Cho `A=(\sqrt{x}-2)/(\sqrt{x}+1) ; B= (x)/((\sqrt{x}+3)(\sqrt{x}-3)`
Với `x>9`, tính min `M=A.B`
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NT
Nguyễn thành Đạt
CTVHS
14 tháng 9 2023
Trước tiên ta cần phải rút gọn biểu thức A trước.
Ta có : \(A\text{=}\dfrac{\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}}{\sqrt{x+\sqrt{2x-1}}+\sqrt{x-\sqrt{2x-1}}}\)
\(A\text{=}\dfrac{\sqrt{x-1+2\sqrt{x-1}+1}+\sqrt{x-1-2\sqrt{x-1}+1}}{\sqrt{x+\sqrt{2x-1}+\sqrt{x-\sqrt{2x-1}}}}\)
\(A\text{=}\dfrac{\sqrt{\left(\sqrt{x-1}+1\right)^2}+\sqrt{\left(\sqrt{x-1}-1\right)^2}}{\sqrt{x+\sqrt{2x+1}+\sqrt{x-\sqrt{2x+1}}}}\)
\(A\text{=}\dfrac{\sqrt{x-1}+1+|\sqrt{x-1}-1|}{\sqrt{x+\sqrt{2x-1}+\sqrt{x-\sqrt{2x-1}}}}\)
\(A\text{=}\dfrac{\sqrt{x-1}+1+\sqrt{x-1}-1}{\sqrt{x+\sqrt{2x-1}+\sqrt{x-\sqrt{2x-1}}}}\left(x\ge2\right)\)
\(A\text{=}\dfrac{2\sqrt{x-1}}{\sqrt{x+\sqrt{2x-1}+\sqrt{x-\sqrt{2x-1}}}}\)
\(A\text{=}\dfrac{2\sqrt{2\left(x-1\right)}}{\sqrt{2x-1+2\sqrt{2x-1}+1}+\sqrt{2x-1-2\sqrt{2x-1}+1}}\)
\(A\text{=}\dfrac{2\sqrt{2\left(x-1\right)}}{\sqrt{\left(\sqrt{2x-1}+1\right)^2}+\sqrt{\left(\sqrt{2x-1}-1\right)^2}}\)
\(A\text{=}\dfrac{2\sqrt{2\left(x-1\right)}}{\sqrt{2x-1}+1+\sqrt{2x-1}-1}\left(x\ge2\right)\)
\(A\text{=}\dfrac{\sqrt{2x-2}}{\sqrt{2x-1}}\)
Xét tử thức và mẫu thức của A ta thấy :
\(\sqrt{2x-2}< \sqrt{2x-1}\left(x\ge2\right)\)
\(\Rightarrow A< 1\left(đpcm\right)\)
H
18 tháng 9 2023
\(a,B=\dfrac{\sqrt{x}+1}{\sqrt{x}-2}+\dfrac{2}{\sqrt{x}+3}-\dfrac{9\sqrt{x}-3}{x+\sqrt{x}-6}\left(x>0;x\ne6\right)\\ =\dfrac{\sqrt{x}+1}{\sqrt{x}-2}+\dfrac{2}{\sqrt{x}+3}-\dfrac{9\sqrt{x}-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\\ =\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}+\dfrac{2\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}-\dfrac{9\sqrt{x}-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\\ =\dfrac{x+3\sqrt{x}+\sqrt{x}+3+2\sqrt{x}-4-9\sqrt{x}+3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\\ =\dfrac{x-3\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\\\)
\(=\dfrac{x-\sqrt{x}-2\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\\ =\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)-2\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\\ =\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\\ =\dfrac{\sqrt{x}-1}{\sqrt{x}+3}\)
`b,` Tớ tính mãi ko ra, xl cậu nha=')
H9
HT.Phong (9A5)
CTVHS
19 tháng 9 2023
b) Xét hiệu:
\(\dfrac{\sqrt{x}-1}{\sqrt{x}+3}-3\)
\(=\dfrac{\sqrt{x}-1}{\sqrt{x}+3}-\dfrac{3\left(\sqrt{x}+3\right)}{\sqrt{x}+3}\)
\(=\dfrac{\sqrt{x}-1-3\sqrt{x}-9}{\sqrt{x}+3}\)
\(=\dfrac{-2\sqrt{x}-10}{\sqrt{x}+3}\)
\(=\dfrac{-2\left(\sqrt{x}+5\right)}{\sqrt{x}+3}\)
Mà: \(x>0\Rightarrow\left\{{}\begin{matrix}\sqrt{x}+5\ge5>0\\\sqrt{x}+3\ge3>0\end{matrix}\right.\)
\(\Rightarrow\dfrac{\sqrt{x}+5}{\sqrt{x}+3}>0\)
\(\Rightarrow\dfrac{-2\left(\sqrt{x}+5\right)}{\sqrt{x}+3}< 0\)
Vậy: \(\dfrac{\sqrt{x}-1}{\sqrt{x}+3}< 3\forall x>0\)
(giúp cậu nó nha)
3 tháng 7 2023
Sửa đề: \(A=\left(\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}-\dfrac{\sqrt{x}-2}{x-1}+4\sqrt{x}\right):\dfrac{\sqrt{x}}{\sqrt{x}+1}\)
\(=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)+4\sqrt{x}\left(x-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
\(=\dfrac{x+\sqrt{x}-2-x+\sqrt{x}+2+4\sqrt{x}\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)\sqrt{x}}\)
\(=\dfrac{2\sqrt{x}\left[1+2\left(x+2\sqrt{x}+1\right)\left(\sqrt{x}-1\right)\right]}{\sqrt{x}\left(x-1\right)}\)
\(=\dfrac{2\left[1+2\left(x\sqrt{x}-x+2x-2\sqrt{x}+\sqrt{x}-1\right)\right]}{x-1}\)
\(=\dfrac{2\left[1+2x\sqrt{x}+2x-2\sqrt{x}-2\right]}{x-1}=\dfrac{2\left(2x\sqrt{x}+2x-2\sqrt{x}-1\right)}{x-1}\)
H9
HT.Phong (9A5)
CTVHS
18 tháng 10 2023
1) \(A=\sqrt{x}\left(\sqrt{x}+1\right)-\left(\sqrt{x}-1\right)^2-2\)
\(A=\sqrt{x}\cdot\sqrt{x}+\sqrt{x}-\left(x-2\sqrt{x}+1\right)-2\)
\(A=x+\sqrt{x}-\left(x-2\sqrt{x}+1\right)-2\)
\(A=x+\sqrt{x}-x+2\sqrt{x}-1-2\)
\(A=3\sqrt{x}-3\)
Thay \(x=9\) vào A ta có:
\(A=3\cdot\sqrt{9}-3=3\cdot3-3=9-3=6\)
31 tháng 10 2021
a) ĐKXĐ: \(x>0,x\ne1\)
\(A=\dfrac{x+2+x-\sqrt{x}-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{2\sqrt{x}}{\sqrt{x}-1}\)
\(=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{2\sqrt{x}}{\sqrt{x}-1}\)
\(=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{2\sqrt{x}}{\sqrt{x}-1}=\dfrac{2\sqrt{x}}{x+\sqrt{x}+1}\)
31 tháng 8 2021
a: Thay \(x=6-2\sqrt{5}\) vào A, ta được:
\(A=1-\dfrac{\sqrt{5}-1}{\sqrt{5}-1+1}=1-\dfrac{\sqrt{5}-1}{\sqrt{5}}=\dfrac{\sqrt{5}}{5}\)
b: Ta có: P=A:B
\(=\left(1-\dfrac{\sqrt{x}}{\sqrt{x}+1}\right):\left(\dfrac{\sqrt{x}-1}{\sqrt{x}-2}-\dfrac{\sqrt{x}+2}{\sqrt{x}-3}+\dfrac{5\sqrt{x}-10}{x-5\sqrt{x}+6}\right)\)
\(=\dfrac{1}{\sqrt{x}+1}:\dfrac{x-4\sqrt{x}+3-x+4+5\sqrt{x}-10}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{1}{\sqrt{x}+1}:\dfrac{1}{\sqrt{x}-2}\)
\(=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\)
NV
Nguyễn Việt Lâm
Giáo viên
1 tháng 7 2019
ĐKXĐ: \(\left\{{}\begin{matrix}-1\le x\le1\\x\ne0\end{matrix}\right.\)
\(A=\frac{\sqrt{1+x}}{\sqrt{1+x}-\sqrt{1-x}}+\frac{\sqrt{1-x}^2}{\sqrt{1-x}\left(\sqrt{1+x}-\sqrt{1-x}\right)}-\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\)
\(=\frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}}-\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\)
\(=\frac{1+x+1-x+2\sqrt{1-x^2}-\left(1+x+1-x-2\sqrt{1-x^2}\right)}{2x}\)
\(=\frac{2\sqrt{1-x^2}}{x}\)
\(\sqrt{1-x^2}=\sqrt{1-\frac{4+2\sqrt{3}}{8}}=\sqrt{\frac{4-2\sqrt{3}}{8}}=\frac{\sqrt{3}-1}{2\sqrt{2}}\)
\(\Rightarrow A=\frac{\sqrt{3}-1}{\sqrt{2}}.\frac{2\sqrt{2}}{\sqrt{3}+1}=\frac{2\left(\sqrt{3}-1\right)^2}{2}=4-2\sqrt{3}\)