Ta chứng minh bđt phụ \(\dfrac{a}{1+a^2}\le\dfrac{3}{10}+\dfrac{18}{25}\left(a-\dfrac{1}{3}\right)\)

Thật vậy bđt trên \(\Leftrightarrow\dfrac{-3a^2+10a-3}{10\left(1+a^2\right)}-\dfrac{18}{25}\left(a-\dfrac{1}{3}\right)\le0\)

\(\Leftrightarrow\left(a-\dfrac{1}{3}\right)\left[\dfrac{3\left(3-a\right)}{10\left(1+a^2\right)}-\dfrac{18}{25}\right]\le0\)

\(\Leftrightarrow-\dfrac{36\left(a-\dfrac{1}{3}\right)^2\left(\dfrac{3}{4}+a\right)}{50\left(1+a^2\right)}\le0\) ( luôn đúng với mọi \(a\)\(\ge\)0)

Tương tự cũng có:\(\dfrac{b}{1+b^2}\le\dfrac{3}{10}+\dfrac{18}{25}\left(b-\dfrac{1}{3}\right)\); \(\dfrac{c}{1+c^2}\le\dfrac{3}{10}+\dfrac{18}{25}\left(c-\dfrac{1}{3}\right)\)

Cộng vế với vế => VT\(\le\dfrac{9}{10}+\dfrac{18}{25}\left(a+b+c-1\right)=\dfrac{9}{10}\)

Dấu = xảy ra khi \(a=b=c=\dfrac{1}{3}\)

Pt có hai nghiệm trái dấu khi \(ac< 0\Leftrightarrow16m< 0\Leftrightarrow m< 0\)

Theo định lí viet: \(\left\{{}\begin{matrix}x_1+x_2=\dfrac{15}{4}\\x_1x_2=m\end{matrix}\right.\)

kết hợp với giả thiết ta có hệ: \(\left\{{}\begin{matrix}x_1+x_2=\dfrac{15}{4}\\x_1^2=x_2\\x_1x_2=m\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x_1+x_1^2=\dfrac{15}{4}\\x_1^2=x_2\\x_1x_2=m\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x_1=-\dfrac{5}{2}\\x_2=\dfrac{25}{4}\\x_1x_2=m\end{matrix}\right.\\\left\{{}\begin{matrix}x_1=\dfrac{3}{2}\\x_2=\dfrac{9}{4}\\x_1x_2=m\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}m=-\dfrac{125}{8}\left(TM\right)\\m=\dfrac{27}{8}L\end{matrix}\right.\)

Vậy \(m=-\dfrac{125}{8}\) ý B

\(x^2-2\left(2m+1\right)x+m\left(m+9\right)< 0\) vô nghiệm

\(\Leftrightarrow\)\(x^2-2\left(2m+1\right)x+m\left(m+9\right)\ge0\forall x\in R\)

\(\Leftrightarrow\left\{{}\begin{matrix}a>0\\\Delta\le0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}1>0\left(lđ\right)\\3x^2-5x+1\le0\end{matrix}\right.\)

\(\Rightarrow m\in\left[\dfrac{5-\sqrt{13}}{6};\dfrac{5+\sqrt{13}}{6}\right]\)

Vậy \(\Rightarrow m\in\left[\dfrac{5-\sqrt{13}}{6};\dfrac{5+\sqrt{13}}{6}\right]\) làm cho bpt vô nghiệm

\(gt\Leftrightarrow\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}+\dfrac{1}{\sqrt{z}}=1\)

\(P=\dfrac{1}{xyz}\left(x\sqrt{2y^2+yz+2z^2}+y\sqrt{2x^2+xz+2z^2}+z\sqrt{2y^2+xy+2x^2}\right)\)

\(=\dfrac{1}{xyz}\left(x\sqrt{\dfrac{5}{4}\left(y+z\right)^2+\dfrac{3}{4}\left(y-z\right)^2}+y\sqrt{\dfrac{5}{4}\left(x+z\right)^2+\dfrac{3}{4}\left(x-z\right)^2}+z\sqrt{\dfrac{5}{4}\left(x+y\right)^2+\dfrac{3}{4}\left(x-y\right)^2}\right)\)

\(\ge\dfrac{1}{xyz}\left[x.\dfrac{\sqrt{5}\left(z+y\right)}{2}+y.\dfrac{\sqrt{5}\left(x+z\right)}{2}+z.\dfrac{\sqrt{5}\left(x+y\right)}{2}\right]\)

\(=\dfrac{\sqrt{5}\left(z+y\right)}{2yz}+\dfrac{\sqrt{5}\left(x+z\right)}{2xz}+\dfrac{\sqrt{5}\left(x+y\right)}{2xy}\)

\(=\dfrac{\sqrt{5}}{3}\left(1+1+1\right)\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\ge\dfrac{\sqrt{5}}{3}\left(\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}+\dfrac{1}{\sqrt{z}}\right)^2=\dfrac{\sqrt{5}}{3}\) (bunhia)

Dấu = xảy ra khi \(x=y=z=9\)

b)Xét pt hoành độ giao điểm của (P) và (d) có:

\(\dfrac{1}{2}x^2=mx-m+1\) 

\(\Leftrightarrow x^2-2mx+2m-2=0\)

Có \(\Delta=4m^2-4\left(2m-2\right)=4\left(m^2-2m+1\right)+4=4\left(m-1\right)^2+4>0\forall m\)

=> (d) luôn cắt (P) tại hai điểm phân biệt

Theo định lí viet: \(\left\{{}\begin{matrix}x_1+x_2=2m\\x_1x_2=2m-2\end{matrix}\right.\)

Vì \(A;B\in\left(P\right)\Rightarrow\left\{{}\begin{matrix}y_1=\dfrac{1}{2}x_1^2\\y_2=\dfrac{1}{2}x_2^2\end{matrix}\right.\) 

\(\Rightarrow y_1+y_2=\dfrac{1}{2}x_1^2+\dfrac{1}{2}x_2^2=\dfrac{1}{2}\left(x_1+x_2\right)^2-x_1x_2\)\(=\dfrac{1}{2}.\left(2m\right)^2-\left(2m-2\right)=2m^2-2m+2\)

Vậy...

Vì \(\pi< \alpha< \dfrac{3\pi}{2}\) \(\Rightarrow\dfrac{\pi}{2}< \dfrac{\alpha}{2}< \dfrac{3\pi}{4}\)

\(\Rightarrow sin\dfrac{\alpha}{2}>0;cos\dfrac{\alpha}{2}< 0\)

\(\pi< \alpha< \dfrac{3\pi}{2}\Rightarrow cos\alpha< 0\)

\(\Rightarrow cos\alpha=-\sqrt{1-sin^2\alpha}=-\dfrac{3}{5}\)

Có \(sin^2\dfrac{\alpha}{2}=\dfrac{1-cosa}{2}=\dfrac{4}{5}\Rightarrow sin\dfrac{\alpha}{2}=\sqrt{\dfrac{4}{5}}=\dfrac{2\sqrt{5}}{5}\)

\(cos^2\dfrac{\alpha}{2}=\dfrac{1+cosa}{2}=\dfrac{1}{5}\Rightarrow cos\dfrac{\alpha}{2}=-\sqrt{\dfrac{1}{5}}=-\dfrac{\sqrt{5}}{5}\)

\(tan\dfrac{\alpha}{2}=\dfrac{sin\dfrac{\alpha}{2}}{cos\dfrac{\alpha}{2}}=-2\)

\(cot\dfrac{\alpha}{2}=-\dfrac{1}{2}\)

Để pt có hai nghiệm \(x_1;x_2\Leftrightarrow\Delta\ge0\)

 \(\Leftrightarrow4-m^2\ge0\) \(\Leftrightarrow m\in\left[-2;2\right]\)

Theo định lí viet: \(\left\{{}\begin{matrix}x_1+x_2=m\\x_1x_2=\dfrac{m^2-2}{2}\end{matrix}\right.\)

\(H=2x_1x_2-x_1-x_2+9=m^2-2-m+9\)\(=m^2-m+7\)

Ta thấy H là một parabol và m nằm trong \(\left[-2;2\right]\) ,max của chúng sẽ chỉ ở vị trí m=-2 hoặc m=2 

Tại m=-2 thì H=13

Tại m=2 thì H=9
Vậy maxH=132 khi m=-2 

(Mình chỉ biets trình bày cách này thôi, nếu bạn biết vẽ bảng biến thiên sẽ dễ hơn)

a. Thay m=1 vào pt ta được: \(x^2+2x=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)

b, Để pt có hai nghiệm pb \(\Leftrightarrow\Delta>0\)

\(\Leftrightarrow4-4\left(m-1\right)>0\Leftrightarrow m< 2\)

Theo hệ thức viet: \(\left\{{}\begin{matrix}x_1+x_2=-2\\x_1x_2=m-1\end{matrix}\right.\)

Có \(x_1^3+x_2^3-6x_1x_2=4\left(m-m^2\right)\)

\(\Leftrightarrow\left(x_1+x_2\right)^3-3x_1x_2\left(x_1+x_2\right)-6x_1x_2=4\left(m-m^2\right)\)

\(\Leftrightarrow-8+6\left(m-1\right)-6\left(m-1\right)=4\left(m-m^2\right)\)

\(\Leftrightarrow4m^2-4m-8=0\) 

<=>\(\left[{}\begin{matrix}m=2\left(L\right)\\m=-1\left(Tm\right)\end{matrix}\right.\)

Vậy m=-1