Rút gọn:
a, A = \(\sqrt{\left(1-x\right)^2}-1\) với x < 1
b, B = \(\frac{3-\sqrt{x}}{x-9}\) với x ≥ 0 và x ≠ 9
c, C = \(\frac{x-5\sqrt{x}+6}{\sqrt{x}-3}\) với x ≥ 0 và x ≠ 9
d, D = 5 - 3x - \(\sqrt{25-10x+x^2}\) với x < 5
e, E = \(\sqrt{3a}.\sqrt{27a}\) với a ≥ 0
f, F = \(\frac{1}{a-1}\sqrt{9\left(a-1\right)^2}\) với a > 1
a, \(A=\sqrt{\left(1-x\right)^2}-1=\left|1-x\right|-1=1-x-1\)(vì x<1)
<=> A=\(-x\)
b,B=\(\frac{3-\sqrt{x}}{x-9}\left(x\ge0,x\ne9\right)\)
=\(\frac{-\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=-\frac{1}{\sqrt{x}+3}\)
Vậy \(B=-\frac{1}{\sqrt{x}+3}\)
c, C=\(\frac{x-5\sqrt{x}+6}{\sqrt{x}-3}\left(x\ge0,x\ne9\right)\)
=\(\frac{x-2\sqrt{x}-3\sqrt{x}+6}{\sqrt{x}-3}\)=\(\frac{\sqrt{x}\left(\sqrt{x}-2\right)-3\left(\sqrt{x}-2\right)}{\sqrt{x}-3}\)=\(\frac{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}{\sqrt{x}-3}\)=\(\sqrt{x}-2\)
Vậy C= \(\sqrt{x}-2\)
d, D=\(5-3x-\sqrt{25-10x+x^2}\left(x< 5\right)\)
= \(5-3x-\sqrt{\left(5-x\right)^2}\)=\(5-3x-\left|5-x\right|\)=\(5-3x-5+x\) (vì x<5)=-2x
Vậy D=-2x
e, E=\(\sqrt{3a}.\sqrt{27a}\) (đk \(a\ge0\))
=\(\sqrt{3.27.a^2}=\sqrt{3^4}.a=9a\)
Vậy E=9a
f, F=\(\frac{1}{a-1}\sqrt{9\left(a-1\right)^2}\) (đk :a>1)
= \(\frac{1}{a-1}.3\left|a-1\right|\)=\(\frac{1}{a-1}.3\left(a-1\right)\) (vì a>1)=3
Vậy F=3
