Question

\(A=\dfrac{\sqrt{X}-2}{\sqrt{X}-1};B=\dfrac{\sqrt{X}}{\sqrt{X}+1}-\dfrac{\sqrt{X}-4}{1-X}\left(X\ge1;X\ne1\right)\)

a)     Tính giá trị của biểu thức A khi x = 25

b)    Rút gọn biểu thức B

c)     Tìm x để A: B <1/2

Answer 1

a: Khi x=25 thì \(A=\dfrac{5-2}{5-1}=\dfrac{3}{4}\)

b: \(B=\dfrac{\sqrt{x}}{\sqrt{x}+1}-\dfrac{\sqrt{x}-4}{1-x}\)

\(=\dfrac{\sqrt{x}}{\sqrt{x}+1}+\dfrac{\sqrt{x}-4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)+\sqrt{x}-4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(=\dfrac{x-\sqrt{x}+\sqrt{x}-4}{x-1}=\dfrac{x-4}{x-1}\)

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

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

P<1/2

=>P-1/2<0

=>\(\dfrac{\sqrt{x}+1}{\sqrt{x}+2}-\dfrac{1}{2}< 0\)

=>\(\dfrac{2\sqrt{x}+2-\sqrt{x}-2}{2\left(\sqrt{x}+2\right)}< 0\)

=>\(\dfrac{\sqrt{x}}{2\left(\sqrt{x}+2\right)}< 0\)

=>\(x\in\varnothing\)