Bài 1:
a:
x<0
=>|x|=-x
\(P=xy^2\cdot\sqrt{\dfrac{5}{x^2y^4}}\)
\(=xy^2\cdot\dfrac{\sqrt{5}}{\sqrt{x^2y^4}}\)
\(=xy^2\cdot\dfrac{\sqrt{5}}{\left|x\right|\cdot y^2}=\sqrt{5}\cdot\dfrac{x}{\left|x\right|}=\sqrt{5}\cdot\dfrac{x}{-x}=-\sqrt{5}\)
b:
a<4
=>a-4<0
\(Q=\sqrt{\dfrac{36\left(a-4\right)^2}{144}}\)
\(=\sqrt{\dfrac{36}{144}}\cdot\sqrt{\left(a-4\right)^2}\)
\(=\sqrt{\dfrac{1}{4}}\cdot\left|a-4\right|\)
\(=\dfrac{1}{2}\left(4-a\right)\)
Bài 2:
a: \(A=\sqrt{\dfrac{\sqrt{a}-1}{\sqrt{b}+1}}:\sqrt{\dfrac{\sqrt{b}-1}{\sqrt{a}+1}}\)
\(=\sqrt{\dfrac{\sqrt{a}-1}{\sqrt{b}+1}:\dfrac{\sqrt{b}-1}{\sqrt{a}+1}}\)
\(=\sqrt{\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\left(\sqrt{b}-1\right)\left(\sqrt{b}+1\right)}}=\sqrt{\dfrac{a-1}{b-1}}\)
b: Thay a=7,25; b=3,25 vào A, ta được:
\(A=\sqrt{\dfrac{7,25-1}{3,25-1}}=\sqrt{\dfrac{6.25}{2.25}}=\dfrac{2.5}{1.5}=\dfrac{5}{3}\)
