cho a , b , c đôi một khác nhau , chứng minh
\(\left(\frac{a-b}{b-c}\right)^2+\left(\frac{b-c}{c-a}\right)^2+\left(\frac{c-a}{a-b}\right)\ge5\)
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Để y đồng biến trên R thì
\(2m^2-4m+7>0\)
<=> \(2\left(m^2-2m+1\right)+5>0\)
<=> \(2\left(m-1\right)^2+5>0\)( Phương trình có ngiệm với mọi m)
Vậy hàm số luôn đồng biến trên R
Đặt \(\sqrt{x^2-5x+14}=a\) và \(\sqrt{x^2-5x+10}=b\) \(\left(a,b>0\right)\)
\(\Rightarrow a-b=2\)
\(\Rightarrow a^2-b^2=x^2-5x+14-x^2+5x-10=4\)
\(\Leftrightarrow\left(a-b\right)\left(a+b\right)=4\)
\(\Leftrightarrow a-b=2\)
\(\Leftrightarrow\sqrt{x^2-5x+14}+\sqrt{x^2-5x+10}=2\left(đpcm\right)\)
Ta có:
x2 +4x + 4= (x + 2)2
X2 - 4x + 4 = (x - 2)2
Suy ra, ta có Q = x + 2 + x - 2 = 2x
\(Q=\sqrt{x^2+4x+4}+\sqrt{x^2-4x+4}\)
\(=\sqrt{\left(x+2\right)^2}+\sqrt{\left(x-2\right)^2}\)
\(=|x+2|+|x-2|\)
\(=|x+2|+|2-x|\ge|x+2+2-x|=4\)
\(\Rightarrow Q_{min}=4\)\(\Leftrightarrow\left(x+2\right)\left(2-x\right)\ge0\)
Th1 : \(\hept{\begin{cases}x+2\ge0\\2-x\ge0\end{cases}\Rightarrow\hept{\begin{cases}x\ge-2\\x\le2\end{cases}\Rightarrow-2\le x\le}2}\)
Th2 : \(\hept{\begin{cases}x+2< 0\\2-x< 0\end{cases}\Rightarrow\hept{\begin{cases}x< -2\\x>2\end{cases}\Rightarrow}x\in\varnothing}\)
Vậy \(Q_{min}=4\Leftrightarrow-2\le x\le2\)
\(S=\sqrt{x-2}+\sqrt{y-3}\)
\(\Rightarrow S^2=\left(\sqrt{x-2}+\sqrt{y-3}\right)^2\)
\(\Rightarrow S^2=x-2+2\sqrt{\left(x-2\right)\left(y-3\right)}+y-3\)
\(\Rightarrow S^2=x+y-5+2\sqrt{\left(x-2\right)\left(y-3\right)}\)
\(\Rightarrow S^2=1+2\sqrt{\left(x-2\right)\left(y-3\right)}\)
Vì \(2\sqrt{\left(x-2\right)\left(y-3\right)}\ge0\)
\(\Rightarrow1+2\sqrt{\left(x-2\right)\left(y-3\right)}\ge1\)
\(\Rightarrow S^2\ge1\Leftrightarrow\orbr{\begin{cases}S\ge1\left(tm\right)\\S\le-1\left(ktm\right)\end{cases}}\)
\(\Rightarrow S_{min}=1\Leftrightarrow2\sqrt{\left(x-2\right)\left(y-3\right)}=0\)
TH1 : \(x-2=0\Leftrightarrow x=2\Rightarrow y=6-2=4\)
Th2 : \(y-3=0\Rightarrow y=3\Rightarrow x=6-3=3\)
Vậy \(S_{min}=1\Leftrightarrow\hept{\begin{cases}x=2\\y=4\end{cases}}\)hoặc \(x=y=3\)
Áp dụng bđt Bu-nhi-a-cốp-xki ta có
\(S^2=\left(\sqrt{x-2}+\sqrt{y-3}\right)^2\le\left(1+1\right)\left(x+y-5\right)=2\left(6-5\right)=2\)(vì \(x+y=6\) )
\(\Rightarrow S^2\le2\)
\(\Leftrightarrow-\sqrt{2}\le S\le\sqrt{2}\)
\(\Rightarrow minS=-\sqrt{2}\)
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}\frac{\sqrt{x-2}}{1}=\frac{\sqrt{y-3}}{1}\\x+y=6\end{cases}}\) \(\Leftrightarrow\hept{\begin{cases}x=2,5\\y=3,5\end{cases}}\)
\(DK:2\le x\le3\)
\(\Leftrightarrow6+3\sqrt{x-2}=2x+\sqrt{x+6}\)
\(\Leftrightarrow6-2x=\sqrt{x+6}-3\sqrt{x-2}\)
\(\Leftrightarrow36-24x+4x^2=10x-12-6\sqrt{\left(x+6\right)\left(x-2\right)}\)
\(\Leftrightarrow2x^2-12x+18=5x-6-3\sqrt{\left(x+6\right)\left(x-2\right)}\)
\(\Leftrightarrow2x^2-17x+24=-3\sqrt{\left(x+6\right)\left(x-2\right)}\)
\(\Leftrightarrow4x^4+289x^2+576-68x^3-816x+96x^2=9x^2+36x-108\)
\(\Leftrightarrow4x^4-68x^3+376x^2-852x+684=0\)
\(\Leftrightarrow2x^4-34x^3+188x^2-426x+342=0\)
\(\Leftrightarrow\left(2x^4-6x^3\right)-\left(28x^3-84x^2\right)+\left(104x^2-312x\right)-\left(114x-342\right)=0\)
\(\Leftrightarrow2x^3\left(x-3\right)-28x^2\left(x-3\right)+104x\left(x-3\right)-114\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(2x^3-28x^2+104x-114\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(2x^3-6x^2-22x^2+66x+38x-114\right)=0\)
\(\Leftrightarrow\left(x-3\right)^2\left(2x^2-22x+38\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=3\left(n\right)\\x^2-11x+19=0\left(1\right)\end{cases}}\)
(1) Ta co:\(\Delta\left(-11\right)^2-4.1.19=45\)
Suy ra:\(x_1=\frac{11+3\sqrt{5}}{2}\left(l\right),x_2=\frac{11-3\sqrt{5}}{2}\left(n\right)\)
Vay nghiem cua PT la \(x=3,x=\frac{11-3\sqrt{5}}{2}\)
Em làm bừa thôi, mới học dạng này .
ĐK: \(1\le x\le7\)
Đặt \(\sqrt{6}\ge a=\sqrt{7-x}\ge0;\sqrt{6}\ge b=\sqrt{x-1}\ge0\)
PT<=>\(b^2+2a=2b+ab\left(1\right)\)
(1) \(\Leftrightarrow\left(a-b\right)\left(2-b\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}a=b\\b=2\end{cases}}\). Nếu a = b thì \(\sqrt{7-x}=\sqrt{x-1}\Leftrightarrow7-x=x-1\Leftrightarrow x=4\) (TM)
Nếu b = 2 thì \(\sqrt{x-1}=2\Leftrightarrow x=5\left(TM\right)\)
Vậy...