Question

[Kem Py 2]

Bài 40:Cho A=(√x+2/x+2√x+1. - √x-2/x-1):√x/√x+1 với x>0;x≠1

A.Rút gọn A. B.Tính giá trị của A khi x=√3/2+√3

C.tìm x để A =4/√x +4

Answer 1

[@Nk>↑@]

A.\(A=\left(\frac{\sqrt{x}+2}{x+2\sqrt{x}+1}-\frac{\sqrt{x}-2}{x-1}\right):\frac{\sqrt{x}}{\sqrt{x}+1}\)

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

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

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

\(=\frac{2\sqrt{x}}{x-1}.\frac{1}{\sqrt{x}}\)

\(=\frac{2}{x-1}\)

B.Khi \(x=\frac{\sqrt{3}}{2+\sqrt{3}}\) thì

\(B=\frac{2}{\frac{\sqrt{3}}{2+\sqrt{3}}}=2.\frac{2+\sqrt{3}}{\sqrt{3}}=\frac{4+2\sqrt{3}}{\sqrt{3}}=\frac{4\sqrt{3}+6}{3}\)

C.Vì \(A=\frac{4}{\sqrt{x}+4}\) nên

\(\frac{2}{x-1}=\frac{4}{\sqrt{x}+4}\)

\(\Leftrightarrow\frac{1}{x-1}=\frac{2}{\sqrt{x}+4}\)

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

\(\Leftrightarrow2x-\sqrt{x}-6=0\)

\(\Leftrightarrow2x-4\sqrt{x}+3\sqrt{x}-6=0\)

\(\Leftrightarrow2\sqrt{x}\left(\sqrt{x}-2\right)+3\left(\sqrt{x}-2\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}-2=0\\2\sqrt{x}+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=2\left(c\right)\\\sqrt{x}=-\frac{3}{2}\left(l\right)\end{matrix}\right.\Leftrightarrow x=4\)

*Đoạn này bạn dùng denta tính cũng được nha*