Question

Cho P= \(\dfrac{\sqrt{x}-3}{\sqrt{x}-2}-\dfrac{2\sqrt{x}-1}{\sqrt{x}-1}+\dfrac{x-2}{x-3\sqrt{x}+2}\)

a) Tìm ĐKXĐ

b) Rút gọn

c) Tìm x để P<1

d) Tìm x thuộc số nguyên để P thuộc số nguyên

Answer 1

a) * x≥ 0; x≠4 ; x≠1

b)\(P=\dfrac{\sqrt{x}-3}{\sqrt{x}-2}-\dfrac{2\sqrt{x}-1}{\sqrt{x}-1}+\dfrac{x-2}{x-3\sqrt{x}+2}\)

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

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

=\(\dfrac{x-\sqrt{x}-3\sqrt{x}+3-2x+4\sqrt{x}+\sqrt{x}-2+x-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)

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

c) để P<1 thì \(\sqrt{x}-2< 1\Leftrightarrow\sqrt{x}< 3\Leftrightarrow x< 9\)

mà x ≠ 1 , x≠ 4, x≥ 0

=> x ϵ { 0,2,3,5,6,7,8}

Answer 2

d/ \(P=\dfrac{1}{\sqrt{x}-2}\)

Để P nguyên thì \(\dfrac{1}{\sqrt{x}-2}\in Z\)

=> \(\sqrt{x}-2\inƯ\left(1\right)\)

=> \(\sqrt{x}-2=\left\{-1;1\right\}\)

<=> \(\sqrt{x}=\left\{1;3\right\}\)

<=> \(x=\left\{1;9\right\}\)

Vậy..........

Answer 3