Question

Cho A = \(\dfrac{x+3}{x-9}+\dfrac{2}{\sqrt{x}+3}\) , B = \(\dfrac{1}{\sqrt{x}-3}\)

a) Khi x = 16 thì B = ?

b) Rút gọn P = A - B

c) \(P=\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\) thì x = ?

Answer 1

a: Khi x=16 thì B=1/(4-3)=1

b: P=A-B

\(=\dfrac{x+3+2\sqrt{x}-6-\sqrt{x}-3}{x-9}=\dfrac{x+\sqrt{x}-6}{x-9}=\dfrac{\sqrt{x}-2}{\sqrt{x}-3}\)

 

Answer 2

ĐK: \(x\ge0;x\ne9\)

a) Khi \(x=16\) TMĐKXĐ thì \(B=\dfrac{1}{\sqrt{16}-3}=1\)

b) \(P=A-B\)

\(P=\dfrac{x+3}{x-9}+\dfrac{2}{\sqrt{x}+3}-\dfrac{1}{\sqrt{x}-3}\)

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

\(=\dfrac{x+3+2\sqrt{x}-6-\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(=\dfrac{x-\sqrt{x}-6}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

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

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

\(\Rightarrow\dfrac{\sqrt{x}+2}{\sqrt{x}+3}=\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\)

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

\(\Leftrightarrow x+2\sqrt{x}+2\sqrt{x}+4=x+3\sqrt{x}+\sqrt{x}+3\)

\(\Leftrightarrow4=3\) (Sai)

Vậy \(x\in\varnothing\)

Answer 3

\(a,x=16\Rightarrow B=\dfrac{1}{\sqrt{16}-3}=\dfrac{1}{4-3}=1\)

\(b,\) Rút gọn : \(A=\dfrac{x+3}{x-9}+\dfrac{2}{\sqrt{x}+3}\left(dkxd:x\ne9,x\ge0\right)\)

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

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

\(=\dfrac{x+3+2\sqrt{x}-6}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(=\dfrac{x+2\sqrt{x}-3}{x-9}\)

  Rút gọn \(P=A-B=\dfrac{x+2\sqrt{x}-3}{x-9}-\dfrac{1}{\sqrt{x}-3}\left(dkxd:x\ge0,x\ne9\right)\)

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

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

\(=\dfrac{x+\sqrt{x}-6}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(=\dfrac{x-2\sqrt{x}+3\sqrt{x}-6}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

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

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

\(=\dfrac{\sqrt{x}-2}{\sqrt{x}-3}\)

\(c,P=\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\Rightarrow\dfrac{\sqrt{x}+1}{\sqrt{x}+2}=\dfrac{\sqrt{x}-2}{\sqrt{x}-3}\left(dkxd:x\ne9,x\ne4,x\ge0\right)\)

\(\Leftrightarrow\dfrac{\sqrt{x}+1}{\sqrt{x}+2}-\dfrac{\sqrt{x}-2}{\sqrt{x}-3}=0\)

\(\Leftrightarrow\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}=0\)

\(\Leftrightarrow\dfrac{x-3\sqrt{x}+\sqrt{x}-3-x+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}=0\)

\(\Leftrightarrow-2\sqrt{x}+1=0\) ( Mất mẫu là bạn lấy mẫu nhân ngược vào 0 bên vế phải nha. )

\(\Leftrightarrow-2\sqrt{x}=-1\)

\(\Leftrightarrow\sqrt{x}=\dfrac{1}{2}\)

\(\Leftrightarrow x=\dfrac{1}{4}\)

Vậy khi \(P=\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\) thì \(x=\dfrac{1}{4}\)