Tìm x để P=\(\dfrac{-\sqrt{x+3}}{\sqrt{x}-1}\le1\)
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NV
Nguyễn Việt Lâm
Giáo viên
8 tháng 8 2021
ĐKXĐ: \(x\ge1\)
\(3\sqrt[]{x-1}+m\sqrt[]{x+1}=2\sqrt[4]{\left(x-1\right)\left(x+1\right)}\)
\(\Leftrightarrow3\sqrt[]{\dfrac{x-1}{x+1}}+m=2\sqrt[4]{\dfrac{x-1}{x+1}}\)
Đặt \(\sqrt[4]{\dfrac{x-1}{x+1}}=t\Rightarrow0\le t< 1\)
\(\Rightarrow3t^2+m=2t\Leftrightarrow-3t^2+2t=m\)
Xét \(f\left(t\right)=-3t^2+2t\) trên \([0;1)\)
\(f'\left(t\right)=-6t+2=0\Rightarrow t=\dfrac{1}{3}\)
\(f\left(0\right)=0;f\left(\dfrac{1}{3}\right)=\dfrac{1}{3};f\left(1\right)=-1\)
\(\Rightarrow-1< f\left(t\right)\le\dfrac{1}{3}\)
\(\Rightarrow-1< m\le\dfrac{1}{3}\)
Y
26 tháng 4 2022
\(a,=\dfrac{x+8\sqrt{x}+8-\left(\sqrt{x+2}\right)^2}{\sqrt{x}\left(\sqrt{x}+2\right)}:\dfrac{x+\sqrt{x}+3+\sqrt{x}+2}{\sqrt{x}\left(\sqrt{x}+2\right)}\)
\(=\dfrac{x+8\sqrt{x}+8-x-4\sqrt{x}-4}{\sqrt{x}\left(\sqrt{x}+2\right)}.\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{2\sqrt{x}+x+5}\)
\(=\dfrac{4\sqrt{x}-4}{2\sqrt{x}+x+5}\)
Vậy \(P=\dfrac{4\sqrt{x}-4}{2\sqrt{x}+x+5}\)
NV
Nguyễn Việt Lâm
Giáo viên
2 tháng 3 2021
\(\lim\limits_{x\rightarrow1^+}f\left(x\right)=\lim\limits_{x\rightarrow1^+}\dfrac{\sqrt{x+3}-2}{x-1}=\lim\limits_{x\rightarrow1^+}\dfrac{x-1}{\left(x-1\right)\left(\sqrt{x+3}+2\right)}=\lim\limits_{x\rightarrow1^+}\dfrac{1}{\sqrt{x+3}+2}=\dfrac{1}{4}\)
\(f\left(1\right)=\lim\limits_{x\rightarrow1^-}f\left(x\right)=\lim\limits_{x\rightarrow1^-}\left(mx\right)=m\)
Hàm liên tục tại x=1 khi: \(\lim\limits_{x\rightarrow1^+}f\left(x\right)=\lim\limits_{x\rightarrow1^-}f\left(x\right)=f\left(1\right)\)
\(\Leftrightarrow m=\dfrac{1}{4}\)
22 tháng 11 2021
\(a,\dfrac{x^2+x+2}{\sqrt{x^2+x+1}}=\dfrac{x^2+x+1+1}{\sqrt{x^2+x+1}}=\sqrt{x^2+x+1}+\dfrac{1}{\sqrt{x^2+x+1}}\left(1\right)\)
Áp dụng BĐT cosi: \(\left(1\right)\ge2\sqrt{\sqrt{x^2+x+1}\cdot\dfrac{1}{\sqrt{x^2+x+1}}}=2\)
Dấu \("="\Leftrightarrow x^2+x+1=1\Leftrightarrow x^2+x=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
2 tháng 3 2021
\(\lim\limits_{x\rightarrow1^+}f\left(x\right)=\lim\limits_{x\rightarrow1^+}\dfrac{\sqrt{x+3}-2}{x-1}=\lim\limits_{x\rightarrow1^+}\dfrac{x-1}{\left(x-1\right)\left(\sqrt{x+3}+2\right)}=\lim\limits_{x\rightarrow1^+}\dfrac{1}{\sqrt{x+3}+2}=\dfrac{1}{4}\)
\(f\left(1\right)=\lim\limits_{x\rightarrow1^-}f\left(x\right)=\lim\limits_{x\rightarrow1^-}\left(ax+2\right)=a+2\)
Hàm liên tục tại x=1 khi:
\(a+2=\dfrac{1}{4}\Rightarrow a=-\dfrac{7}{4}\)
AT
13 tháng 7 2021
a) ĐKXĐ: \(x\ge0,x\ne1\)
\(P=\left(\dfrac{3}{\sqrt{x}-1}+\dfrac{\sqrt{x}-3}{x-1}\right):\left(\dfrac{x+2}{x+\sqrt{x}-2}-\dfrac{\sqrt{x}}{\sqrt{x}+2}\right)\)
\(=\left(\dfrac{3}{\sqrt{x}-1}+\dfrac{\sqrt{x}-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\left(\dfrac{x+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}-\dfrac{\sqrt{x}}{\sqrt{x}+2}\right)\)
\(=\dfrac{3\left(\sqrt{x}+1\right)+\sqrt{x}-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}:\dfrac{x+2-\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{4\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}:\dfrac{\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{4\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}:\dfrac{1}{\sqrt{x}-1}=\dfrac{4\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\left(\sqrt{x}-1\right)\)
\(=\dfrac{4\sqrt{x}}{\sqrt{x}+1}\)
b) \(P=\sqrt{x}-1\Rightarrow\dfrac{4\sqrt{x}}{\sqrt{x}+1}=\sqrt{x}-1\Rightarrow4\sqrt{x}=\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\)
\(\Rightarrow4\sqrt{x}=x-1\Rightarrow x-4\sqrt{x}-1=0\)
\(\Delta=\left(-4\right)^2-4.\left(-1\right)=20\Rightarrow\left[{}\begin{matrix}\sqrt{x}=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{4-2\sqrt{5}}{2}=2-\sqrt{5}\\\sqrt{x}=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{4+2\sqrt{5}}{2}=2+\sqrt{5}\end{matrix}\right.\)
mà \(\sqrt{x}\ge0\Rightarrow\sqrt{x}=2+\sqrt{5}\Rightarrow x=9+4\sqrt{5}\)
c) \(P=\dfrac{4\sqrt{x}}{\sqrt{x}+1}=\dfrac{4\left(\sqrt{x}+1\right)-4}{\sqrt{x}+1}=4-\dfrac{4}{\sqrt{x}+1}\)
Để \(P\in Z\Rightarrow4⋮\sqrt{x}+1\Rightarrow\sqrt{x}+1\in\left\{1;2;4\right\}\left(\sqrt{x}+1\ge1\right)\)
\(\Rightarrow x\in\left\{0;1;9\right\}\) mà \(x\ne1\Rightarrow x\in\left\{0;9\right\}\)
AT
13 tháng 7 2021
Từ khúc có \(x-4\sqrt{x}-1=0\)
Ta có: \(\left(2-\sqrt{5}\right)\left(2+\sqrt{5}\right)=4-5=-1\)
Thế vào \(\Rightarrow x-4\sqrt{x}+\left(2-\sqrt{5}\right)\left(2+\sqrt{5}\right)=0\)
\(\Rightarrow x-\sqrt{x}\left(2-\sqrt{5}+2+\sqrt{5}\right)+\left(2-\sqrt{5}\right)\left(2+\sqrt{5}\right)=0\)
\(\Rightarrow x-\left(2-\sqrt{5}\right)\sqrt{x}-\left(2+\sqrt{5}\right)\sqrt{x}+\left(2-\sqrt{5}\right)\left(2+\sqrt{5}\right)=0\)
\(\Rightarrow\sqrt{x}\left(\sqrt{x}-\left(2-\sqrt{5}\right)\right)-\left(2+\sqrt{5}\right)\left(\sqrt{x}-\left(2-\sqrt{5}\right)\right)=0\)
\(\Rightarrow\left(\sqrt{x}-\left(2-\sqrt{5}\right)\right)\left(\sqrt{x}-\left(2+\sqrt{5}\right)\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt{x}=2-\sqrt{5}\\\sqrt{x}=2+\sqrt{5}\end{matrix}\right.\) rồi khúc sau như trên
24 tháng 8 2021
a: Ta có: \(P=\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+3}{x-9}\right):\left(\dfrac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
\(=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}-3}{2\sqrt{x}-2-\sqrt{x}+3}\)
\(=\dfrac{-3\left(\sqrt{x}+1\right)}{\sqrt{x}+3}\cdot\dfrac{1}{\sqrt{x}+1}\)
\(=\dfrac{-3}{\sqrt{x}+3}\)
Tìm x để P=\(\dfrac{-\sqrt{x}+3}{\sqrt[]{x}-1}\le1\) với \(x\ge0;x\ne1\)
\(-\dfrac{\sqrt{x+3}}{\sqrt{x}-1}\le1\\ \Rightarrow-\sqrt{x+3}\le\sqrt{x}-1\\ \Rightarrow\sqrt{x+3}\ge1-\sqrt{x}\\ \Rightarrow x+3\ge1+x-2\sqrt{x}\\ 2\sqrt{x}\ge-2\\ \sqrt{x}\ge-1\)
vì \(x\ge0\:và\:\:x\ne1\) nên: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
vậy...