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a)ĐKXĐ:\(a\ge0;a\ne16\)
\(B=\left[\dfrac{3\sqrt{a}}{\sqrt{a}+4}+\dfrac{\sqrt{a}}{\sqrt{a}-4}+\dfrac{4\left(a+2\right)}{16-a}\right]:\left(1-\dfrac{2\sqrt{a}+5}{\sqrt{a}+4}\right)\)
=\(\dfrac{3\sqrt{a}\left(\sqrt{a}-4\right)+\sqrt{a}\left(\sqrt{a}+4\right)-4\left(a+2\right)}{a-16}:\dfrac{\sqrt{a}+4-2\sqrt{a}-5}{\sqrt{a}+4}=\dfrac{3a-12\sqrt{a}+a+4\sqrt{a}-4a-8}{\left(\sqrt{a}-4\right)\left(\sqrt{a}+4\right)}\cdot\dfrac{\sqrt{a}+4}{-\sqrt{a}-1}=\dfrac{-8\sqrt{a}-8}{\left(\sqrt{a}-4\right)\left(-\sqrt{a}-1\right)}=\dfrac{8\left(-\sqrt{a}-1\right)}{\left(\sqrt{a}-4\right)\left(-\sqrt{a}-1\right)}=\dfrac{8}{\sqrt{a}-4}\)
Vậy...
b)Với \(a\ge0;a\ne16\) thì B=\(\dfrac{8}{\sqrt{a}-4}\)
B=-3 thì \(\dfrac{8}{\sqrt{a}-4}=-3\)
=>\(9=-3\sqrt{a}+24\)
<=>-15=-3\(\sqrt{a}\)
<=>\(\sqrt{a}=5\)
<=>a=25(TM)
Vậy a=25 thì B=-3
c)Với \(a\ge0;a\ne16\) thì B=\(\dfrac{8}{\sqrt{a}-4}\)
| \(\sqrt{a}-4\) | -8 | -4 | -2 | -1 | 1 | 2 | 4 | 8 |
| \(\sqrt{a}\) | -4(L) | 0 | 2 | 3 | 5 | 6 | 8 | 12 |
| \(\sqrt{a}\) | 0 | 2 | 3 | 5 | 6 | 8 | 12 |
| a | 0(TM) | 4(TM) | 9(TM) | 25(TM) | 36(TM) | 64(TM) | 144(TM) |
(BẠN KẺ 1 BẢNG 3 HÀNG THÔI NHA,MÌNH KẺ LỖI NÊN LÀM 2 BẢNG)
Vậy...
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
Cho số tự nhiên n ≥ 4. Nếu \(C_n^4\) = K thì \(A^4_n\) bằng:
A. 24K
B. 4K
C. 16K
D. \(\frac{K}{24}\)
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):
\(x_4\left(q-s\right)+p-x_4=r\)
\(\Rightarrow p-r=x_4\left(1-q+s\right)\Rightarrow x_4=\frac{p-r}{1-q+s}\)
\(\lim\limits\frac{3-16.4^n}{2^n+3.4^n}=\lim\limits\frac{3\left(\frac{1}{4}\right)^n-16}{\left(\frac{2}{4}\right)^n+3}=-\frac{16}{3}\)
5.
\(\lim\limits_{x\rightarrow-\infty}\frac{-3x^5+7x^3-11}{x^5+x^4-3x}=\lim\limits_{x\rightarrow-\infty}\frac{-3+\frac{7}{x^2}-\frac{11}{x^5}}{1+\frac{1}{x}-\frac{3}{x^4}}=\frac{-3}{1}=-3\)
6.
\(\lim\limits_{x\rightarrow-4}\frac{\left(x+4\right)\left(x-1\right)}{x\left(x+4\right)}=\lim\limits_{x\rightarrow-4}\frac{x-1}{x}=\frac{-5}{-4}=\frac{5}{4}\)
7.
Khi \(x< 2\Rightarrow x-2< 0\) mà \(x+2\rightarrow4\Rightarrow\lim\limits_{x\rightarrow2^-}\frac{x+2}{x-2}=\frac{4}{-0}=-\infty\)
8.
\(\lim\limits_{x\rightarrow1}\frac{9-\left(2x+7\right)}{\left(x-1\right)\left(x+1\right)\left(3+\sqrt{2x+7}\right)}=\lim\limits_{x\rightarrow1}\frac{-2\left(x-1\right)}{\left(x-1\right)\left(x+1\right)\left(3+\sqrt{2x+7}\right)}\)
\(=\lim\limits_{x\rightarrow1}\frac{-2}{\left(x+1\right)\left(3+\sqrt{2x+7}\right)}=\frac{-2}{2.\left(3+3\right)}=-\frac{1}{6}\)
9.
\(\lim\limits_{x\rightarrow4}\frac{\left(4-x\right)\left(16-4x+x^2\right)}{4-x}=\lim\limits_{x\rightarrow4}\left(16-4x+x^2\right)=16\)
1.
\(\lim\limits_{x\rightarrow-\infty}\frac{x^2-7x+1-\left(x^2-3x+2\right)}{\sqrt{x^2-7x+1}+\sqrt{x^2-3x+2}}=\lim\limits_{x\rightarrow-\infty}\frac{-4x-1}{\sqrt{x^2-7x+1}+\sqrt{x^2-3x+2}}\)
\(=\lim\limits_{x\rightarrow-\infty}\frac{x\left(-4-\frac{1}{x}\right)}{-x\sqrt{1-\frac{7}{x}+\frac{1}{x^2}}-x\sqrt{1-\frac{3}{x}+\frac{2}{x^2}}}=\frac{-4}{-1-1}=2\)
2.
\(\lim\limits_{x\rightarrow0^+}\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}=\lim\limits_{x\rightarrow0^+}\frac{\sqrt{x}+1}{\sqrt{x}-1}=-1\)
3.
\(\lim\limits_{x\rightarrow-1}\frac{x^2-3}{x^3+2}=\frac{1-3}{-1+2}=-2\) (ko phải dạng vô định, cứ thay số tính)
4.
\(\lim\limits_{x\rightarrow1}f\left(x\right)=\lim\limits_{x\rightarrow1}\frac{2x^2-x-1}{x-1}=\lim\limits_{x\rightarrow1}\frac{\left(x-1\right)\left(2x+1\right)}{x-1}=\lim\limits_{x\rightarrow1}\left(2x+1\right)=3\)
Để hs có giới hạn tại \(x=1\Rightarrow m=3\)
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 \(f'\left(x\right)< 0;\forall x\in\left(0;2\right)\)
\(\Leftrightarrow\left\{{}\begin{matrix}m^2-20< 0\\\left[{}\begin{matrix}m>0\\m< -4\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}-\sqrt{20}< m< \sqrt{20}\\\left[{}\begin{matrix}m>0\\m< -4\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow m=\left\{1;2;3;4\right\}\)
\(a^4+b^4+c^4\ge a^3+b^3+c^3\)
\(\Leftrightarrow a^4+b^4+c^4-\left(a^3+b^3+c^3\right)\ge0\)
\(\Leftrightarrow a^4+b^4+c^4-\left(a^3+b^3+c^3\right)+3-\left(a+b+c\right)\ge0\)
\(\Leftrightarrow a^4-a^3-a+1+b^4-b^3-b+1+c^4-c^3-c+1\ge0\)
\(\Leftrightarrow\left(a-1\right)^2\left(a^2+a+1\right)+\left(b-1\right)^2\left(b^2+b+1\right)+\left(c-1\right)^2\left(c^2+c+1\right)\ge0\) đúng với mọi a, b, c.
\(\Rightarrowđpcm\)
mong giúp mk bài này nữa vs ạ
https://hoc24.vn/cau-hoi/cho-ab-8-va-bge3-cm-27a210b3-945.1638861610920