Đặt \(f\left(x\right)=\sqrt{x^2-ax}-b\) \(\Rightarrow f\left(1\right)=\sqrt{1-a}-b\)
Ta có : \(Lim_{x\rightarrow1}\dfrac{f\left(x\right)}{x-1}=\dfrac{5}{4}\Rightarrow f\left(1\right)=0\) \(\Leftrightarrow\sqrt{1-a}-b=0\Leftrightarrow\sqrt{1-a}=b\)
\(\Leftrightarrow1-a=b^2\Leftrightarrow a=1-b^2\)
Khi đó : \(f\left(x\right)=\sqrt{x^2-\left(1-b^2\right)x}-b\) \(=\dfrac{x^2-\left(1-b^2\right)x-b^2}{\sqrt{x^2-\left(1-b^2\right)x}+b}=\dfrac{x\left(x-1\right)+b^2\left(x-1\right)}{\sqrt{x^2-\left(1-b^2\right)x}+b}\)
\(=\dfrac{\left(x+b^2\right)\left(x-1\right)}{\sqrt{x^2-\left(1-b^2\right)x}+b}\)
\(Lim_{x\rightarrow1}\dfrac{f\left(x\right)}{x-1}=Lim_{x\rightarrow1}\dfrac{x+b^2}{\sqrt{x^2-\left(1-b^2\right)x}+b}=\dfrac{1+b^2}{\sqrt{1-\left(1-b^2\right)}+b}=\dfrac{1+b^2}{b+\left|b\right|}=\dfrac{5}{4}\)\(\Rightarrow4\left(1+b^2\right)=5b+5\left|b\right|\)
b < 0 -> Loại
b \(\ge0\) 0 : \(4+4b^2=10b\Leftrightarrow2b^2-5b+2=0\) \(\Leftrightarrow\left[{}\begin{matrix}b=2\\b=\dfrac{1}{2}\end{matrix}\right.\)
b = 2 \(\Rightarrow a=1-2^2=-3\)
b = 1/2 \(\Rightarrow a=1-\left(\dfrac{1}{2}\right)^2=\dfrac{3}{4}\)
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