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

Ta có:

1m² - 4m + 7 = 2(m² - 2m + 1) + 4

= 2(m - 1)² + 4 > 0 với mọi m

Vậy y luôn đồng biến trên R với m

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...

#)Góp ý :

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