Với x > = 0 ; x khác 4 

\(A=\dfrac{9+2x+\sqrt{x}-10-x+1}{x-\sqrt{x}-2}=\dfrac{x+\sqrt{x}}{x-\sqrt{x}-2}=\dfrac{\sqrt{x}}{\sqrt{x}-2}\)

\(=\dfrac{\sqrt{x}-2+2}{\sqrt{x}-2}=1+\dfrac{2}{\sqrt{x}-2}\Rightarrow\sqrt{x}-2\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)

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

\(=\sqrt{3}+2+\sqrt{2}-\sqrt{3}+2=4+\sqrt{2}\)

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\(=\dfrac{17}{13}-1+\dfrac{1}{3}=\dfrac{17-13}{13}+\dfrac{1}{3}=\dfrac{4}{13}+\dfrac{1}{3}=\dfrac{12+13}{39}=\dfrac{25}{39}\)

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Ta có \(\left(x+1,5\right)^2+\left(2,7-x\right)^{10}\ge0\)

mà \(\left(x+1,5\right)^8+\left(2,7-y\right)^{10}=0\)

Dấu ''='' xảy ra khi x = 1,5 ; y = 2,7 

\(=3x^5-2x^4+x^3+6x^3-4x^2+2x+9x^2-6x+3-3x^4-6x^2-4x^3+4x=3x^5-5x^4+3x^3-x^2+3\)

Vậy bth phụ thuộc vào biến x 

\(\dfrac{x^4-3x^3-3x^2+8x-5}{x-1}=\dfrac{x^3\left(x-1\right)-3x^2+3x+5x-5}{x-1}=\dfrac{x^3\left(x-1\right)-3x\left(x-1\right)+5\left(x-1\right)}{x-1}=x^3-3x+5\)

Thay x = 1 vào ta được 

\(1-m^2+m-7-3m^2+3m+6=0\Leftrightarrow-4m^2+4m=0\Leftrightarrow-4m\left(m-1\right)=0\Leftrightarrow m=0;m=1\)