Rut gọn : A \(=\frac{x-3\sqrt{x}+4}{x-22}-\frac{1}{\sqrt{x}-2}\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
N
24 tháng 3 2019
ĐK: \(x>0,x\ne4\)
1, \(A=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)-\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)^2}.\frac{\left(x-4\right)^2}{\sqrt{x}^3}\)
\(A=\frac{2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)^2}.\frac{\left(x-4\right)^2}{x}\)
\(A=\frac{2}{\sqrt{x}+2}.\frac{x-4}{x}\)
\(A=\frac{2\sqrt{x}-4}{x}\)
2, \(x=\left(\sqrt{3}+1\right)^2\)
Thay \(x=\left(\sqrt{3}+1\right)^2\):
\(A=\frac{2\sqrt{3}-2}{\left(\sqrt{3}+1\right)^2}\)
3, \(A\ge\frac{1}{4}\Rightarrow\)\(\frac{2\sqrt{x}-4}{x}-\frac{1}{4}\ge0\)
\(\Leftrightarrow\frac{8\sqrt{x}-16-x}{4x}\ge0\)
\(\Rightarrow x-8\sqrt{x}+16\le0\)
\(\Leftrightarrow\sqrt{x}=4\Leftrightarrow x=16\left(TM\right)\)
5 tháng 7 2020
a) Ta có: \(A=\frac{\sqrt{x}+1}{\sqrt{x}-2}+\frac{2\sqrt{x}}{\sqrt{x}+2}+\frac{2+5\sqrt{x}}{4-x}\)
\(=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{2\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\frac{2+5\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{x+3\sqrt{x}+2+2x-4\sqrt{x}-2-5\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{3x-6\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{3\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{3\sqrt{x}}{\sqrt{x}+2}\)
29 tháng 7 2019
\(M=\frac{2\sqrt{x}-3}{\sqrt{x}-4}-\frac{\sqrt{x}+2}{\sqrt{x}+1}-\frac{2-3\sqrt{x}}{x-3\sqrt{x}-4}\)
\(=\frac{2\sqrt{x}-3}{\sqrt{x}-4}-\frac{\sqrt{x}+2}{\sqrt{x}+1}\)\(+\frac{3\sqrt{x}-2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-4\right)}\)
\(=\frac{\left(2\sqrt{x}-3\right)\left(\sqrt{x}+1\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}+4\right)+3\sqrt{x}-2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-4\right)}\)
\(=\frac{2x-\sqrt{x}-3-x+2\sqrt{x}+8+3\sqrt{x}-2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-4\right)}\)
\(=\frac{x+4\sqrt{x}+3}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-4\right)}=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-4\right)}\)
\(=\frac{\sqrt{x}+3}{\sqrt{x}-4}\)
5 tháng 8 2019
P=\(\left(\frac{3\sqrt{x}\left(\sqrt{x}-2\right)+\sqrt{x}\left(\sqrt{x}+2\right)-(x-\sqrt{x})}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\right):\left(\frac{3\sqrt{x}}{\sqrt{x}+2}\right)=\left(\frac{3x-6\sqrt{x}+x+2\sqrt{x}-x+\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\right):\left(\frac{3\sqrt{x}}{\sqrt{x}+2}\right)=\left(\frac{3x-3\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\right).\frac{\sqrt{x}+2}{3\sqrt{x}}=\frac{3\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}.\frac{\sqrt{x}+2}{3\sqrt{x}}=\frac{\sqrt{x}-1}{\sqrt{x}-2}\)
