Question

Cho biểu thức Q=\(\left(\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}-\dfrac{\sqrt{x}-2}{x-1}\right).\dfrac{\sqrt{x}+1}{\sqrt{x}}\) ( Với x>0, x\(\ne\)1)

a.Chứng minh Q=\(\dfrac{2}{x-1}\)

b.Tìm số nguyên x lớn nhất để Q có giá trị nguyên

Answer 1

a) Mạn phép ko chép lại đề , mk làm luôn.

ĐKXĐ : x > 0 ; x # 1

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

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

\(Q=\dfrac{2}{x-1}\)

b) Để Q ∈ Z ⇒ x ∈ Z

⇒ x - 1 ∈ Ư(2)

+) x - 1 = 1 ⇔ x = 2 ( TM )

+) x - 1 = - 1 ⇔ x = 0 ( KTM)

+) x - 1 = 2 ⇔ x = 3 ( TM)

+) x - 1 = - 2⇔ x = -1 ( KTM)

KL.....