Đặt \(a=\sqrt[3]{7-x},b=\sqrt[3]{x-5}\Rightarrow a^3+b^3=2,a^3-b^3=12-2x\)

Ta có hệ: 

\(\hept{\begin{cases}\frac{a-b}{a+b}=\frac{a^3-b^3}{2}\\a^3+b^3=2\end{cases}}\Rightarrow\frac{a-b}{a+b}=\frac{a^3-b^3}{a^3+b^3}=\frac{\left(a-b\right)\left(a^2+ab+b^2\right)}{\left(a+b\right)\left(a^2-ab+b^2\right)}\Rightarrow a^2+ab+b^2=a^2-ab+b^2\)

\(\Rightarrow ab=0\)\(\Leftrightarrow\orbr{\begin{cases}a=0\\b=0\end{cases}\Rightarrow\orbr{\begin{cases}x=5\\x=7\end{cases}}}\)(thử lại thỏa mãn).

Lấy điểm \(I\)thỏa mãn: \(\overrightarrow{IA}+\overrightarrow{IB}+5\overrightarrow{IC}=\overrightarrow{0}\). 

\(MA^2+MB^2+5MC^2\)

\(=\overrightarrow{MA}^2+\overrightarrow{MB}^2+5\overrightarrow{MC}^2\)

\(=\left(\overrightarrow{MI}+\overrightarrow{IA}\right)^2+\left(\overrightarrow{MI}+\overrightarrow{IB}\right)^2+5\left(\overrightarrow{MI}+\overrightarrow{IC}\right)^2\)

\(=7MI^2+IA^2+IB^2+5IC^2=4a^2\)

\(\Rightarrow MI=\sqrt{\frac{4a^2-IA^2-IB^2-5IC^2}{7}}=r\)

Suy ra \(M\)thuộc đường tròn tâm \(I\)bán kính \(r\).

Tính tan15^o

\(\cos^215^o=1-\left(\frac{\sqrt{6}-\sqrt{2}}{4}\right)^2=\frac{8+2\sqrt{12}}{16}\)

\(=\frac{\left(\sqrt{6}\right)^2+2\sqrt{6}\sqrt{2}+\left(\sqrt{2}\right)^2}{16}=\frac{\left(\sqrt{6}+\sqrt{2}\right)^2}{16}\)

Vì 15^o<90^o nên cos15^o>0 => cos15^o=\(\frac{\sqrt{6}+\sqrt{2}}{4}\)

tan 15^o \(=\frac{sin15^0}{c\text{os}15^0}=\frac{\sqrt{6}-\sqrt{2}}{\sqrt{6}+\sqrt{2}}=\left(\frac{\left(\sqrt{6}-\sqrt{2}\right)}{6-2}\right)^2=2-\sqrt{3}\)

CM:

\(2\sin15^0\cos15^0=2\frac{\sqrt{6}-\sqrt{2}}{4}.\frac{\sqrt{6}+\sqrt{2}}{4}=\frac{1}{2}=\sin30^0\)

BĐT cần chứng minh tương đương với: 

\(\frac{a}{b}-\frac{a}{b+c}+\frac{b}{c}-\frac{b}{c+a}+\frac{c}{a}-\frac{c}{a+b}\ge\frac{3}{2}\)

\(\Leftrightarrow\frac{ac}{b\left(b+c\right)}+\frac{ba}{c\left(c+a\right)}+\frac{cb}{a\left(a+b\right)}\ge\frac{3}{2}\)

Ta có: 

\(\frac{ac}{b\left(b+c\right)}+\frac{ba}{c\left(c+a\right)}+\frac{cb}{a\left(a+b\right)}\)

\(=\frac{a^2c^2}{abc\left(b+c\right)}+\frac{b^2a^2}{abc\left(c+a\right)}+\frac{c^2b^2}{abc\left(a+b\right)}\)

\(\ge\frac{\left(ab+bc+ca\right)^2}{abc\left(a+b\right)+abc\left(b+c\right)+abc\left(c+a\right)}\)

\(=\frac{\left(ab+bc+ca\right)^2}{2abc\left(a+b+c\right)}\)

Bất đẳng thức cần chứng minh sẽ đúng nếu ta chứng minh được \(\frac{\left(ab+bc+ca\right)^2}{abc\left(a+b+c\right)}\ge3\)

\(\Leftrightarrow\left(ab+bc+ca\right)^2\ge3abc\left(a+b+c\right)\)

Đặt \(ab=x,bc=y,ca=z\)suy ra ta cần chứng minh 

\(\left(x+y+z\right)^2\ge3\left(xy+yz+zx\right)\)

\(\Leftrightarrow x^2+y^2+z^2+2xy+2yz+2zx\ge3xy+3yz+3zx\)

\(\Leftrightarrow x^2+y^2+z^2\ge xy+yz+zx\)(đúng) 

Vậy bất đẳng thức ban đầu là đúng. 

Dấu \(=\)xảy ra khi \(a=b=c\).

\(\Delta'=\left(m+1\right)^2-\left(m^2+2\right)=2m-1\)

Để phương trình có hai nghiệm \(x_1,x_2\)thì \(\Delta'\ge0\Rightarrow2m-1\ge0\Leftrightarrow m\ge\frac{1}{2}\).

Theo Viete ta có: \(\hept{\begin{cases}x_1+x_2=2\left(m+1\right)\\x_1x_2=m^2+2\end{cases}}\)

\(A=\sqrt{2\left(x_1^2+x_2^2\right)+16}-3x_1x_2\)

\(A=\sqrt{2\left(x_1+x_2\right)^2-4x_1x_2+16}-3x_1x_2\)

\(A=\sqrt{2\left(2\left(m+1\right)\right)^2-4\left(m^2+2\right)+16}-3\left(m^2+2\right)\)

\(A=\sqrt{4m^2+16m+16}-3\left(m^2+2\right)\)

\(A=2\sqrt{\left(m+2\right)^2}-3\left(m^2+2\right)\)

\(A=2\left|m+2\right|-3\left(m^2+2\right)\)

\(A=2m+4-3m^2-6\)(vì \(m\ge\frac{1}{2}\)nên \(m+2>0\))

\(A=-3m^2+2m-2\)

Ta lập bảng biến thiên với hàm \(f\left(x\right)=-3x^2+2x-2\)với \(x\ge\frac{1}{2}\).

Kết quả: \(maxA=-\frac{7}{4}\)

suy ra \(a=-7,b=4\Rightarrow2a-3b=-26\)

\(\left(x-y\right)^2=49\Leftrightarrow\orbr{\begin{cases}x-y=7\\x-y=-7\end{cases}}\)

- \(\hept{\begin{cases}x-y=7\\3x+4y=84\end{cases}\Leftrightarrow\hept{\begin{cases}x=16\\y=9\end{cases}}}\)

- \(\hept{\begin{cases}x-y=-7\\3x+4y=84\end{cases}\Leftrightarrow\hept{\begin{cases}x=8\\y=15\end{cases}}}\)