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\(a=\lim\limits_{x\rightarrow1}\frac{\left(\sqrt{3x+1}-\sqrt{x+3}\right)\left(\sqrt{3x+1}+\sqrt{x+3}\right)}{\left(x-1\right)\left(x+1\right)\left(\sqrt{3x+1}+\sqrt{x+3}\right)}=\lim\limits_{x\rightarrow1}\frac{2\left(x-1\right)}{\left(x-1\right)\left(x+1\right)\left(\sqrt{3x+1}+\sqrt{x+3}\right)}\)
\(=\lim\limits_{x\rightarrow1}\frac{2}{\left(x+1\right)\left(\sqrt{3x+1}+\sqrt{x+3}\right)}=\frac{2}{2.4}=\frac{1}{4}\)
\(b=\frac{3}{0}=+\infty\)
\(c=\frac{-13}{0}=-\infty\)
Lời giải:
Ta có:
\(f'(x)=3x^2+2(a-1)x+2\)
Theo định lý về dấu của tam thức bậc 2, để \(f'(x)>0\) với mọi \(x\in\mathbb{R}\) thì \(\Delta'=(a-1)^2-6<0\)
\(\Leftrightarrow -\sqrt{6}< a-1< \sqrt{6}\)
\(\Leftrightarrow 1-\sqrt{6}< a< 1+\sqrt{6}\)
Đáp án B
\(\frac{\sqrt{ax+1}\left(\sqrt[3]{bx+1}-1\right)+\sqrt{ax+1}-1}{x}=\frac{\frac{bx\sqrt{ax+1}}{\sqrt[3]{\left(bx+1\right)^2}+\sqrt[3]{bx+1}+1}+\frac{ax}{\sqrt{ax+1}+1}}{x}=\frac{b\sqrt{ax+1}}{\sqrt[3]{\left(bx+1\right)^2}+\sqrt[3]{bx+1}+1}+\frac{a}{\sqrt{ax+1}+1}\)
\(\Rightarrow\lim\limits_{x\rightarrow0}f\left(x\right)=a+b\Rightarrow a+b=1\)
\(a^2+b^2\ge\frac{1}{2}\left(a+b\right)^2=\frac{1}{2}\)
Dấu "=" xảy ra khi \(a=b=\frac{1}{2}\)
Mình sử dụng L'Hopital nhé, 2 loại căn thế này tìm liên hợp kép dài lắm :D
\(\lim\limits_{x\rightarrow1}\frac{\left(6x-5\right)^{\frac{1}{3}}-\left(4x-3\right)^{\frac{1}{2}}}{\left(x-1\right)^2}=\lim\limits_{x\rightarrow1}\frac{2\left(6x-5\right)^{-\frac{2}{3}}-2\left(4x-3\right)^{-\frac{1}{2}}}{2\left(x-1\right)}\)
\(=\lim\limits_{x\rightarrow1}\frac{-4\left(6x-5\right)^{-\frac{5}{3}}+2\left(4x-3\right)^{-\frac{3}{2}}}{1}=-2\)
Nếu ko bạn tách liên hợp như vầy:
\(\frac{\left(\sqrt[3]{6x-5}-2x+1\right)+\left(2x-1-\sqrt{4x-3}\right)}{\left(x-1\right)^2}\)
Sẽ khử được \(\left(x-1\right)^2\)
16.
\(y'=\frac{\left(cos2x\right)'}{2\sqrt{cos2x}}=\frac{-2sin2x}{2\sqrt{cos2x}}=-\frac{sin2x}{\sqrt{cos2x}}\)
17.
\(y'=4x^3-\frac{1}{x^2}-\frac{1}{2\sqrt{x}}\)
18.
\(y'=3x^2-2x\)
\(y'\left(-2\right)=16;y\left(-2\right)=-12\)
Pttt: \(y=16\left(x+2\right)-12\Leftrightarrow y=16x+20\)
19.
\(y'=-\frac{1}{x^2}=-x^{-2}\)
\(y''=2x^{-3}=\frac{2}{x^3}\)
20.
\(\left(cotx\right)'=-\frac{1}{sin^2x}\)
21.
\(y'=1+\frac{4}{x^2}=\frac{x^2+4}{x^2}\)
22.
\(lim\left(3^n\right)=+\infty\)
11.
\(\lim\limits_{x\rightarrow1^+}\frac{-2x+1}{x-1}=\frac{-1}{0}=-\infty\)
12.
\(y=cotx\Rightarrow y'=-\frac{1}{sin^2x}\)
13.
\(y'=2020\left(x^3-2x^2\right)^{2019}.\left(x^3-2x^2\right)'=2020\left(x^3-2x^2\right)^{2019}\left(3x^2-4x\right)\)
14.
\(y'=\frac{\left(4x^2+3x+1\right)'}{2\sqrt{4x^2+3x+1}}=\frac{8x+3}{2\sqrt{4x^2+3x+1}}\)
15.
\(y'=4\left(x-5\right)^3\)
lim (x-->0) \(\frac{\sqrt[3]{ax+1}-\sqrt{1-bx}}{x}=2\)
<=> lim ( x-->0) \(\left(\frac{\sqrt[3]{ax+1}-1}{x}+\frac{1-\sqrt{1-bx}}{x}\right)=2\)
<=> lim (x-->0)\(\left(\frac{a}{\sqrt[3]{\left(ax+1\right)^2}+\sqrt[3]{ax+1}+1}+\frac{b}{\sqrt{1-bx}+1}\right)=2\)
<=> \(\frac{a}{3}+\frac{b}{2}=2\)
mà a + 3b = 3
=> a= 3; b = 2
=> A là đáp án sai.