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\(x>1\)
\(f'\left(x\right)=\left(2x+2\right)\sqrt{x-1}+\frac{x^2+2x}{2\sqrt{x-1}}=\frac{5x^2+2x-4}{2\sqrt{x-1}}\)
\(f'\left(x\right)\ge0\Leftrightarrow\frac{5x^2+2x-4}{\sqrt{x-1}}\ge0\Leftrightarrow5x^2+2x-4\ge0\)
\(\Rightarrow x>1\)
\(\lim\limits_{x\rightarrow1^-}f\left(x\right)=\lim\limits_{x\rightarrow1^-}\frac{\sqrt[3]{5x+3}-2+2-\sqrt{2x+2}}{x-1}=\lim\limits_{x\rightarrow1^-}\frac{\frac{5\left(x-1\right)}{\sqrt[3]{\left(5x+3\right)^2}+2\sqrt[3]{5x+3}+4}-\frac{2\left(x-1\right)}{2+\sqrt{2x+2}}}{x-1}\)
\(=\lim\limits_{x\rightarrow1^-}\left(\frac{5}{\sqrt[3]{\left(5x+3\right)^2}+2\sqrt[3]{5x+3}+4}-\frac{2}{2+\sqrt{2x+2}}\right)=-\frac{1}{12}\)
\(\lim\limits_{x\rightarrow1^+}=\lim\limits_{x\rightarrow1^+}m.sin\left(\frac{\pi x}{2}+2019\right)=\)
Đến đây lại thêm vấn đề nữa, \(sin\left(\frac{\pi x}{2}+2019\right)\) hay \(sin\left(\frac{\pi x}{2}+2019\pi\right)\) bạn?
Bạn ghi đề sai thì phải, nhìn hàm khi \(x< 1\) thì \(\lim\limits_{x\rightarrow1^-}f\left(x\right)\) không tồn tại (ko phải dạng vô định \(\frac{0}{0}\), khi thay x=1 vào tử số ra khác 0)
\(\Leftrightarrow\left(2\sin x+1\right)\left(\sqrt{3}\sin x+2\cos^2x-1\right)-\sin2x-\cos x=0\Leftrightarrow\left(2\sin x+1\right)\left(\sqrt{3}\sin x+2\cos^2x-1-2\cos^2x+1-\cos x\right)=0\Leftrightarrow\left(2\sin x+1\right)\left(\sqrt{3}\sin x-\cos x\right)=0\Rightarrow\left[{}\begin{matrix}2\sin x+1=0\\\sqrt{3}\sin x-\cos x=0\end{matrix}\right.\)
1/ \(y'=\frac{\sqrt{9-x^2}-x\left(\sqrt{9-x^2}\right)'}{9-x^2}=\frac{\sqrt{9-x^2}+\frac{x^2}{\sqrt{9-x^2}}}{9-x^2}=\frac{9}{\left(9-x^2\right)\sqrt{9-x^2}}\)
2/ \(y'=\frac{\left(\sqrt{x^2+x+3}\right)'.\left(2x+1\right)-2\sqrt{x^2+x+3}}{\left(2x+1\right)^2}=\frac{\frac{\left(2x+1\right)}{2\sqrt{x^2+x+3}}.\left(2x+1\right)-2\sqrt{x^2+x+3}}{\left(2x+1\right)^2}\)
\(=\frac{\left(2x+1\right)^2-4\left(x^2+x+3\right)}{2\left(2x+1\right)^2\sqrt{x^2+x+3}}=\frac{-11}{2\left(2x+1\right)^2\sqrt{x^2+x+3}}\)
3/ \(y'=3\left(1+tan^23x\right)=3+3tan^23x\)
4/ \(y'=\frac{\left(cosx-sinx\right)\left(sinx-cosx\right)-\left(cosx+sinx\right)\left(sinx+cosx\right)}{\left(sinx-cosx\right)^2}\)
\(=-\frac{\left(sinx-cosx\right)^2+\left(sinx+cosx\right)^2}{\left(sinx-cosx\right)^2}=-\frac{sin^2x+cos^2x-2sinxcosx+sin^2x+cos^2x+2sinxcosx}{sin^2x+cos^2x-2sinxcosx}\)
\(=\frac{-2}{1-sin2x}\)
5/ \(y'=4x+\frac{1}{2\sqrt{x}}-\frac{\pi}{2}cos\left(\frac{\pi x}{2}\right)\)
6/ \(y'=3sin^2\left(1-3x\right).\left(sin\left(1-3x\right)\right)'=3sin^2\left(1-3x\right).cos\left(1-3x\right).\left(1-3x\right)'\)
\(=-9sin^2\left(1-3x\right).cos\left(1-3x\right)\)
Câu 1:
Đặt \(f\left(x\right)=x^3+mx^2+\left(m-3\right)x-1\)
Ta có \(f\left(0\right)=-1\) ; \(f\left(-1\right)=1\)
\(\Rightarrow f\left(0\right).f\left(-1\right)< 0\Rightarrow f\left(x\right)\) có ít nhất 1 nghiệm thuộc \(\left(-1;0\right)\)
Mặt khác \(\left\{{}\begin{matrix}f\left(0\right)=-1< 0\\\lim\limits_{x\rightarrow+\infty}=+\infty\end{matrix}\right.\) \(\Rightarrow f\left(x\right)\) có ít nhất 1 nghiệm thuộc \(\left(0;+\infty\right)\)
\(\left\{{}\begin{matrix}f\left(-1\right)=1>0\\\lim\limits_{x\rightarrow-\infty}=-\infty\end{matrix}\right.\) \(\Rightarrow f\left(x\right)\) có ít nhất 1 nghiệm thuộc \(\left(-\infty;-1\right)\)
Vậy pt đã cho có 3 nghiệm phân biệt với mọi m
Câu 2:
\(f'\left(x\right)=x^2+2\left(m-1\right)x+m+1\)
Để \(f'\left(x\right)\ge0\) \(\forall x\) \(\Leftrightarrow\Delta'=\left(m-1\right)^2-\left(m+1\right)\le0\)
\(\Leftrightarrow m^2-3m\le0\Leftrightarrow0\le m\le3\)
Câu 3:
Nhận thấy \(x=0\) không phải nghiệm
\(\Leftrightarrow2x^3+3x^2-2=-mx\)
\(\Leftrightarrow\frac{2x^3+3x^2-2}{x}=-m\)
Đặt \(f\left(x\right)=\frac{2x^3+3x^2-2}{x}\Rightarrow f'\left(x\right)=\frac{\left(6x^2+6x\right)x-\left(2x^3+3x^2-2\right)}{x^2}=\frac{4x^3+3x^2+2}{x^2}\)
\(f'\left(x\right)=\frac{4x^2\left(x+1\right)+2-x^2}{x^2}\Rightarrow f'\left(x\right)>0\) \(\forall x\in\left(-1;1\right)\)
\(\Rightarrow f\left(x\right)\) đồng biến trên \(\left(-1;1\right)\)
\(\lim\limits_{x\rightarrow0^-}f\left(x\right)=+\infty\) ; \(\lim\limits_{x\rightarrow0^+}f\left(x\right)=-\infty\)
\(\Rightarrow y=-m\) luôn cắt đồ thị \(y=f\left(x\right)\) hay phương trình đã cho luôn có ít nhất 1 nghiệm trong khoảng \(\left(-1;1\right)\) với mọi m
a) \(\lim\limits_{x\rightarrow0}\frac{\sqrt{1+2x}-1}{2x}=\lim\limits_{x\rightarrow0}\frac{2x}{2x\left(\sqrt{1+2x}+1\right)}=\lim\limits_{x\rightarrow0}\frac{1}{\sqrt{1+2x}+1}=\frac{1}{2}\)
b) \(\lim\limits_{x\rightarrow0}\frac{4x}{\sqrt{9+x}-3}=\lim\limits_{x\rightarrow0}\frac{4x\left(\sqrt{9+x}+3\right)}{x}=\lim\limits_{x\rightarrow0}[4\left(\sqrt{9+x}+3\right)=24\)
c) \(\lim\limits_{x\rightarrow2}\frac{\sqrt{x+7}-3}{x-2}=\lim\limits_{x\rightarrow2}\frac{x-2}{\left(x-2\right)\left(\sqrt{x+7}+3\right)}=\lim\limits_{x\rightarrow2}\frac{1}{\sqrt{x+7}+3}=\frac{1}{6}\)
d) \(\lim\limits_{x\rightarrow1}\frac{3x-2-\sqrt{4x^2-x-2}}{x^2-3x+2}=\lim\limits_{x\rightarrow1}\frac{\left(3x-2\right)^2-\left(4x^2-4x-2\right)}{(x^2-3x+2)\left(3x-2+\sqrt{4x^2-x-2}\right)}=\lim\limits_{x\rightarrow1}\frac{\left(x-1\right)\left(5x-6\right)}{\left(x-1\right)\left(x-2\right)\left(3x-2+\sqrt{4x^2-x-2}\right)}=\frac{1}{2}\\ \\\\ \\ \\ \\ \)
e)\(\lim\limits_{x\rightarrow1}\frac{\sqrt{2x+7}+x-4}{x^3-4x^2+3}=\lim\limits_{x\rightarrow1}\frac{2x+7-\left(x^2-8x+16\right)}{\left(x-1\right)\left(x^2-3x-3\right)\left(\sqrt{2x+7}-x+4\right)}=\lim\limits_{x\rightarrow1}\frac{\left(x-1\right)\left(x-9\right)}{\left(x-1\right)\left(x^2-3x-3\right)\left(\sqrt{2x+7}-x+4\right)}=\lim\limits_{x\rightarrow1}\frac{x-9}{\left(x^2-3x-3\right)\left(\sqrt{2x+7}-x+4\right)}=-8\)
f) \(\lim\limits_{x\rightarrow1}\frac{\sqrt{2x+7}-3}{2-\sqrt{x+3}}=\lim\limits_{x\rightarrow1}\frac{(2x-2)\left(2+\sqrt{x+3}\right)}{\left(1-x\right)\left(\sqrt{2x+7}+3\right)}=\lim\limits_{x\rightarrow1}\frac{-2\left(2+\sqrt{x+3}\right)}{\sqrt{2x+7}+3}=\frac{-4}{3}\)
g) \(\lim\limits_{x\rightarrow0}\frac{\sqrt{x^2+1}-1}{\sqrt{x^2+16}-4}=\lim\limits_{x\rightarrow0}\frac{x^2\left(\sqrt{x^2+16}+4\right)}{x^2\left(\sqrt{x^2+1}+1\right)}=4\)
h)
\(\lim\limits_{x\rightarrow4}\frac{\sqrt{x+5}-\sqrt{2x+1}}{x-4}=\lim\limits_{x\rightarrow4}\frac{\sqrt{x+5}-3}{x-4}+\lim\limits_{x\rightarrow4}\frac{3-\sqrt{2x+1}}{x-4}=\lim\limits_{x\rightarrow4}\frac{1}{\sqrt{x+5}+4}+\lim\limits_{x\rightarrow4}\frac{8-2x}{\left(x-4\right)\left(3+\sqrt{2x+1}\right)}=\frac{1}{7}-\frac{1}{3}=\frac{-4}{21}\)
k) \(\lim\limits_{x\rightarrow0}\frac{\sqrt{x+1}+\sqrt{x+4}-3}{x}=\lim\limits_{x\rightarrow0}\frac{\sqrt{x+1}-1}{x}+\lim\limits_{x\rightarrow0}\frac{\sqrt{x+4}-2}{x}=\lim\limits_{x\rightarrow0}\frac{1}{\sqrt{x+1}+1}+\lim\limits_{x\rightarrow0}\frac{1}{\sqrt{x+4}+2}=\frac{1}{2}+\frac{1}{4}=\frac{3}{4}\)
:v bn ns v là bn bik hết là dạng gì rr mà lm ko đc á :))
\(C'=0\) với mọi hằng số C
nguyen thi khanh nguyen
\(f'\left(x\right)=6x^2-2x\)
\(g'\left(x\right)=3x^2+x\)
\(f'\left(x\right)>g'\left(x\right)\Leftrightarrow6x^2-2x>3x^2+x\)
\(\Leftrightarrow3x^2-3x>0\Rightarrow\left[{}\begin{matrix}x>1\\x< 0\end{matrix}\right.\)
Ta có:f'(x)=4x-1
=>f'(x)\(\sqrt{x^2+1}=2x^2+2x+1\)
<=>(4x-1)\(\sqrt{x^2+1}=2x^2+2x+1\)
Nhận xét: vế phải > 0 nên đk để phương trình có nghiệm:x>\(\dfrac{1}{4}\)
Từ điều kiện trên phương trình
<=>(16x2-8x+1)(x2+1)=4x4+8x3+8x2+4x+1
<=>16x4+16x2-8x3-8x+x2+1=4x4+8x3+8x2+4x+1
<=>12x4-16x3+9x2-12x=0
<=>x(12x3-16x2+9x-12)=0
<=>x(3x-4)(4x2+3)=0
<=>x=0 hoặc x=\(\dfrac{4}{3}\)(do 4x2+3>0)
Vậy...
x=0 loại