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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.
\(y'=4\left(x-5\right)^3\)
3.
\(x-2y+1=0\Leftrightarrow y=\frac{1}{2}x+\frac{1}{2}\)
\(y'=\frac{2}{\left(x+1\right)^2}\Rightarrow\frac{2}{\left(x+1\right)^2}=\frac{1}{2}\)
\(\Rightarrow\left(x+1\right)^2=4\Rightarrow\left[{}\begin{matrix}x=1\Rightarrow y=1\\x=-3\Rightarrow y=3\end{matrix}\right.\)
Có 2 tiếp tuyến: \(\left[{}\begin{matrix}y=\frac{1}{2}\left(x-1\right)+1\\y=\frac{1}{2}\left(x+3\right)+3\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}y=\frac{1}{2}x+\frac{1}{2}\left(l\right)\\y=\frac{1}{2}x+\frac{9}{2}\end{matrix}\right.\)
4.
\(\lim\limits\frac{\sqrt{2n^2+1}-3n}{n+2}=\lim\limits\frac{\sqrt{2+\frac{1}{n^2}}-3}{1+\frac{2}{n}}=\sqrt{2}-3\)
\(\Rightarrow\left\{{}\begin{matrix}a=2\\b=3\end{matrix}\right.\)
5.
\(\lim\limits_{x\rightarrow a}\frac{2\left(x^2-a^2\right)+a\left(a+1\right)-\left(a+1\right)x}{\left(x-a\right)\left(x+a\right)}=\lim\limits_{x\rightarrow a}\frac{\left(x-a\right)\left(2x+2a\right)-\left(a+1\right)\left(x-a\right)}{\left(x-a\right)\left(x+a\right)}\)
\(=\lim\limits_{x\rightarrow a}\frac{\left(x-a\right)\left(2x+a-1\right)}{\left(x-a\right)\left(x+a\right)}=\lim\limits_{x\rightarrow a}\frac{2x+a-1}{x+a}=\frac{3a-1}{2a}\)
1.
\(f'\left(x\right)=-3x^2+6mx-12=3\left(-x^2+2mx-4\right)=3g\left(x\right)\)
Để \(f'\left(x\right)\le0\) \(\forall x\in R\) \(\Leftrightarrow g\left(x\right)\le0;\forall x\in R\)
\(\Leftrightarrow\Delta'=m^2-4\le0\Rightarrow-2\le m\le2\)
\(\Rightarrow m=\left\{-1;0;1;2\right\}\)
2.
\(f'\left(x\right)=\frac{m^2-20}{\left(2x+m\right)^2}\)
Để \(f'\left(x\right)< 0;\forall x\in\left(0;2\right)\)
\(\Leftrightarrow\left\{{}\begin{matrix}m^2-20< 0\\\left[{}\begin{matrix}m>0\\m< -4\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}-\sqrt{20}< m< \sqrt{20}\\\left[{}\begin{matrix}m>0\\m< -4\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow m=\left\{1;2;3;4\right\}\)
giống câu này thôi : https://hoc24.vn/hoi-dap/question/649217.html
a/ \(y'=42\left(2x+3\right)^{20}\left(x-4\right)^{23}+23\left(x-4\right)^{22}\left(2x+3\right)^{21}\)
b/ \(y=\frac{1}{x\sqrt{x}}=\frac{1}{\sqrt{x^3}}=x^{-\frac{3}{2}}\Rightarrow y'=-\frac{3}{2}x^{-\frac{5}{2}}=-\frac{3}{2x^2\sqrt{x}}\)
c/ \(y'=\frac{\left(x+\frac{1}{x}\right)'}{2\sqrt{\frac{x^2+1}{x}}}=\frac{1-\frac{1}{x^2}}{2\sqrt{\frac{x^2+1}{x}}}=\frac{\left(x^2-1\right)\sqrt{x}}{2x^2\sqrt{x^2+1}}\)
d/ \(y=x^2+x^{\frac{3}{2}}+1\Rightarrow y'=2x+\frac{3}{2}x^{\frac{1}{2}}=2x+\frac{3}{2}\sqrt{x}\)
e/ \(y'=\frac{\sqrt{1-x}+\frac{1+x}{2\sqrt{1-x}}}{1-x}=\frac{3-x}{2\left(1-x\right)\sqrt{1-x}}\)
f/ \(y'=\frac{\sqrt{a^2-x^2}+\frac{x^2}{\sqrt{a^2-x^2}}}{a^2-x^2}=\frac{a^2}{a^2-x^2}\)
\(sin\left(\frac{2x}{3}-\frac{\pi}{3}\right)=0\Rightarrow\frac{2x}{3}-\frac{\pi}{3}=k\pi\Rightarrow\frac{2x}{3}=\frac{\pi}{3}+k\pi\)
\(\Rightarrow x=\frac{\pi}{2}+\frac{k3\pi}{2}\)
ĐKXĐ: \(cos\left(x+\frac{\pi}{3}\right)\ne0\Rightarrow x+\frac{\pi}{3}\ne\frac{\pi}{2}+k\pi\)
\(\Rightarrow x\ne\frac{\pi}{6}+k\pi\)
\(\Rightarrow D=R\backslash\left\{\frac{\pi}{6}+k\pi;k\in Z\right\}\)
2.
a. ĐKXĐ: \(x\ne\frac{\pi}{2}+k\pi\)
Miền xác định đối xứng
\(f\left(-x\right)=\frac{-x+tan\left(-x\right)}{\left(-x\right)^2+1}=\frac{-x-tanx}{x^2+1}=-\frac{x+tanx}{x^2+1}=-f\left(x\right)\)
Hàm lẻ
b. \(f\left(-x\right)=\frac{5\left(-x\right).cos\left(-5x\right)}{sin^2\left(-x\right)+2}=\frac{-5x.cos5x}{sin^2x+2}=-f\left(x\right)\)
Hàm lẻ
c. \(f\left(-x\right)=\left(-2x-3\right)sin\left(-4x\right)=\left(2x+3\right)sin4x\)
Hàm không chẵn không lẻ
d. \(f\left(-x\right)=sin^4\left(-2x\right)+cos^4\left(-2x-\frac{\pi}{6}\right)\)
\(=sin^42x+cos^4\left(2x+\frac{\pi}{6}\right)\)
Hàm ko chẵn ko lẻ
1. ĐKXĐ:
a.
\(cos\left(x-\frac{\pi}{4}\right)\ne0\)
\(\Leftrightarrow x-\frac{\pi}{4}\ne\frac{\pi}{2}+k\pi\)
\(\Leftrightarrow x\ne\frac{3\pi}{4}+k\pi\)
b.
\(x^2-1\ne0\Leftrightarrow x\ne\pm1\)
c.
Hàm xác định trên R
d.
\(cosx\ne0\Leftrightarrow x\ne\frac{\pi}{2}+k\pi\)
2 : cho ab=cd(a,b,c,d≠0)ab=cd(a,b,c,d≠0) và đôi 1 khác nhau, khác đôi nhau
Chứng minh :
a) C1: Đặt \(\frac{a}{b}=\frac{c}{d}=k\)\(\Rightarrow\left\{{}\begin{matrix}a=kb\\c=kd\end{matrix}\right.\)
\(\frac{a-b}{a+b}=\frac{kb-b}{kb+b}=\frac{b\left(k-1\right)}{b\left(k+1\right)}=\frac{k-1}{k+1}\)
\(\frac{c-d}{c+d}=\frac{kd-d}{kd+d}=\frac{d\left(k-1\right)}{d\left(k+1\right)}\frac{k-1}{k+1}\)
Bài 1:
a: Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{2}=\dfrac{y}{\dfrac{3}{2}}=\dfrac{z}{\dfrac{4}{3}}=\dfrac{x-y}{2-\dfrac{3}{2}}=\dfrac{15}{\dfrac{1}{2}}=30\)
Do đó: x=60; y=45; z=40
b: Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{10}=\dfrac{y}{15}=\dfrac{z}{21}=\dfrac{x+y+z}{10+15+21}=\dfrac{92}{46}=2\)
Do đó: x=20; y=30; z=42