Question

Bài 5: Cho P= \(\left(\dfrac{x\sqrt{x}-1}{x-\sqrt{x}}-\dfrac{x\sqrt{x}+1}{x+\sqrt{x}}\right):\dfrac{x+2}{x-2}\)

a) Rút gọn P

b) Tìm x nguyên để P nguyên

Bài 6: Cho Q= \(\left(\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}-\dfrac{\sqrt{x}-2}{x-1}\right).\left(x+\sqrt{x}\right)\)

a) Rút gọn Q

b) Tìm x nguyên để Q nguyên

giải chi tiết giúp mk vs ạ! mk đang cần gấp

Answer 1

Bài 1: 

a: \(P=\dfrac{x+\sqrt{x}+1-x+\sqrt{x}-1}{\sqrt{x}}\cdot\dfrac{x-2}{x+2}=\dfrac{2x-4}{x+2}\)

b: Để P là số nguyên thì 2x+4-8 chia hết cho x+2

=>-8 chia hết cho x+2

=>\(x+2\in\left\{1;-1;2;-2;4;-4;8;-8\right\}\)

hay \(x\in\left\{2;6\right\}\)

Answer 2

Bài 6: 

a) Q= \(\left[\dfrac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}-\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right].\left(x+\sqrt{x}\right)\) (ĐK: x ≥ 0; x ≠ \(\pm1\))

Q= \(\left[\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}-\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}\right].\left(x+\sqrt{x}\right)\)

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

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

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

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

Q= \(\dfrac{2x}{x-1}\)

b) Q= \(\dfrac{2x}{x-1}\)

Q= \(\dfrac{x+x-1+1}{x-1}\)

Q= \(\dfrac{x-1}{x-1}+\dfrac{x+1}{x-1}\)

Q= \(1+\dfrac{x+1}{x-1}\)

Q= \(1+\dfrac{x-1+2}{x-1}\)

Q= \(2+\dfrac{2}{x-1}\)

để Q nguyên thì \(2⋮\left(x-1\right)\)

⇒ \(x-1\inƯ\left(2\right)\)

⇒ \(x-1\in\left\{-2;-1;1;2\right\}\)

vì \(x-1\ge-1\)

⇒ \(x-1\in\left\{-1;1;2\right\}\)

Cho x lần lượt bằng -1;1;2

⇒ \(x\in\left\{0;2;3\right\}\)

vậy để Q nguyên thì \(x\in\left\{0;2;3\right\}\)