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Áp dụng bất đẳng thức Cô-si liên tục 2 lần ta có :
\(\frac{1}{a+b-c}+\frac{1}{b+c-a}\ge\frac{2}{\sqrt{\left(a+b-c\right)\left(b+c-a\right)}}\ge\frac{2}{\frac{\left(a+b-c\right)+\left(b+c-a\right)}{2}}=\frac{2}{\frac{2b}{2}}=\frac{2}{b}\)
Chứng minh tương tự ta cũng có :
\(\frac{1}{a+b-c}+\frac{1}{c+a-b}\ge\frac{2}{a};\frac{1}{b+c-a}+\frac{1}{c+a-b}\ge\frac{2}{c}\)
Cộng theo vế của 3 bất đẳng thức trên ta được :
\(2\cdot\left(\frac{1}{a+b-c}+\frac{1}{b+c-a}+\frac{1}{c+a-b}\right)\ge2\cdot\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
Hay ta có đpcm
Dấu "=" xảy ra \(\Leftrightarrow a=b=c\) hay tam giác ABC đều
Để giá trị của giới hạn là một số thực xác định thì biểu thức trên tử số ít nhất phải có nghiệm kép \(x=1\)
Đặt \(f\left(x\right)=\sqrt{3x-2}+\sqrt[3]{3x+5}+ax+b\)
\(f\left(1\right)=a+b+3=0\Rightarrow b=-3-a\)
Thay ngược lại vào \(f\left(x\right)\)
\(f\left(x\right)=\sqrt{3x-2}+\sqrt[3]{3x+5}+ax-3-a\)
\(f\left(x\right)=\frac{3\left(x-1\right)}{\sqrt{3x-2}+1}+\frac{3\left(x-1\right)}{\sqrt[3]{\left(3x+5\right)^2}+2\sqrt[3]{3x+5}+4}+a\left(x-1\right)\)
\(f\left(x\right)=\left(x-1\right)\left(\frac{3}{\sqrt{3x-2}+1}+\frac{3}{\sqrt[3]{\left(3x+5\right)^2}+2\sqrt[3]{3x+5}+4}+a\right)\)
\(\Rightarrow\) Để \(f\left(x\right)\) có nghiệm kép \(x=1\) thì
\(g\left(x\right)=\frac{3}{\sqrt{3x-2}+1}+\frac{3}{\sqrt[3]{\left(3x+5\right)^2}+2\sqrt[3]{3x+5}+4}+a\) có ít nhất một nghiệm \(x=1\)
\(g\left(1\right)=\frac{3}{2}+\frac{3}{4+4+4}+a=0\Rightarrow a=-\frac{7}{4}\Rightarrow b=-\frac{5}{4}\)
\(\Rightarrow\lim\limits_{x\rightarrow1}\frac{\sqrt{3x-2}+\sqrt[3]{3x+5}-\frac{7}{4}x-\frac{5}{4}}{x^2-2x+1}=-\frac{37}{32}\)
\(\Rightarrow P=\frac{-\frac{7}{4}-\frac{5}{4}}{-\frac{37}{32}}=\frac{96}{37}\)
Chỉ cần viết tử số thôi nhé, ta quy đồng 4 lên rồi đưa 4 xuông mẫu, sau đó tách tử số thành
\(\frac{1}{4}\left(4\sqrt{3x-2}-2\left(3x-1\right)+4\sqrt[3]{3x+5}-\left(x+7\right)\right)\)
\(=\frac{1}{4}\left(\frac{2\left[4\left(3x-2\right)-\left(3x-1\right)^2\right]}{2\sqrt{3x-2}+3x-1}+\frac{4^3\left(3x+5\right)-\left(x+7\right)^3}{16\sqrt[3]{\left(3x+5\right)^2}+4\sqrt[3]{3x+5}\left(x+7\right)+\left(x+7\right)^2}\right)\)
\(=\frac{1}{4}\left(\frac{2\left(18x-9x^2-9\right)}{2\sqrt{3x-2}+3x-1}+\frac{45x-x^3-21x^2-23}{16\sqrt[3]{\left(3x+5\right)^2}+4\sqrt[3]{3x+5}\left(x+7\right)+\left(x+7\right)^2}\right)\)
\(=\frac{1}{4}\left(\frac{-18\left(x^2-2x+1\right)}{2\sqrt{3x-2}+3x-1}+\frac{-\left(x+23\right)\left(x^2-2x+1\right)}{16\sqrt[3]{\left(3x+5\right)^2}+4\sqrt[3]{3x+5}\left(x+7\right)+\left(x+7\right)^2}\right)\)
\(=\frac{\left(x^2-2x+1\right)}{4}\left(\frac{-18}{2\sqrt{3x-2}+3x-1}-\frac{x+23}{16\sqrt[3]{\left(3x+5\right)^2}+4\sqrt[3]{3x+5}\left(x+7\right)+\left(x+7\right)^2}\right)\)
Rút gọn \(x^2-2x+1\) với mẫu số và thay \(x=1\) vào
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.
\(y'=4\left(x-5\right)^3\)
Áp dụng Viet với lưu ý \(tanA+tanB+tanC=tanA.tanB.tanC\) ta có:
\(x_4+tanA+tanB+tanC=p\) (1)
\(x_4\left(tanA+tanB+tanC\right)+tanA.tanB+tanB.tanC+tanC.tanA=q\) (2)
\(x_4\left(tanA.tanB+tanB.tanC+tanC.tanA\right)+tanA.tanB.tanC=r\)(3)
\(x_4.tanA.tanB.tanC=s\) (4)
\(\left(1\right)\Rightarrow tanA+tanB+tanC=tanA.tanB.tanC=p-x_4\)
\(\left(4\right)\Rightarrow x_4\left(p-x_4\right)=s\)
Thế vào (2):
\(x_4\left(p-x_4\right)+tanA.tanB+tanB.tanC+tanC.tanA=q\)
\(\Rightarrow tanA.tanB+tanB.tanC+tanC.tanA=q-x_4\left(p-x_4\right)=q-s\)
Thế vào (3):
\(x_4\left(q-s\right)+p-x_4=r\)
\(\Rightarrow p-r=x_4\left(1-q+s\right)\Rightarrow x_4=\frac{p-r}{1-q+s}\)
Câu 1.
\(y = \dfrac{{n + \sin 2n}}{{n + 5}} = \dfrac{{\dfrac{n}{n} + \dfrac{{\sin 2n}}{n}}}{{\dfrac{n}{n} + \dfrac{5}{n}}} = \dfrac{{1 + \dfrac{{2.\sin 2n}}{{2n}}}}{{1 + \dfrac{5}{n}}}\\ \Rightarrow \lim y = \dfrac{{1 + 0}}{{1 + 0}} = 1 \)
Câu 2.
\(\lim \dfrac{{3\sin n + 4\cos n}}{{n + 1}}\)
Vì \( - 1 \le \sin n \le 1; - 1 \le \cos n \le 1 \Rightarrow \) khi \(x \to \infty \) thì \(3\sin n + 4{\mathop{\rm cosn}\nolimits} = const \)
\(\Rightarrow T = \lim \dfrac{{3\sin n + 4\cos n}}{{n + 1}} = 0 \)
Chú thích: $const$ là kí hiệu hằng số, giống như dạng giới hạn L/vô cùng.
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 \(f'\left(x\right)< 0;\forall x\in\left(0;2\right)\)
\(\Leftrightarrow\left\{{}\begin{matrix}m^2-20< 0\\\left[{}\begin{matrix}m>0\\m< -4\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}-\sqrt{20}< m< \sqrt{20}\\\left[{}\begin{matrix}m>0\\m< -4\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow m=\left\{1;2;3;4\right\}\)
Lời giải:
Từ $a+b+c=2; \frac{1}{a+b}+\frac{1}{a+c}+\frac{1}{b+c}=2,5$
$\Rightarrow (a+b+c)\left(\frac{1}{a+b}+\frac{1}{a+c}+\frac{1}{b+c}\right)=5$
\(\Leftrightarrow \frac{a}{a+b}+\frac{a}{a+c}+\frac{a}{b+c}+\frac{b}{a+b}+\frac{b}{a+c}+\frac{b}{b+c}+\frac{c}{a+b}+\frac{c}{a+c}+\frac{c}{b+c}=5\)
\(\Leftrightarrow \frac{a+b}{a+b}+\frac{b+c}{b+c}+\frac{c+a}{c+a}+\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=5\)
\(\Leftrightarrow \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=2\)
Khi đó:
\(A=\frac{a-(b+c)}{b+c}+\frac{b-(c+a)}{c+a}+\frac{c-(a+b)}{a+b}=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}-3\)
\(=2-3=-1\)
Vậy $A=-1$