Tìm GTNN của phân thức \(A=\dfrac{2030x^2+8x+1}{x^2}\left(x\ne0\right)\)
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19 tháng 11 2021
\(a,VT=\dfrac{3y\cdot2x}{4\cdot2x}=\dfrac{6xy}{8x}=VP\\ b,VT=\dfrac{\left(x+y\right)\cdot3a\left(x+y\right)}{3a\cdot3a\left(x+y\right)}=\dfrac{3a\left(x+y\right)^2}{9a^2\left(x+y\right)}=VP\)
TN
1
14 tháng 3 2021
\(P-\dfrac{2}{3}=\dfrac{x^2-6x+9}{3x^2}=\dfrac{\left(x-3\right)^2}{3x^2}\ge0\Rightarrow P\ge\dfrac{2}{3}\).
Dấu "=" xảy ra khi x = 3.
H9
HT.Phong (9A5)
CTVHS
24 tháng 9 2023
a) \(P=\left(\dfrac{4\sqrt{x}}{\sqrt{x}+2}+\dfrac{8x}{4-x}\right):\left(\dfrac{\sqrt{x}-1}{x-2\sqrt{x}}-\dfrac{2}{\sqrt{x}}\right)\)
\(P=\left(\dfrac{4\sqrt{x}}{\sqrt{x}+2}-\dfrac{8x}{x-4}\right):\left[\dfrac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-2\right)}-\dfrac{2\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\right]\)
\(P=\left[\dfrac{4\sqrt{x}}{\sqrt{x}+2}-\dfrac{8x}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\right]:\dfrac{\sqrt{x}-1-2\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(P=\left[\dfrac{4\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}-\dfrac{8x}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\right]:\dfrac{-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(P=\dfrac{4x-8\sqrt{x}-8x}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}:\dfrac{-\left(\sqrt{x}-3\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(P=\dfrac{-4x-8\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}:\dfrac{-\left(\sqrt{x}-3\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(P=\dfrac{-4\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{-\left(\sqrt{x}-3\right)}\)
\(P=\dfrac{-4\sqrt{x}\cdot\sqrt{x}}{-\left(\sqrt{x}-3\right)}\)
\(P=\dfrac{4x}{\sqrt{x}-3}\)
b) \(P=\dfrac{4x}{\sqrt{x}-3}\)
\(P=4\left(\sqrt{x}-3\right)+\dfrac{36}{\sqrt{x}-3}+24\)
Theo BĐT côsi ta có:
\(P\ge\sqrt{\dfrac{4\left(\sqrt{x}-3\right)\cdot36}{\sqrt{x}-3}}+24=36\)
Vậy: \(P_{min}=36\Leftrightarrow x=36\)
25 tháng 2 2021
`a,ĐKXĐ:x-4 ne 0,2x+2 ne 0`
`<=>x ne 4,x me -1`
`b,ĐKXĐ:4x^2-25 ne 0`
`<=>(2x-5)(2x+5) ne 0`
`<=>x ne +-5/2`
`c,ĐKXĐ:8x^3+27 ne 0`
`<=>8x^3 ne -27`
`<=>2x ne -3`
`<=>x ne -3/2`
`d,2x+2 ne 0,4y^2-9 ne 0`
`<=>2x ne -2,(2y-3)(2y+3) ne 0`
`<=>x ne -1,y ne +-3/2`
25 tháng 2 2021
b) ĐKXĐ: \(x\notin\left\{\dfrac{5}{2};-\dfrac{5}{2}\right\}\)
c) ĐKXĐ: \(x\ne-\dfrac{3}{2}\)
d) ĐKXĐ: \(\left\{{}\begin{matrix}x\ne-1\\y\notin\left\{\dfrac{3}{2};-\dfrac{3}{2}\right\}\end{matrix}\right.\)
P
1
NV
Nguyễn Việt Lâm
Giáo viên
25 tháng 8 2021
Đặt \(x+3=t\ne0\Rightarrow x=t-3\)
\(A=\dfrac{\left(t+2\right)\left(t-4\right)}{t^2}=\dfrac{t^2-2t-8}{t^2}=-\dfrac{8}{t^2}-\dfrac{2}{t}+1=-8\left(\dfrac{1}{t}+\dfrac{1}{8}\right)^2+\dfrac{9}{8}\le\dfrac{9}{8}\)
\(A_{max}=\dfrac{9}{8}\) khi \(t=-8\) hay \(x=-11\)
\(A=2030+\dfrac{8}{x}+\dfrac{1}{x^2}=\left(\dfrac{1}{x}\right)^2+8.\dfrac{1}{x}+16+2014\)
\(\Rightarrow A=\left(\dfrac{1}{x}+4\right)^2+2014\ge2014\)
\(\Rightarrow A_{min}=2014\) khi \(\dfrac{1}{x}+4=0\Rightarrow x=-\dfrac{1}{4}\)
\(A=\dfrac{2030x^2+8x+1}{x^2}\\ =\dfrac{2030x^2}{x^2}+\dfrac{8x}{x^2}+\dfrac{1}{x^2}\\ =2030+\dfrac{8}{x}+\dfrac{1}{x^2}\\ =\left(\dfrac{1}{x}\right)^2+2\cdot\dfrac{1}{x}\cdot4+16+2014\\ =\left(\dfrac{1}{x}+4\right)^2+2014\)
Do \(\left(\dfrac{1}{x}+4\right)^2\ge0,2014>0\)
\(\Rightarrow\left(\dfrac{1}{x}+4\right)^2+2014\ge2014\)
\(\Rightarrow Min\left(A\right)=2014\Leftrightarrow\dfrac{1}{x}+4=0\Rightarrow x=\dfrac{-1}{4}\)