Question

[Bảo]

Rút gọn các biểu thức:

a) \(\dfrac{\sqrt{16a^4b^6}}{\sqrt{128a^6b^6}}\) ( a <0 ; b # 0 )

b) \(\sqrt{\dfrac{x-2\sqrt{x}+1}{x+2\sqrt{x}+1}}\) ( x lớn hơn hoặc = 0)

c) \(\sqrt{\dfrac{\left(x-2\right)^2}{\left(3-x\right)^2}}+\dfrac{x^2-1}{x-3}\) ( x<3 tại x = 0,5)

d) \(\dfrac{x-1}{\sqrt{y}-1}.\sqrt{\dfrac{\left(y-2\sqrt{y}+1^2\right)}{\left(x-1\right)^4}}\) ( x # 1; y >= 0, y #1)

e) \(4x-\sqrt{8}+\dfrac{\sqrt{x^3+2x^2}}{\sqrt{x+2}}\) ( x > -2 tại x = -\(\sqrt{2}\))

Answer 1

[qwerty]

a) \(\dfrac{\sqrt{16a^4b^6}}{\sqrt{128a^6b^6}}\)

\(=\dfrac{4a^2b^3}{8\sqrt{2}a^3b^3}\)

\(=\dfrac{1}{2\sqrt{2}a}\)

\(=\dfrac{\sqrt{2}}{4a}\)

b) \(\sqrt{\dfrac{x-2\sqrt{x}+1}{x+2\sqrt{x}+1}}\)

chịu đấy :v

c) \(\sqrt{\dfrac{\left(x-2\right)^2}{\left(3-x\right)^2}}+\dfrac{x^2-1}{x-3}\)

\(=\dfrac{x-2}{3-x}+\dfrac{x^2-1}{x-3}\)

\(=\dfrac{x-2}{-\left(x-3\right)}+\dfrac{x^2-1}{x-3}\)

\(=-\dfrac{x-2}{x-3}+\dfrac{x^2-1}{x-3}\)

\(=\dfrac{-\left(x-2\right)+x^2-1}{x-3}\)

\(=\dfrac{-x+1+x^2}{x-3}\)

d) \(\dfrac{x-1}{\sqrt{y}-1}\cdot\sqrt{\dfrac{\left(y-2\sqrt{y}+1^2\right)}{\left(x-1\right)^4}}\)

\(=\dfrac{x-1}{\sqrt{y}-1}\cdot\sqrt{\dfrac{y-2\sqrt{y}+1}{\left(x-1\right)^4}}\)

\(=\dfrac{x-1}{\sqrt{y}-1}\cdot\dfrac{\sqrt{y-2\sqrt{y}+1}}{\left(x-1\right)^2}\)

\(=\dfrac{1}{\sqrt{y}-1}\cdot\dfrac{\sqrt{y-2\sqrt{y}+1}}{x-1}\)

\(=\dfrac{\sqrt{y-2\sqrt{y}+1}}{\left(\sqrt{y}-1\right)\left(x-1\right)}\)

\(=\dfrac{\sqrt{y-2\sqrt{y}+1}}{x\sqrt{y}-\sqrt{y}-x+1}\)

e) \(4x-\sqrt{8}+\dfrac{\sqrt{x^3+2x^2}}{\sqrt{x+2}}\)

\(=4x-2\sqrt{2}+\dfrac{\sqrt{x^2\cdot\left(x+2\right)}}{\sqrt{x+2}}\)

\(=4x-2\sqrt{2}+\sqrt{x^2}\)

\(=4x-2\sqrt{x}+x\)

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

Answer 2

[Bảo]

bạn ơi phần c mình sai đề bài.. bạn giúp mk giải lại đc k \(\sqrt{\dfrac{\left(x-2\right)^4}{\left(3-x\right)^2}}+\dfrac{x^2-1}{x-3}\)