\sqrt{32-2.4\sqrt{2}.5+25}=4\sqrt{2}-5\)

\(d.\sqrt{\left(5-2\sqrt{6}\right)\left(4-2\sqrt{3}\right)}=\sqrt{3-2\sqrt{3}.\sqrt{2}+2}.\sqrt{3-2\sqrt{3}+1}=\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}-1\right)\) \(e.\sqrt{21+6\sqrt{6}}+\sqrt{21-6\sqrt{6}}=\sqrt{18+2.3\sqrt{2}.\sqrt{3}+3}+\sqrt{18-2.3\sqrt{2}.\sqrt{3}+3}=3\sqrt{2}+\sqrt{3}+3\sqrt{2}-\sqrt{3}=6\sqrt{2}\) \(g.\sqrt{7-4\sqrt{3}}-\sqrt{7+4\sqrt{3}}=\sqrt{4-2.2\sqrt{3}+3}-\sqrt{4+2.2\sqrt{3}+3}=2-\sqrt{3}-2-\sqrt{3}=-2\sqrt{3}\)

1 tháng 8 2018

a)

=\(\sqrt{18-2.3\sqrt{2}.1+1}\)

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

\(=3\sqrt{2}-1\)

b)

=\(\sqrt{12+2.2\sqrt{3}.3+9}\)

=\(\sqrt{\left(2\sqrt{3}+3\right)^2}\)

=\(2\sqrt{3}+3\)

c)

=\(\sqrt{25-2.5.4\sqrt{2}+32}\)

=\(\sqrt{\left(5-4\sqrt{2}\right)^2}\)

=\(4\sqrt{2}-5\)

d)

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

e)

\(=\sqrt{18+2.3\sqrt{2}.\sqrt{3}+3}+\sqrt{18-2.3\sqrt{2}.\sqrt{3}+3}\\ =\sqrt{\left(3\sqrt{2}+\sqrt{3}\right)^2}+\sqrt{\left(3\sqrt{2}-\sqrt{3}\right)^2}\\ =3\sqrt{2}+\sqrt{3}+3\sqrt{2}-\sqrt{3}\\ =6\sqrt{2}\)

g)

\(=\sqrt{4-2.2.\sqrt{3}+3}-\sqrt{4+2.2.\sqrt{3}+3}\\ =\sqrt{\left(2-\sqrt{3}\right)^2}-\sqrt{\left(2+\sqrt{3}\right)^2}\\ =2-\sqrt{3}-2-\sqrt{3}\\ =-2\sqrt{3}\)

11 tháng 7 2018

B1:

1. \(\sqrt{12.5}\cdot\sqrt{0.2}\cdot\sqrt{0.1}\) \(=\sqrt{12.5\cdot0.2\cdot0.1}\) \(=\sqrt{0.25}=0.5\)

2.\(\sqrt{48.4}\cdot\sqrt{5}\cdot\sqrt{0.5}\) = \(\sqrt{48.4\cdot5\cdot0.5}\) =\(\sqrt{121}=11\)

B2:

a, \(\left(\sqrt{7}+\sqrt{3}\right)^2=7+2\cdot\sqrt{7}\cdot\sqrt{3}+3=7+2\cdot\sqrt{21}+3\)\(=10+2\sqrt{21}\)

b,\(\left(\sqrt{11}-\sqrt{5}\right)^2=11-2\sqrt{55}+5=16-2\sqrt{55}\)

c,\(\left(\sqrt{x}+\sqrt{y}\right) ^2=x+2\sqrt{xy}+y\)

d.\(\left(\sqrt{13}+\sqrt{7}\right)^2=13+2\sqrt{7}+7=20+2\sqrt{7}\)

e,\(\left(\sqrt{a}-\sqrt{b}\right)^2=a-2\sqrt{ab}+b\)

f,\(\left(\sqrt{3}-1\right)^2=3-2\sqrt{3}+1=4-2\sqrt{3}\)