bài 1 :

a) ta có : \(\left(x-3\right)\left[x^2+\left(x-1\right)x+k^2\right]=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\2x^2-x+k=0\end{matrix}\right.\) để phương trình có 3 nghiệm phân biệt

\(\Leftrightarrow2x^2-x+k\) có 2 nghiệm và 2 nghiệm này phải khác 3

\(\Leftrightarrow\left\{{}\begin{matrix}2.3^2-3+k\ne0\\1^2-4.2.k>0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}k\ne-15\\k< \dfrac{1}{8}\end{matrix}\right.\)

vậy ...

b) tương tự

2) sữa đề

ta có : \(x^2+3\left(m-3x^2\right)^2=m\)

\(\Leftrightarrow x^2+3\left(m^2-6mx^2+9x^4\right)=m\)

\(\Leftrightarrow27x^4-\left(18m-1\right)x^2-3m^2-m=0\)

phương trình có nghiệm khi phương trình \(27t^2-\left(18m-1\right)t-3m^2-m=0\) có ít nhất 1 nghiệm dương

->...

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a. ĐKXĐ \(x\ge0\)và \(x\ne9\)

Ta có \(K=\left(\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right):\left(\frac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)

\(=\frac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-3\sqrt{x}-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}:\frac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\)

\(=\frac{2x-6\sqrt{x}+x+3\sqrt{x}-3\sqrt{x}-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}:\frac{\sqrt{x}+1}{\sqrt{x}-3}\)

\(=\frac{3x-6\sqrt{x}-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\frac{\sqrt{x}-3}{\sqrt{x}+1}=\frac{3\left(x-2\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\frac{\sqrt{x}-3}{\sqrt{x}+1}\)

\(=\frac{3\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\frac{\sqrt{x}-3}{\sqrt{x}+1}=\frac{3\left(\sqrt{x}-3\right)}{\sqrt{x}+3}\)

b. Để \(K< -1\Rightarrow\frac{3\sqrt{x}-9+\sqrt{x}+3}{\sqrt{x}+3}< 0\Rightarrow\frac{4\sqrt{x}-6}{\sqrt{x}+3}< 0\Rightarrow4\sqrt{x}-6< 0\)vì \(\sqrt{x}+3\ge3\)

\(\Rightarrow0\le x< \frac{9}{4}\left(tm\right)\)

Vậy với \(0\le x< \frac{9}{4}\)thì K<-1

c. \(K=\frac{3\sqrt{x}-9}{\sqrt{x}+3}=3+\frac{-18}{\sqrt{x}+3}\)

Ta có \(\sqrt{x}+3\ge3\Rightarrow\frac{1}{\sqrt{x}+3}\le\frac{1}{3}\Rightarrow-\frac{18}{\sqrt{x}+3}\ge-6\Rightarrow3+\frac{-18}{\sqrt{x}+3}\ge-3\)

\(\Rightarrow K\ge-3\)

Vậy \(MinK=-3\Leftrightarrow\sqrt{x}=0\Leftrightarrow x=0\)

ĐKXĐ: a > 0

a/ \(K=\left[\frac{\sqrt{a}}{\sqrt{a}-1}-\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right]:\left[\frac{1}{\sqrt{a}-1}+\frac{2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right]\)

       \(=\left[\frac{a-1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right]:\left[\frac{\sqrt{a}+3}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right]\)

        \(=\left[\frac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}\right].\left[\frac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{\sqrt{a}+3}\right]\)  \(=\frac{\left(\sqrt{a}+1\right)^2\left(\sqrt{a}-1\right)}{\sqrt{a}\left(\sqrt{a}+3\right)}\)

b/ Ta có: \(\sqrt{a}=\sqrt{3+2\sqrt{2}}=\sqrt{\left(\sqrt{2}+1\right)^2}=\sqrt{2}+1\)

     \(K=\frac{\left(\sqrt{a}+1\right)^2\left(\sqrt{a}-1\right)}{\sqrt{a}\left(\sqrt{a}+3\right)}=\frac{\left(\sqrt{2}+2\right)\sqrt{2}}{\left(\sqrt{2}+1\right)\left(\sqrt{2}+4\right)}=\frac{2\left(\sqrt{2}+1\right)}{\sqrt{2}\left(\sqrt{2}+1\right)\left(2\sqrt{2}+1\right)}\)

            \(=\frac{\sqrt{2}}{2\sqrt{2}+1}\)

c/ \(K< 0\Leftrightarrow\frac{\left(\sqrt{a}+1\right)^2\left(\sqrt{a}-1\right)}{\sqrt{a}\left(\sqrt{a}+3\right)}< 0\)\(\Rightarrow\left(\sqrt{a}+1\right)^2\left(\sqrt{a}-1\right)< 0\)

       \(\Rightarrow\sqrt{a}-1< 0\) (vì \(\left(\sqrt{a}+1\right)^2>0\))    \(\Rightarrow\sqrt{a}< 1\Rightarrow a< 1\)

               Vậy \(0< a< 1\) thì K < 0

a) \(K=\left(\frac{\sqrt{a}}{\sqrt{a}-1}-\frac{1}{a-\sqrt{a}}\right):\left(\frac{1}{\sqrt{a}+1}+\frac{2}{a-1}\right)\)

\(=\left(\frac{\sqrt{a}}{\sqrt{a}-1}-\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\left(\frac{1}{\sqrt{a}+1}+\frac{2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right)\)

\(=\left(\frac{\sqrt{a}\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}-\frac{1}{\sqrt{a}\left(\sqrt{a-1}\right)}\right):\left(\frac{\sqrt{a}-1}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}+\frac{2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\right)\)

\(=\left(\frac{a-1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\left(\frac{\sqrt{a}-1+2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right)\)

\(=\left(\frac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\left(\frac{\sqrt{a}+1}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right)\)

\(=\left(\frac{\sqrt{a}+1}{\sqrt{a}}\right):\left(\frac{1}{\sqrt{a}-1}\right)\)

\(=\frac{\sqrt{a}+1}{\sqrt{a}}\cdot\sqrt{a}-1\)

\(=\frac{a-1}{\sqrt{a}}\)

b) thay \(a=3+2\sqrt{2}\) vào bt K được:

\(\frac{3+2\sqrt{2}-1}{\sqrt{3+2\sqrt{2}}}\) \(=\frac{2+2\sqrt{2}}{\sqrt{2+2\sqrt{2}+1}}\) \(=\frac{2\left(1+\sqrt{2}\right)}{\sqrt{\left(\sqrt{2}+1\right)^2}}\) \(=\frac{2\left(1+\sqrt{2}\right)}{1+\sqrt{2}}\) \(=2\)

c) để K>0 thì:

\(\frac{a-1}{\sqrt{a}}>0\)

\(\Rightarrow a-1>0\)

\(\Rightarrow a>1\)