A5

giúp mk với ạ

NT
4 tháng 9 2023 lúc 16:52

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

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

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

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

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

b: Khi x=9 thì \(P=\dfrac{-\sqrt{9}}{9+\sqrt{9}+1}=\dfrac{-3}{9+3+1}=-\dfrac{3}{13}\)

 

Bình luận (0)
H9
4 tháng 9 2023 lúc 16:55

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

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

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

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

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

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

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

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

b) Thay x=9 vào P ta có:

\(P=\dfrac{-\sqrt{9}}{9+\sqrt{9}+1}=\dfrac{-3}{9+3+1}=-\dfrac{3}{13}\)

Bình luận (0)

Các câu hỏi tương tự
H24
Xem chi tiết
KT
Xem chi tiết
19
Xem chi tiết
NH
Xem chi tiết
HL
Xem chi tiết
H24
Xem chi tiết
H24
Xem chi tiết
NP
Xem chi tiết
QT
Xem chi tiết