Question

Rút gọn các biểu thức sau:

a. $A = (\sqrt{12}-2\sqrt5)\sqrt3 + \sqrt{60}$.

b. $B = \dfrac{\sqrt{4x}}{x-3}.\sqrt{\dfrac{x^2-6x+9}x}$ với $0<x<3$.

Answer 1

a, \(A=\left(\sqrt{12}-2\sqrt{5}\right)\sqrt{3}+\sqrt{60}\)

\(=\left(2\sqrt{3}-2\sqrt{5}\right)\sqrt{3}+2\sqrt{15}\)

\(=2\sqrt{9}-2\sqrt{15}+2\sqrt{15}=2\sqrt{9}\)

b, \(B=\frac{\sqrt{4x}}{x-3}\sqrt{\frac{x^2-6x+9}{x}}=\frac{2\sqrt{x}}{x-3}.\sqrt{\frac{\left(x-3\right)^2}{x}}\)

\(=\frac{2\sqrt{x}}{x-3}.\frac{x-3}{\sqrt{x}}=2\)

Answer 2

em thiếu, giờ mới nhìn lại \(2\sqrt{9}=2.3=6\)

Answer 3

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Answer 7

a, A=\((\sqrt{12}-2\sqrt{5})\sqrt{3}+\sqrt{60}\)

      =\((2\sqrt{3}-2\sqrt{5})\sqrt{3}+\sqrt{60}\)

      =\(2\sqrt{3}(\sqrt{3}-\sqrt{5})+2\sqrt{15}\)

      =\(6-2\sqrt{15}+2\sqrt{15}\)

      =6

b,B=\(\dfrac{\sqrt{x}}{x-3}.\sqrt{\dfrac{x^2-6x+9}{x}}\) với $0<x<3$.

      =\(\dfrac{\sqrt{x}}{x-3}.\sqrt{\dfrac{(x-3)^{2}}{x}}\)

     =1

Answer 8

A = 6

B = -2

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Answer 13

1.

A = (\sqrt{12}-2\sqrt5)\sqrt3 + \sqrt{60} = \sqrt{36} - 2\sqrt{15} + 2\sqrt{15} = 6.A=(12​−25​)3​+60​=36​−215​+215​=6.

2.

Với $0<x<3$ ta có

B = \dfrac{\sqrt{4x}}{x-3}.\sqrt{\dfrac{x^2-6x+9}x} = \dfrac{2\sqrt x}{x-3}.\sqrt{\dfrac{(x-3)^2}x} = \dfrac{-2\sqrt x}{3-x}.\dfrac{|x-3|}{\sqrt x} = \dfrac{-2\sqrt x(3-x)}{(3-x)\sqrt x} = -2.B=x−34x​​.xx2−6x+9​​=x−32x​​.x(x−3)2​​=3−x−2x​​.x​∣x−3∣​=(3−x)x​−2x​(3−x)​=−2.

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Answer 26

_{\(\left(2\sqrt{3}-2\sqrt{5}\right)\sqrt{3}+2\sqrt{15}\)}

_{\(=6-2\sqrt{15}+2\sqrt{15}\)}

_\(=6\)

 

Answer 27