N
1
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
NV
Nguyễn Việt Lâm
Giáo viên
2 tháng 6 2019
ĐKXĐ: ....
\(A=\left(\frac{3\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\frac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{8\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right):\left(\frac{2\left(\sqrt{x}+2\right)-2\sqrt{x}-3}{\sqrt{x}+2}\right)\)
\(=\left(\frac{3x-6\sqrt{x}-x-2\sqrt{x}+8\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right):\left(\frac{2\sqrt{x}+4-2\sqrt{x}-3}{\sqrt{x}+2}\right)\)
\(=\frac{2x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}.\frac{\left(\sqrt{x}+2\right)}{1}=\frac{2x}{\sqrt{x}-2}\)
b/ \(A=\frac{2x}{\sqrt{x}-2}=2\sqrt{x}+4+\frac{8}{\sqrt{x}-2}=2\left(\sqrt{x}-2\right)+\frac{8}{\sqrt{x}-2}+8\ge2\sqrt{\frac{16\left(\sqrt{x}-2\right)}{\sqrt{x}-2}}+8=16\)
\(\Rightarrow A_{min}=16\) khi \(\left(\sqrt{x}-2\right)^2=4\Rightarrow x=16\)
NV
Nguyễn Việt Lâm
Giáo viên
6 tháng 11 2019
\(A=\frac{\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)}-\frac{14\sqrt{x}+4}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)}-\frac{\left(2\sqrt{x}+3\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{3x+7\sqrt{x}-6}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)}-\frac{14\sqrt{x}+4}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)}-\frac{2x+5\sqrt{x}+3}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{x-12\sqrt{x}-13}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)}=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-13\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)}=\frac{\sqrt{x}-13}{\sqrt{x}+3}\)
\(A=\frac{\sqrt{x}+3-16}{\sqrt{x}+3}=1-\frac{16}{\sqrt{x}+3}\ge1-\frac{16}{3}=-\frac{13}{3}\)
\(A_{min}=-\frac{13}{3}\) khi \(x=0\)
PN
31 tháng 7 2017
\(A=\frac{3\sqrt{x}\left(\sqrt{x}-2\right)-\sqrt{x}\left(\sqrt{x}+2\right)+8\sqrt{x}}{x-4}:\frac{2\left(\sqrt{x}+2\right)-2\sqrt{x}-3}{\sqrt{x}+2}\)
\(A=\frac{2x}{x-4}.\left(\sqrt{x}+2\right)=\frac{2x\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(A=\frac{2x}{\sqrt{x}-2}\)
12 tháng 1 2021
\(A=\frac{\sqrt{x}+1}{\sqrt{x}-1}+\frac{\sqrt{x}-1}{\sqrt{x}+1}+\frac{3\sqrt{x}+1}{1-x}\)
a) ĐKXĐ : \(\hept{\begin{cases}x\ge0\\x\ne1\end{cases}}\)
\(=\frac{\sqrt{x}+1}{\sqrt{x}-1}+\frac{\sqrt{x}-1}{\sqrt{x}+1}-\frac{3\sqrt{x}+1}{x-1}\)
\(=\frac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\frac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{3\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{x+2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\frac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{3\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{x+2\sqrt{x}+1+x-2\sqrt{x}+1-3\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{2x-3\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{2x-2\sqrt{x}-\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{2\sqrt{x}\left(\sqrt{x}-1\right)-\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{\left(\sqrt{x}-1\right)\left(2\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{2\sqrt{x}-1}{\sqrt{x}+1}\)
b) Với x = 4 thỏa mãn ĐKXĐ
\(A=\frac{2\sqrt{4}-1}{\sqrt{4}+1}=\frac{4-1}{2+1}=\frac{3}{3}=1\)
c) Chưa nghĩ ra :<
\(A=\frac{x-2+3\sqrt{x-2}+2}{x-2+4\sqrt{x-2}+3}\)
Đặt: \(t=\sqrt{x-2}\ge0\)
\(A=\frac{\left(t+1\right)\left(t+2\right)}{\left(t+1\right)\left(t+3\right)}=\frac{t+2}{t+3}=1-\frac{1}{t+3}\ge1-\frac{1}{0+3}=\frac{2}{3}\)
Dấu "=" xảy ra <=> t = 0 hay x = 2
Vậy min A =2/3 tại x =2