Question

[Nguyễn Gia Huy]

Giúp mk vs ạ phần 1

Answer 1

[Khang Diệp Lục]

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

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

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

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

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

P=\(\sqrt{x}+4\)

b)  \(P=\dfrac{x}{4}+5\)

⇔\(\sqrt{x}+4=\dfrac{x}{4}+5\)

⇔\(\dfrac{x}{4}-\sqrt{x}+1=0\)

⇔\(x-4\sqrt{x}+4=0\)

⇔\(\left(\sqrt{x}-2\right)^2=0\)

⇔\(\sqrt{x}-2=0\)

⇔\(\sqrt{x}=2\)

⇔\(x=4\) 

Vậy x=4 thì P=\(\dfrac{x}{4}+5\)

Answer 2

[Nguyễn Lê Phước Thịnh]

Bài 1: 

a) Ta có: \(P=\dfrac{\sqrt{x}+4}{\sqrt{x}-1}:\left(\dfrac{3\sqrt{x}+1}{x+2\sqrt{x}-3}-\dfrac{2}{\sqrt{x}+3}\right)\)

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

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

\(=\sqrt{x}+4\)

b) Ta có: \(P=\dfrac{x}{4}+5\)

\(\Leftrightarrow\sqrt{x}+4=\dfrac{1}{4}x+5\)

\(\Leftrightarrow\dfrac{1}{4}x-\sqrt{x}+1=0\)

\(\Leftrightarrow x-4\sqrt{x}+4=0\)

\(\Leftrightarrow x=4\)