1. Tim x de \(\dfrac{\sqrt{x}-2}{3\sqrt{x}}>\dfrac{1}{6}\) voi \(x>0\), \(x\ne1\), \(x\ne2\)
2. So sanh \(\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\) voi \(\dfrac{1}{3}\)
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
28 tháng 8 2018
\(a.A=\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}+\dfrac{1}{1-\sqrt{x}}\right):\dfrac{\sqrt{x}-1}{2}=\dfrac{x+2+x-\sqrt{x}-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{2}{\sqrt{x}-1}=\dfrac{2\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)^2\left(x+\sqrt{x}+1\right)}=\dfrac{2}{x+\sqrt{x}+1}\left(x\ge0;x\ne1\right)\)
Để : \(A=\dfrac{2}{7}\Leftrightarrow\dfrac{2}{x+\sqrt{x}+1}=\dfrac{2}{7}\)
\(\Leftrightarrow x+\sqrt{x}-6=0\)
\(\Leftrightarrow x-2\sqrt{x}+3\sqrt{x}-6=0\)
\(\Leftrightarrow\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)=0\)
\(\Leftrightarrow x=4\left(TM\right)\)
\(b.A^2=\left(\dfrac{2}{x+\sqrt{x}+1}\right)^2=\dfrac{4}{\left(x+\sqrt{x}+1\right)^2}\left(1\right)\)
\(2A=2.\dfrac{2}{x+\sqrt{x}+1}=\dfrac{4}{x+\sqrt{x}+1}\left(2\right)\)
Mà : \(x+\sqrt{x}+1\le\left(x+\sqrt{x}+1\right)^2\left(3\right)\)
Từ \(\left(1;2;3\right)\Rightarrow2A\ge A^2\)
AT
30 tháng 6 2021
a) ĐKXĐ có thêm \(x\ne4\)
\(A=\left(\dfrac{x-\sqrt{x}+2}{x-\sqrt{x}-2}-\dfrac{x}{x-2\sqrt{x}}\right):\dfrac{1-\sqrt{x}}{2-\sqrt{x}}\)
\(=\left(\dfrac{x-\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}-\dfrac{x}{\sqrt{x}\left(\sqrt{x}-2\right)}\right).\dfrac{2-\sqrt{x}}{1-\sqrt{x}}\)
\(=\dfrac{\sqrt{x}\left(x-\sqrt{x}+2\right)-x\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}.\dfrac{2-\sqrt{x}}{1-\sqrt{x}}\)
\(=\dfrac{-2x+2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}.\dfrac{\sqrt{x}-2}{\sqrt{x}-1}\)
\(=\dfrac{-2\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}.\dfrac{\sqrt{x}-2}{\sqrt{x}-1}=\dfrac{-2}{\sqrt{x}+1}\)
\(B=\left(\dfrac{x}{\sqrt{x}+3}+\dfrac{1}{\sqrt{x}+3}\right):\left(1-\dfrac{2}{\sqrt{x}}+\dfrac{6}{x+3\sqrt{x}}\right)\)
\(=\dfrac{x+1}{\sqrt{x}+3}:\left(1-\dfrac{2}{\sqrt{x}}+\dfrac{6}{\sqrt{x}\left(\sqrt{x}+3\right)}\right)\)
\(=\dfrac{x+1}{\sqrt{x}+3}:\dfrac{x+3\sqrt{x}-2\left(\sqrt{x}+3\right)+6}{\sqrt{x}\left(\sqrt{x}+3\right)}\)
\(=\dfrac{x+1}{\sqrt{x}+3}:\dfrac{x+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+3\right)}=\dfrac{x+1}{\sqrt{x}+3}.\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x+1}{\sqrt{x}+1}\)
16 tháng 10 2018
Cho \(5\sqrt{x}7\) mk viet nham
Sua lai thanh \(5\sqrt{x}-7\)
19 tháng 10 2022
a: \(A=\left(\dfrac{2}{\sqrt{x}-2}+\dfrac{3}{2\sqrt{x}+1}-\dfrac{5\sqrt{x}-7}{\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\right)\cdot\dfrac{5\sqrt{x}\left(\sqrt{x}-2\right)}{2\sqrt{x}+3}\)
\(=\dfrac{4\sqrt{x}+2+3\sqrt{x}-6-5\sqrt{x}+7}{\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{5\sqrt{x}\left(\sqrt{x}-2\right)}{2\sqrt{x}+3}\)
\(=\dfrac{2\sqrt{x}+3}{\left(2\sqrt{x}+1\right)}\cdot\dfrac{5\sqrt{x}}{2\sqrt{x}+3}=\dfrac{5\sqrt{x}}{2\sqrt{x}+1}\)
b: Để A là số nguyên thì \(5\sqrt{x}⋮2\sqrt{x}+1\)
=>10 căn x+5-5 chia hết cho 2 căn x+1
=>\(2\sqrt{x}+1\in\left\{1;5\right\}\)
hay \(x\in\varnothing\)
HP
1 tháng 9 2021
1.
\(A=\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{x+9\sqrt{x}}{9-x}\)
\(=\dfrac{2\sqrt{x}}{\sqrt{x}+3}-\dfrac{x+9\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}-\dfrac{x+9\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{2x-6\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}-\dfrac{x+9\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{x-15\sqrt{x}}{x-9}\)
HP
1 tháng 9 2021
2.
\(B=\dfrac{3}{\sqrt{x}-3}+\dfrac{2}{\sqrt{x}+3}+\dfrac{x-5\sqrt{x}-3}{x-9}\)
\(=\dfrac{3\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\dfrac{2\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\dfrac{x-5\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{3\sqrt{x}+9+2\sqrt{x}-6+x-5\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{x}{x-9}\)
1 tháng 12 2023
\(C=\dfrac{2}{\sqrt{5}+1}+\sqrt{\dfrac{2}{3-\sqrt{5}}}\)
\(=\dfrac{2}{\sqrt{5}+1}+\sqrt{\dfrac{2\left(3+\sqrt{5}\right)}{9-5}}\)
\(=\dfrac{2}{\sqrt{5}+1}+\sqrt{\dfrac{6+2\sqrt{5}}{4}}\)
\(=\dfrac{2\left(\sqrt{5}-1\right)}{5-1}+\dfrac{\sqrt{6+2\sqrt{5}}}{2}\)
\(=\dfrac{\sqrt{5}-1}{2}+\dfrac{\sqrt{\left(\sqrt{5}+1\right)^2}}{2}\)
\(=\dfrac{\sqrt{5}-1}{2}+\dfrac{\sqrt{5}+1}{2}=\dfrac{\sqrt{5}-1+\sqrt{5}+1}{2}=\dfrac{2\sqrt{5}}{2}=\sqrt{5}\)
\(D=\dfrac{1}{x-\sqrt{x}}-\dfrac{2\sqrt{x}}{x-1}+\dfrac{1}{x+\sqrt{x}}\)
\(=\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}-\dfrac{2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\dfrac{1}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}+1-2x+\sqrt{x}-1}{\left(\sqrt{x}-1\right)\cdot\left(\sqrt{x}+1\right)\cdot\sqrt{x}}\)
\(=\dfrac{-2x+2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{-2\left(x-\sqrt{x}\right)}{\left(x-\sqrt{x}\right)\left(\sqrt{x}+1\right)}=-\dfrac{2}{\sqrt{x}+1}\)
ND
1 tháng 8 2018
\(\dfrac{2\sqrt{X}-9}{x-5\sqrt{X}+6}-\dfrac{\sqrt{X}+3}{\sqrt{X}-2}-\dfrac{2\sqrt{X}+1}{3-\sqrt{X}}\) \(\left(X\ne2;X\ne3,X\ge0\right)\)
\(=\dfrac{2\sqrt{X}-9-\left(\sqrt{X}+3\right)\left(\sqrt{X}-3\right)+\left(2\sqrt{X}+1\right)\left(\sqrt{X}-2\right)}{\left(\sqrt{X}-2\right)\left(\sqrt{X}-3\right)}\)
\(=\dfrac{2\sqrt{X}-9-X+9+2X-4\sqrt{X}+\sqrt{X}-2}{\left(\sqrt{X}-2\right)\left(\sqrt{X}-3\right)}\)
\(=\dfrac{X-\sqrt{X}-2}{\left(\sqrt{X}-2\right)\left(\sqrt{X}-3\right)}=\dfrac{X-2\sqrt{X}+\sqrt{X}-2}{\left(\sqrt{X}-2\right)\left(\sqrt{X}-3\right)}\)
\(=\dfrac{\sqrt{X}\left(\sqrt{X}-2\right)+\left(\sqrt{X}-2\right)}{\left(\sqrt{X}-2\right)\left(\sqrt{X}-3\right)}=\dfrac{\left(\sqrt{X}-2\right)\left(\sqrt{X}+1\right)}{\left(\sqrt{X}-2\right)\left(\sqrt{X}-3\right)}=\dfrac{\sqrt{X}+1}{\sqrt{X}-3}\)
\(C=\dfrac{\sqrt{X}+1}{\sqrt{X}-3}< 1\)
\(\Rightarrow\dfrac{\sqrt{X}+1-\sqrt{X}+3}{\sqrt{X}-3}< 0\)
\(\Rightarrow\dfrac{4}{\sqrt{X}+3}< 0\) ( VÔ LÍ)
⇒ Không có X thỏa mãn
1 tháng 8 2018
sai dấu tỷ ới , hình như là \(\dfrac{4}{\sqrt{X}-3}< 0\) mà
NV
Nguyễn Việt Lâm
Giáo viên
19 tháng 1
Đặt \(P=\dfrac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\dfrac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\dfrac{x+2+x-1-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\)
\(P-\dfrac{1}{3}=\dfrac{\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{1}{3}=-\dfrac{x-2\sqrt{x}+1}{x+\sqrt{x}+1}=-\dfrac{\left(\sqrt{x}-1\right)^2}{x+\sqrt{x}+1}\le0;\forall x\ge0\)
\(\Rightarrow P\le\dfrac{1}{3}\)
Dấu "=" xảy ra khi \(x=1\) ko thỏa mãn ĐKXĐ nên \(P< \dfrac{1}{3}\)
28 tháng 6 2021
`a)(sqrtx-3)/(sqrtx-2)-(2sqrtx-1)/(sqrtx-1)+(x-2)/(x-3sqrtx+2)`
`=(x-4sqrtx+3-(2sqrtx-1)(sqrtx-2)+x-2)/(x-3sqrtx+2)`
`=(2x-4sqrtx+1-2x+5sqrtx-2)/(x-3sqrtx+2)`
`=(sqrtx-1)/(x-3sqrtx+2)`
`=1/(sqrtx-2)`
`b)((x+2)/(xsqrtx-1)-sqrtx/(x+sqrtx+1)+1/(1-sqrtx)):(sqrtx-1)/2`
`=((x+2-x+sqrtx-x-sqrtx-1)/(xsqrtx-1))*2/(sqrtx-1)`
`=(1-x)/(xsqrtx-1)*2/(sqrtx-1)`
`=(-2(sqrtx+1))/(x+sqrtx+1)`
28 tháng 6 2021
a) Ta có: \(A=\dfrac{\sqrt{x}-3}{\sqrt{x}-2}-\dfrac{2\sqrt{x}-1}{\sqrt{x}-1}+\dfrac{x-2}{x-3\sqrt{x}+2}\)
\(=\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}-\dfrac{\left(2\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}+\dfrac{x-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{x-4\sqrt{x}+3-2x+4\sqrt{x}+\sqrt{x}-2+x-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{\sqrt{x}-1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{1}{\sqrt{x}-2}\)
b) Ta có: \(\left(\dfrac{x+2}{x\sqrt{x}-1}-\dfrac{\sqrt{x}}{x+\sqrt{x}+1}+\dfrac{1}{1-\sqrt{x}}\right):\dfrac{\sqrt{x}-1}{2}\)
\(=\dfrac{x+2-\sqrt{x}\left(\sqrt{x}-1\right)-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{2}{\sqrt{x}-1}\)
\(=\dfrac{-\sqrt{x}+1-x+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{2}{\sqrt{x}-1}\)
\(=\dfrac{-\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{2}{\sqrt{x}-1}\)
\(=\dfrac{-2\sqrt{x}-2}{x\sqrt{x}-1}\)
3 tháng 9 2022
a: \(M=\dfrac{x+\sqrt{x}+1-x+\sqrt{x}-1}{\sqrt{x}}+\dfrac{x+1}{\sqrt{x}}\)
\(=\dfrac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}}\)
b: Để M=9/2 thì \(\dfrac{x+2\sqrt{x}+1}{\sqrt{x}}=\dfrac{9}{2}\)
=>\(2x+4\sqrt{x}+2-9\sqrt{x}=0\)
=>2x-5 căn x+2=0
=>(2 căn x-1)(căn x-2)=0
=>x=4 hoặc x=1/4
c: \(M-4=\dfrac{x-2\sqrt{x}+1}{\sqrt{x}}=\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}}>0\)
=>M>4
1 )Ta có :
\(\dfrac{\sqrt{x}-2}{3\sqrt{x}}>\dfrac{1}{6}\)
\(\Rightarrow6\left(\sqrt{x}-2\right)>3\sqrt{x}\)
\(\Rightarrow6\sqrt{x}-3\sqrt{x}-2>0\)
\(\Rightarrow3\sqrt{x}>2\)
\(\Rightarrow\sqrt{x}>\dfrac{2}{3}\)
\(\Rightarrow x>\dfrac{4}{9}\)
2)
Giả sử
\(\dfrac{\sqrt{x}}{x+\sqrt{x}+1}>\dfrac{1}{3}\)
=> \(3\sqrt{x}>x+\sqrt{x}+1\)
\(\Rightarrow x+\sqrt{x}+1-3\sqrt{x}< 0\)
\(\Rightarrow\left(x-2\sqrt{x}+1\right)< 0\Leftrightarrow\left(\sqrt{x-1}\right)^2< 0\) ( vô lí )
Bất đẳng thức trên là sai, mà các phép biến dổi là tương đương
\(\Rightarrow\dfrac{\sqrt{x}}{x+\sqrt{x}+1}< \dfrac{1}{3}\)
câu 2 tớ nhầm chỗ kết luận, phải là :
\(\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\le\dfrac{1}{3}\) nhé, chỗ dòng cuối cùng đấy, còn bên trên thì không ảnh hưởng gì cả