Bài 1: Căn bậc hai

LK

cho C =(\(\frac{\sqrt{x}}{3+\sqrt{x}}+\frac{x+9}{9-x}\))/(\(\frac{3\sqrt{x}+1}{x-3\sqrt{x}}-\frac{1}{\sqrt{x}}\))

a,rút gọn C

b,tìm x sao cho C<-1

HV
24 tháng 10 2019 lúc 19:38

a) \(C=\left(\frac{\sqrt{x}}{3+\sqrt{x}}+\frac{x+9}{9-x}\right):\left(\frac{3\sqrt{x}+1}{x-3\sqrt{x}}-\frac{1}{\sqrt{x}}\right)\)

\(=\left(\frac{\sqrt{x}}{3+\sqrt{x}}+\frac{x+9}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}\right):\left(\frac{3\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-3\right)}-\frac{1}{\sqrt{x}}\right)\)

\(=\left(\frac{\sqrt{x}\left(3-\sqrt{x}\right)+x+9}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}\right):\left(\frac{3\sqrt{x}+1-\left(\sqrt{x}-3\right)}{\sqrt{x}\left(\sqrt{x}-3\right)}\right)\)

\(=\left(\frac{3\sqrt{x}-x+x+9}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}\right):\left(\frac{3\sqrt{x}+1-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-3\right)}\right)\)

\(=\left(\frac{3\sqrt{x}+9}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}\right):\left(\frac{2\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-3\right)}\right)\)

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

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

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

b) Để C<-1 thì

\(\frac{3\sqrt{x}}{2\sqrt{x}+4}< -1\)

\(\Rightarrow3\sqrt{x}< -\left(2\sqrt{x}+4\right)\)

\(\Rightarrow3\sqrt{x}< -2\sqrt{x}-4\)

\(\Rightarrow3\sqrt{x}+2\sqrt{x}< -4\)

\(\Rightarrow5\sqrt{x}< -4\)

\(\Rightarrow\sqrt{x}< -\frac{4}{5}\)

\(\Rightarrow x< -\frac{16}{25}\)

Bình luận (0)
 Khách vãng lai đã xóa

Các câu hỏi tương tự
H24
Xem chi tiết
NL
Xem chi tiết
LD
Xem chi tiết
NH
Xem chi tiết
MT
Xem chi tiết
HT
Xem chi tiết
KH
Xem chi tiết
SC
Xem chi tiết
HT
Xem chi tiết