Dễ thấy hàm \(f\left(x\right)=\left(1-m\right)x^5+9mx^2-16x-m\) liên tục trên R với mọi giá trị của m

Ta có:

\(f\left(-2\right)=\left(1-m\right).\left(-2\right)^5+9m.\left(-2\right)^2-16.\left(-2\right)-m\)

           \(=-32\left(1-m\right)+4.9m+32-m=67m\)

\(f\left(0\right)=-m\)

\(f\left(2\right)=\left(1-m\right).2^5+9m.2^2-16.2-m\)

        \(=32\left(1-m\right)+4.9m-32-m=3m\)

Nếu \(m=0\) thì ta có đpcm

Nếu \(m\ne0\) thì

    \(\left\{{}\begin{matrix}f\left(-2\right).f\left(0\right)=-67m^2< 0\\f\left(0\right).f\left(2\right)=-3m^2< 0\end{matrix}\right.\)

Do đó pt đã cho có ít nhất một nghiệm trên mỗi khoảng \(\left(-2;0\right)\) và \(\left(0;2\right)\)

\(\Rightarrowđpcm\)

Vậy ta có điều phải chứng minh

Chuẩn hoá \(a+b+c=1\)

Ta có:

\(VT=a\left(\dfrac{1}{a+b}+\dfrac{3}{c+b}\right)+c\left(\dfrac{3}{a+b}+\dfrac{1}{c+b}\right)+\dfrac{4b}{c+a}\)

\(\ge a.\dfrac{16}{a+b+3\left(c+b\right)}+c.\dfrac{16}{3\left(a+b\right)+c+b}+\dfrac{4b}{c+a}\)

\(=8\left(\dfrac{2a}{a+4b+3c}+\dfrac{2c}{3a+4b+c}\right)+\dfrac{4b}{c+a}\)

\(=8\left(\dfrac{2a}{a+4b+3c}+1+\dfrac{2c}{3a+4b+c}+1\right)+\dfrac{4b}{c+a}-8.2\)

\(=8\left(3a+4b+3c\right)\left(\dfrac{1}{a+4b+3c}+\dfrac{1}{3a+4b+c}\right)+\dfrac{4b}{c+a}-8.2\)

\(\ge8\left(3a+4b+3c\right).\dfrac{1}{a+2b+c}+\dfrac{4b}{c+a}-8.2\)

\(=8\left(3+b\right)\dfrac{1}{1+b}+\dfrac{4b}{1-b}-8.2\)

\(=4\left[\dfrac{2\left(3+b\right)}{1+b}+\dfrac{b}{1-b}\right]-8.2\)

Ta có \(\dfrac{2\left(3+b\right)}{1+b}+\dfrac{b}{1-b}\ge\dfrac{11}{2}\left(\text{*}\right)\) với \(0< b< 1\)

Thật vậy \(\left(\text{*}\right)\Leftrightarrow4\left(3+b\right)\left(1-b\right)+2b\left(1+b\right)\ge11\left(1-b\right)\left(1+b\right)\)

\(\Leftrightarrow9b^2-6b+1\ge0\Leftrightarrow\left(3b-1\right)^2\ge0\), đúng

\(\Rightarrow VT\ge\dfrac{4.11}{2}-8.2=6\)

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

Mình không thạo vẽ hình trên này nên bạn tự vẽ hình nhé.

Gọi K là hình chiếu vuông góc của S trên BC.

Giả sử \(\overrightarrow{CK}=x\overrightarrow{CB}\left(0< x< 1\right)\)

Đặt \(SC=ka\Rightarrow\left\{{}\begin{matrix}BC=a\sqrt{k^2+4}\\AC=a\sqrt{k^2+8}\end{matrix}\right.\)

Ta có: \(\dfrac{1}{SK^2}=\dfrac{1}{SB^2}+\dfrac{1}{SC^2}=\dfrac{1}{\left(2a\right)^2}+\dfrac{1}{\left(ka\right)^2}\)

\(\Rightarrow SK=\dfrac{2ka}{\sqrt{k^2+4}}\)

Ta có:

\(\left(\left(SBC\right);\left(ABC\right)\right)=45^0\)

\(\Rightarrow\left(AB;SK\right)=45^0\)

\(\Leftrightarrow\dfrac{\overrightarrow{AB}.\overrightarrow{SK}}{AB.SK}=cos45^0\Leftrightarrow\dfrac{\overrightarrow{AB}.\overrightarrow{SK}}{AB.SK}=\dfrac{\sqrt{2}}{2}\)

Lại có:

\(\overrightarrow{AB}.\overrightarrow{SK}=\left(\overrightarrow{SB}-\overrightarrow{SA}\right).\left[x\overrightarrow{SB}+\left(1-x\right)\overrightarrow{SC}\right]\)

\(=xSB^2-x\overrightarrow{SA}.\overrightarrow{SB}+\left(x-1\right).\overrightarrow{SC}.\overrightarrow{SA}\)

\(=x.4a^2-x.4a^2.\dfrac{1}{2}+\left(x-1\right).\dfrac{4a^2+k^2a^2-a^2\left(k^2+8\right)}{2}\)

\(=2xa^2+\left(x-1\right).\left(-2a^2\right)=2a^2\)

\(\Rightarrow\dfrac{\sqrt{2}}{2}=\dfrac{2a^2}{2a.\dfrac{2ka}{\sqrt{k^2+4}}}\Leftrightarrow k=2\)

Do đó:

\(\left\{{}\begin{matrix}SC=2a\\BC=2a\sqrt{2}\\AC=2a\sqrt{3}\end{matrix}\right.\)

Ta có:

\(R=\sqrt{R_{SAB}^2+R_{ABC}^2-\dfrac{AB^2}{4}}\)

\(=\sqrt{\left(\dfrac{2a\sqrt{3}}{3}\right)^2+\left(a\sqrt{3}\right)^2-\dfrac{\left(2a\right)^2}{4}}=\dfrac{a\sqrt{30}}{3}\)

\(\Rightarrow S=4\pi R^2=4\pi.\dfrac{10}{3}a^2=\dfrac{40}{3}\pi a^2\)

b)Ta có:

\(\dfrac{1}{a^2-a+2}\le\dfrac{3}{4}-\dfrac{1}{4}a\left(\text{*}\right)\) với mọi \(a\) thoả mãn  \(0< a< 2\)

Thật vậy, BĐT \(\left(\text{*}\right)\) tương đương với:

\(4\le\left(3-a\right)\left(a^2-a+2\right)\Leftrightarrow a^3-4a^2+5a-2\le0\)

\(\Leftrightarrow\left(a-2\right)\left(a-1\right)^2\le0\), đúng với mọi \(a\) thoả mãn  \(0< a< 2\)

Ta có: với \(a,b\) là các số thực dương thoả mãn \(a+b=2\) thì \(0< a,b< 2\)

Áp dụng BĐT \(\left(\text{*}\right)\), ta có:

\(VT=\dfrac{1}{a^2-a+2}+\dfrac{1}{b^2-b+2}\le\dfrac{3}{2}-\dfrac{1}{4}\left(a+b\right)=1\) 

Đẳng thức xảy ra khi và chỉ khi \(a=b=1\)

Đặt \(u_n+\dfrac{5}{4}=v_n\)

\(GT\Rightarrow\left\{{}\begin{matrix}v_1=\dfrac{9}{4};v_2=\dfrac{13}{4}\\v_{n+2}=2v_{n+1}+3v_n\end{matrix}\right.\)

Ta có CTTQ của dãy \(\left(v_n\right)\) là:

\(v_n=\dfrac{11}{24}.3^n-\dfrac{7}{8}.\left(-1\right)^n\)

(Bạn tự chứng minh theo quy nạp)

\(\Rightarrow u_n=\dfrac{11}{24}.3^n-\dfrac{7}{8}\left(-1\right)^n-\dfrac{5}{4}\) với \(\forall n\in N\text{*}\)

\(\Rightarrow S=2\left(u_1+u_2+...+u_{100}\right)+u_{101}\)

\(=\left[\dfrac{11}{12}\left(3^1+3^2+...+3^{100}\right)-\dfrac{7}{4}\left(-1+1-...+1\right)-\dfrac{5}{2}.100\right]+\dfrac{11}{24}.3^{101}-\dfrac{7}{8}.\left(-1\right)^{101}-\dfrac{5}{4}\)

\(=\dfrac{11}{12}.\dfrac{3^{101}-3}{2}-250+\dfrac{11}{24}.3^{101}+\dfrac{7}{8}\)

\(=\dfrac{11}{24}.\left(2.3^{101}-3\right)-\dfrac{1993}{8}\)

\(=\dfrac{11}{4}.3^{100}-\dfrac{501}{2}\)

Đặt \(\dfrac{u_n}{n+1}=v_n\)

\(GT\Rightarrow\left\{{}\begin{matrix}v_1=\dfrac{u_1}{1+1}=1\\v_{n+1}=\dfrac{1}{4}v_n,\forall n\in N\text{*}\end{matrix}\right.\)

\(\Rightarrow v_n=\dfrac{1}{4}^{n-1},\forall n\in N\text{*}\)

\(\Rightarrow u_n=\left(n+1\right).\dfrac{1}{4}^{n-1},\forall n\in N\text{*}\)