cho \(C=\dfrac{x}{\sqrt{x}-3}\)
tìm GTLN của \(N=\dfrac{C}{\sqrt{x}-1+C}\)
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T
3 tháng 9 2023
\(\dfrac{M}{N}=\left(\dfrac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{\sqrt{x}+2}{3-\sqrt{x}}\right):\left(\dfrac{\sqrt{x}+2}{\sqrt{x}-3}\right)\) (ĐKXĐ: \(x\ge0;x\ne4;x\ne9\))
\(=\left[\dfrac{2\sqrt{x}-9}{x-2\sqrt{x}-3\sqrt{x}+6}-\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}+\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right]\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+2}\)\(=\left[\dfrac{2\sqrt{x}-9}{\sqrt{x}\left(\sqrt{x}-2\right)-3\left(\sqrt{x}-2\right)}-\dfrac{x-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}+\dfrac{x-4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right]\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+2}\)
\(=\left[\dfrac{2\sqrt{x}-9-x+9+x-4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\right]\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+2}\)
\(=\dfrac{2\sqrt{x}-4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+2}\)
\(=\dfrac{2\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)}\cdot\dfrac{1}{\sqrt{x}+2}\)
\(=\dfrac{2}{\sqrt{x}+2}\)
\(\Rightarrow P=\dfrac{M}{N}+1=\dfrac{2}{\sqrt{x}+2}+1\)
Ta thấy: \(\sqrt{x}\ge0\forall x\)
\(\Rightarrow\sqrt{x}+2\ge2\forall x\)
\(\Rightarrow\dfrac{2}{\sqrt{x}+2}\le1\forall x\)
\(\Rightarrow\dfrac{2}{\sqrt{x}+2}+1\le2\forall x\)
\(\Rightarrow Max_P=2\Leftrightarrow\dfrac{2}{\sqrt{x}+2}+1=2\)
\(\Leftrightarrow\dfrac{2}{\sqrt{x}+2}=1\)
\(\Leftrightarrow\sqrt{x}+2=2\)
\(\Leftrightarrow\sqrt{x}=0\)
\(\Leftrightarrow x=0\left(tm\right)\)
#Urushi☕
3 tháng 9 2023
Bạn tự rút gọn nha .
c) Ta có : \(P\text{=}\dfrac{M}{N}+1\text{=}\dfrac{2}{\sqrt{x}+2}+1\)
Để P có giá trị lớn nhất.
\(\Leftrightarrow\dfrac{2}{\sqrt{x}+2}cóGTLN\)
\(\Leftrightarrow\sqrt{x}+2cóGTNN\)
Mà : \(\sqrt{x}+2\ge2\)
\(\Rightarrow\) Để : \(\left(\sqrt{x}+2\right)_{min}\) \(\Leftrightarrow\sqrt{x}\text{=}0\Leftrightarrow x\text{=}0\)
Vậy............
28 tháng 4 2019
a) Để biểu thức M có nghĩa thì \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
b) \(M=\frac{2}{\sqrt{x}-1}+\frac{2\left(\sqrt{x}+1\right)}{x+\sqrt{x}+1}+\frac{x-10\sqrt{x}+3}{\sqrt{x^3}-1}=\frac{2\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\frac{2\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\frac{x-10\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\frac{2x+2\sqrt{x}+2+2x-2+x-10\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\frac{5x-8\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\frac{\left(\sqrt{x}-1\right)\left(5\sqrt{x}-3\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\frac{5\sqrt{x}-3}{x+\sqrt{x}+1}\)c) Ta có \(M=\frac{5\sqrt{x}-3}{x+\sqrt{x}+1}\Leftrightarrow Mx+M\sqrt{x}+M-5\sqrt{x}+3=0\Leftrightarrow Mx+\left(M-5\right)\sqrt{x}+\left(M+3\right)=0\)Để phương trình có nghiệm( hay có giá trị x) thì \(\left(M-5\right)^2-4.M.\left(M+3\right)\ge0\Leftrightarrow M^2-10M+25-4M^2-12M\ge0\Leftrightarrow3M^2+22M-25\le0\Leftrightarrow\left(M-1\right)\left(3M+25\right)\le0\Leftrightarrow\)\(-\frac{25}{3}\le M\le1\)
Vậy M có GTLN khi \(\frac{5\sqrt{x}-3}{x+\sqrt{x}+1}=1\Leftrightarrow x+\sqrt{x}+1=5\sqrt{x}-3\Leftrightarrow x-4\sqrt{x}+4=0\Leftrightarrow\left(\sqrt{x}-2\right)^2=0\Leftrightarrow\sqrt{x}-2=0\Leftrightarrow x=4\)
Vậy để biểu thức M có GTLN là 1 thì x=4
AT
12 tháng 7 2021
a) \(P=\dfrac{3x+3\sqrt{x}-9}{x+\sqrt{x}-2}+\dfrac{\sqrt{x}+3}{\sqrt{x}+2}-\dfrac{\sqrt{x}-2}{\sqrt{x}-1}\left(x\ge0,x\ne1\right)\)
\(=\dfrac{3x+3\sqrt{x}-9}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}+\dfrac{\sqrt{x}+3}{\sqrt{x}+2}-\dfrac{\sqrt{x}-2}{\sqrt{x}-1}\)
\(=\dfrac{3x+3\sqrt{x}-9+\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{3x+5\sqrt{x}-8}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}=\dfrac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+8\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}=\dfrac{3\sqrt{x}+8}{\sqrt{x}+2}\)
b) \(P=\dfrac{3\sqrt{x}+8}{\sqrt{x}+2}=\dfrac{3\sqrt{x}+6+2}{\sqrt{x}+2}=3+\dfrac{2}{\sqrt{x}+2}\)
Để \(P\in Z\Rightarrow2⋮\sqrt{x}+2\Rightarrow\sqrt{x}+2=2\left(\sqrt{x}+2\ge2\right)\)
\(\Rightarrow x=0\)
c) Ta có: \(\sqrt{x}\ge0\Rightarrow\sqrt{x}+2\ge2\Rightarrow\dfrac{2}{\sqrt{x}+2}\le1\Rightarrow3+\dfrac{2}{\sqrt{x}+2}\le4\)
\(\Rightarrow P_{max}=4\) khi \(x=0\)
NT
9 tháng 7 2021
a) Rút gọn P
ĐKXĐ: \(x\ge0;x\ne1\)
\(P=\dfrac{15\sqrt{x}-11}{x+2\sqrt{x}-3}-\dfrac{3\sqrt{x}-2}{\sqrt{x}-1}-\dfrac{2\sqrt{x}+3}{\sqrt{x}+3}\)
\(=\dfrac{15\sqrt{x}-11}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)\(-\dfrac{\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)\(-\dfrac{\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{15\sqrt{x}-11-\left(3x+9\sqrt{x}-2\sqrt{x}-6\right)-\left(2x-2\sqrt{x}+3\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{15\sqrt{x}-11-3x-7\sqrt{x}+6-2x-\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)\(=\dfrac{\left(-5\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{-5\sqrt{x}+2}{\sqrt{x}+3}\)\(=\dfrac{2-5\sqrt{x}}{\sqrt{x}+3}\)
b) Tìm GTLN
\(P=\dfrac{2-5\sqrt{x}}{\sqrt{x}+3}=\dfrac{17-5\left(\sqrt{x}+3\right)}{\sqrt{x}+3}=\dfrac{17}{\sqrt{x}+3}-5\)
Ta có: \(\sqrt{x}\ge0\Rightarrow\sqrt{x}+3\ge3\Rightarrow P=\dfrac{17}{\sqrt{x}+3}-5\le\dfrac{17}{3}-5=\dfrac{2}{3}\)
Dấu "=" xảy ra khi \(x=0\)
Vậy \(P_{max}=\dfrac{2}{3}\) khi \(x=0\)
25 tháng 9 2021
1) \(P=\dfrac{5-3\sqrt{x}}{\sqrt{x}-1}\left(đk:x\ge0,x\ne1\right)\)
\(=\dfrac{-3\left(\sqrt{x}-1\right)+2}{\sqrt{x}-1}=-3+\dfrac{2}{\sqrt{x}-1}\in Z\)
\(\Rightarrow\sqrt{x}-1\inƯ\left(2\right)=\left\{-2;-1;1;2\right\}\)
Do \(x\ge0,x\ne1\) và x là số chính phương
\(\Rightarrow x\in\left\{0;4;9\right\}\)
2) \(3x^2-5x+1=3\left(x^2-\dfrac{5}{3}x+\dfrac{25}{36}\right)-\dfrac{13}{12}=3\left(x-\dfrac{5}{6}\right)^2-\dfrac{13}{12}\ge-\dfrac{13}{12}\)
\(\Rightarrow C=\dfrac{2022}{3x^2-5x+1}\le2022:\left(-\dfrac{13}{12}\right)=-\dfrac{24264}{13}\)
\(minC=-\dfrac{24624}{13}\Leftrightarrow x=\dfrac{5}{6}\)
20 tháng 8 2023
a: \(y=\sqrt{2}sin\left(x+\dfrac{pi}{4}\right)\)
\(-1< =sin\left(x+\dfrac{pi}{4}\right)< =1\)
=>\(-\sqrt{2}< =y< =\sqrt{2}\)
\(y_{min}=-\sqrt{2}\) khi sin(x+pi/4)=-1
=>x+pi/4=-pi/2+k2pi
=>x=-3/4pi+k2pi
\(y_{max}=\sqrt{2}\) khi sin(x+pi/4)=1
=>x+pi/4=pi/2+k2pi
=>x=pi/4+k2pi
b: \(y=sinx\cdot cos\left(\dfrac{pi}{3}\right)+cosx\cdot sin\left(\dfrac{pi}{3}\right)+3\)
\(=sin\left(x+\dfrac{pi}{3}\right)+3\)
-1<=sin(x+pi/3)<=1
=>-1+3<=sin(x+pi/3)+3<=4
=>2<=y<=4
y min=2 khi sin(x+pi/3)=-1
=>x+pi/3=-pi/2+k2pi
=>x=-5/6pi+k2pi
y max=4 khi sin(x+pi/3)=1
=>x+pi/3=pi/2+k2pi
=>x=pi/6+k2pi
c: \(y=2\cdot\left(sin2x\cdot\dfrac{\sqrt{3}}{2}-cos2x\cdot\dfrac{1}{2}\right)\)
\(=2sin\left(2x-\dfrac{pi}{6}\right)\)
-1<=sin(2x-pi/6)<=1
=>-2<=y<=2
y min=-2 khi sin(2x-pi/6)=-1
=>2x-pi/6=-pi/2+k2pi
=>2x=-1/3pi+k2pi
=>x=-1/6pi+kpi
y max=2 khi sin(2x-pi/6)=1
=>2x-pi/6=pi/2+k2pi
=>2x=2/3pi+k2pi
=>x=1/3pi+kpi
15 tháng 12 2022
a: \(C=\dfrac{3x+3\sqrt{x}-3-x+1}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}+\dfrac{\sqrt{x}-2}{\sqrt{x}}\cdot\dfrac{1-1+\sqrt{x}}{1-\sqrt{x}}\)
\(=\dfrac{2x+3\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}+\dfrac{\sqrt{x}-2}{1-\sqrt{x}}\)
\(=\dfrac{2\sqrt{x}-1-\sqrt{x}+2}{\sqrt{x}-1}=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
b: Để \(C=\sqrt{x}\) thì \(x-\sqrt{x}=\sqrt{x}+1\)
=>\(x-2\sqrt{x}-1=0\)
=>\(\Leftrightarrow x=3+2\sqrt{2}\)
c: |2x-5|=3
=>2x-5=3 hoặc 2x-5=-3
=>2x=2 hoặc 2x=8
=>x=4(nhận) hoặc x=1(loại)
Khi x=4 thì \(C=\dfrac{2+1}{2-1}=3\)
30 tháng 8 2021
a: ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne9\end{matrix}\right.\)
Ta có: \(A=\dfrac{\sqrt{x}}{\sqrt{x}+3}+\dfrac{2\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+9}{x-9}\)
\(=\dfrac{x-3\sqrt{x}+2x+6\sqrt{x}-3x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{3\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{3}{\sqrt{x}+3}\)
b: Thay x=16 vào A, ta được:
\(A=\dfrac{3}{4+3}=\dfrac{3}{7}\)