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\(4x^3-3x+1=\left(2x-1\right)^2\left(x+1\right)\) có nghiệm kép \(x=\dfrac{1}{2}\)
\(\Rightarrow\sqrt{1+ax^2}-bx-2=0\) có nhiều hơn 1 nghiệm \(x=\dfrac{1}{2}\)
\(\Rightarrow\sqrt{1+\dfrac{a}{4}}=\dfrac{b}{2}+2\Rightarrow\sqrt{a+4}=b+4\) (\(b\ge-4\))
\(\Rightarrow a=b^2+8b+12\)
\(\Rightarrow\sqrt{1+\left(b^2+8b+12\right)x^2}=bx+2\)
\(\Rightarrow1+\left(b^2+8b+12\right)x^2=b^2x^2+4bx+4\)
\(\Rightarrow\left(8b+12\right)x^2-4bx-3=0\)
\(\Rightarrow\left(2x-1\right)\left[\left(4b+6\right)x+3\right]=0\)
\(\Rightarrow\left(4b+6\right)x+3=0\) có nghiệm \(x=\dfrac{1}{2}\)
\(\Rightarrow2b+3+3=0\Rightarrow b=-3\) \(\Rightarrow a=-3\)
Khi đó:
\(\lim\limits_{x\rightarrow\dfrac{1}{2}}\dfrac{\sqrt{1-3x^2}+3x-2}{4x^3-3x+1}=\lim\limits_{x\rightarrow\dfrac{1}{2}}\dfrac{-12\left(2x-1\right)^2}{\left(x+1\right)\left(2x-1\right)^2\left(\sqrt{1-3x^2}+2-3x\right)}\)
\(=\lim\limits_{x\rightarrow\dfrac{1}{2}}\dfrac{-12}{\left(x+1\right)\left(\sqrt{1-3x^2}+2-3x\right)}=-8\)
\(\Rightarrow c=-8\)
Lời giải:
\(\lim\limits_{x\to 0,5}\frac{\sqrt{1+ax^2}-bx-2}{4x^3-3x+1}=\lim\limits_{x\to 0,5}\frac{\sqrt{1+ax^2}-bx-2}{(x+1)(2x-1)^2}\)
Để giới hạn hàm đã cho hữu hạn thì $f(x)=\sqrt{1+ax^2}-bx-2$ có nhân tử là $(2x-1)^2$
$f(x)$ có nhân tử $2x-1 \Leftrightarrow f(\frac{1}{2})=0\Leftrightarrow b=\sqrt{4+a}-4$
Khi đó:
$\sqrt{1+ax^2}-bx-2=(2x-1)(2-\frac{2x+1}{\sqrt{1+ax^2}+x\sqrt{4+a}})$
Giờ ta cần xác định $a,b$ để $2-\frac{2x+1}{\sqrt{1+ax^2}+x\sqrt{4+a}}=0$ với $x=\frac{1}{2}$
$\Leftrightarrow \sqrt{4+a}=1\Leftrightarrow a=-3$
$b=\sqrt{4+a}-4=-3$
\(\lim\limits_{x\to 0,5}\frac{\sqrt{1-3x^2}+3x-2}{4x^3-3x+1}=\lim\limits_{x\to 0,5}\frac{-3(2x-1)^2(2x+1)}{(2\sqrt{1-3x^2}+1)(\sqrt{1-3x^2}+x)(2x-1)^2(x+1)}\)
\(=\lim\limits_{x\to 0,5}\frac{-3(2x+1)}{(2\sqrt{1-3x^2}+1)(\sqrt{1-3x^2}+x)(x+1)}=-2=c\)
Giới hạn đã cho hữu hạn nên \(x^2+2ax-b=0\) có nghiệm \(x=2\)
\(\Rightarrow4+4a-b=0\Rightarrow b=4a+4\)
\(\Rightarrow\lim\limits_{x\rightarrow2}\dfrac{x^2+2ax-4a-4}{x^2-4}=\lim\limits_{x\rightarrow2}\dfrac{\left(x-2\right)\left(x+2a+2\right)}{\left(x-2\right)\left(x+2\right)}\)
\(=\lim\limits_{x\rightarrow2}\dfrac{x+2a+2}{x+2}=\dfrac{2a+4}{4}=4\)
\(\Rightarrow a=6\Rightarrow b=28\)
\(\lim\limits_{x\rightarrow-\infty}\dfrac{-\sqrt{\dfrac{x^2}{x^2}-\dfrac{3x}{x^2}}+\dfrac{ax}{x}}{\dfrac{bx}{x}-\dfrac{1}{x}}=\dfrac{a-1}{b}=3\)
=> A
Lời giải:
$x\to -2$ thì $2x+1\to -3<0$
$x\to -2$ thì $(x+2)^2\to 0$
$\Rightarrow \lim\limits_{x\to -2}\frac{2x+1}{(x+2)^2}=-\infty$
\(\lim\limits_{x\rightarrow-1}\dfrac{\sqrt{4x+5}-2x-3}{\left(x+1\right)^2}\)
\(=\lim\limits_{x\rightarrow-1}\dfrac{4x+5-\left(2x+3\right)^2}{\sqrt{4x+5}+2x+3}\cdot\dfrac{1}{\left(x+1\right)^2}\)
\(=\lim\limits_{x\rightarrow-1}\left(\dfrac{4x+5-4x^2-12x-9}{\left(\sqrt{4x+5}+2x+3\right)\cdot\left(x+1\right)^2}\right)\)
\(=\lim\limits_{x\rightarrow-1}\left(\dfrac{-4x^2-8x-4}{\left(\sqrt{4x+5}+2x+3\right)\cdot\left(x+1\right)^2}\right)\)
\(=\lim\limits_{x\rightarrow-1}\left(\dfrac{-4\left(x^2+2x+1\right)}{\left(x+1\right)^2\cdot\left(\sqrt{4x+5}+2x+3\right)}\right)\)
\(=\lim\limits_{x\rightarrow-1}\dfrac{-4}{\sqrt{4x+5}+2x+3}\)
\(=\dfrac{-4}{\sqrt{-4+5}-2+3}=\dfrac{-4}{1+1}=-\dfrac{4}{2}=-2\)
\(\lim\limits_{x\rightarrow\left(-1\right)^+}\left(x^3+1\right)\cdot\sqrt{\dfrac{3x}{x^2-1}}\)
\(=\lim\limits_{x\rightarrow\left(-1\right)^+}\left(x^2-x+1\right)\left(x+1\right)\cdot\dfrac{\sqrt{3x}}{\sqrt{\left(x-1\right)\left(x+1\right)}}\)
\(=\lim\limits_{x\rightarrow\left(-1\right)^-}\sqrt{x+1}\cdot\left(x^2-x+1\right)\cdot\sqrt{\dfrac{3x}{x-1}}\)
\(=\sqrt{-1+1}\left[\left(-1\right)^2-\left(-1\right)+1\right]\cdot\sqrt{\dfrac{3\left(-1\right)}{-1-2}}\)
=0
\(\lim\limits_{x\rightarrow+\infty}\left(\dfrac{\left(a+1\right)x^3+bx^2-2ax-2b+1}{x^2-2}\right)\)
Giới hạn hữu hạn khi \(a+1=0\Rightarrow a=-1\)
\(\lim\limits_{x\rightarrow+\infty}\left(\dfrac{bx^2+2x-2b+1}{x^2-2}\right)=\lim\limits_{x\rightarrow+\infty}\left(\dfrac{b+\dfrac{2}{x}-\dfrac{2b-1}{x^2}}{1-\dfrac{2}{x^2}}\right)=b\)
\(\Rightarrow b=10\)
Lời giải:\(\lim\limits_{x\to +\infty}\left(\frac{x^3+1}{x^2-2}+ax+b\right)=\lim\limits_{x\to +\infty}\frac{x^3(a+1)+bx^2-2ax+(1-2b)}{x^2-2}\)
Nếu $a\neq -1$ thì bậc của tử lớn hơn bậc của mẫu nên giới hạn tiến vô cùng chứ không phải hữu hạn $(10)$
Do đó $a=-1$
Khi đó: \(\lim\limits_{x\to +\infty}(\frac{x^3+1}{x^2-2}+ax+b)=\lim\limits_{x\to +\infty}\frac{bx^2+2x+(1-2b)}{x^2-2}=\lim\limits_{x\to +\infty}\frac{b+\frac{2}{x}+\frac{1-2b}{x^2}}{1-\frac{2}{x^2}}=b\)
Do đó $b=10$.
\(\lim\limits_{x\rightarrow-\infty}\dfrac{-a\sqrt{1+\dfrac{1}{x^2}}+\dfrac{2017}{x}}{1+\dfrac{2018}{x}}=-a\Rightarrow a=-\dfrac{1}{2}\)
\(\lim\limits_{x\rightarrow+\infty}\dfrac{bx+1}{\sqrt{x^2+bx+1}+x}=\lim\limits_{x\rightarrow+\infty}\dfrac{b+\dfrac{1}{x}}{\sqrt{1+\dfrac{b}{x}+\dfrac{1}{x^2}}+1}=\dfrac{b}{2}=2\Rightarrow b=4\)
\(\Rightarrow P=2\)
Giới hạn đã cho hữu hạn nên \(ax^3+bx^2+4=0\) có nghiệm \(x=-2\)
\(\Rightarrow-8a+4b+4=0\Rightarrow b=2a-1\)
\(\lim\limits_{x\rightarrow-2}\dfrac{ax^3+\left(2a-1\right)x^2+4}{\left(x-1\right)^2\left(x+2\right)}=\lim\limits_{x\rightarrow-2}\dfrac{\left(x+2\right)\left(ax^2-x+2\right)}{\left(x-1\right)^2\left(x+2\right)}\)
\(=\lim\limits_{x\rightarrow-2}\dfrac{ax^2-x+2}{\left(x-1\right)^2}=\dfrac{4a+4}{9}=2\Rightarrow a=\dfrac{7}{2}\) \(\Rightarrow b=6\)