Đặt \(g\left(x\right)=f\left(x\right)-x-1\Rightarrow g\left(2\right)=g\left(3\right)=g\left(4\right)=0\)

\(\Rightarrow g\left(x\right)\) có 3 nghiệm 2;3;4

\(\Rightarrow g\left(x\right)=a\left(x-2\right)\left(x-3\right)\left(x-4\right)\)

\(\Rightarrow f\left(x\right)=g\left(x\right)+x+1=a\left(x-2\right)\left(x-3\right)\left(x-4\right)+x+1\)

\(f\left(5\right)=10\Rightarrow a\left(5-2\right)\left(5-3\right)\left(5-4\right)+5+1=10\)

\(\Rightarrow a=\dfrac{2}{3}\)

\(\Rightarrow f\left(x\right)=\dfrac{2}{3}\left(x-2\right)\left(x-3\right)\left(x-4\right)+x+1\)

\(\Rightarrow f\left(6\right)=\dfrac{2}{3}.4.3.2+6+1=...\)

Xét tử \(\left|4-x\right|+\left|x+2\right|\ge0\)

Xét mẫu \(\left|x+5\right|+\left|x-3\right|\ge0\)

Do đó \(\frac{\left|4-x\right|+\left|x+2\right|}{\left|x+5\right|+\left|x-3\right|}\ge0\)

Nhưng đề bài cho \(\frac{\left|4-x\right|+\left|x+2\right|}{\left|x+5\right|+\left|x-3\right|}=-\frac{1}{2}<0\) nên không có giá trị nào của x thỏa mãn.

Đặt \(\left\{{}\begin{matrix}x-4=a\\y-3=b\end{matrix}\right.\) \(\Rightarrow a^2+b^2=5\)

\(Q=\sqrt{\left(a+5\right)^2+b^2}+\sqrt{\left(a+3\right)^2+\left(b+4\right)^2}\)

\(\Rightarrow Q\le\sqrt{2\left[\left(a+5\right)^2+b^2+\left(a+3\right)^2+\left(b+4\right)^2\right]}\) (Bunhiacopxki)

\(\Rightarrow Q\le\sqrt{4\left(a^2+8a+b^2+4b+25\right)}\)

\(\Rightarrow Q\le\sqrt{4\left(a^2+2.4a+b^2+2.2b+25\right)}\)

\(\Rightarrow Q\le\sqrt{4\left(a^2+2\left(a^2+4\right)+b^2+2\left(b^2+1\right)+25\right)}\)

\(\Rightarrow Q\le\sqrt{4\left(3a^2+3b^2+35\right)}\le\sqrt{4\left(3.5+35\right)}=10\sqrt{2}\)

Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}a=2\\b=1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=6\\y=4\end{matrix}\right.\)

Đặt \(H\left(x\right)=P\left(x\right)-\left(x^2+2\right)\)

\(\Rightarrow H\left(1\right)=H\left(3\right)=H\left(5\right)=0\)

\(\Rightarrow H\left(x\right)\) có 3 nghiệm 1; 3; 5

\(\Rightarrow H\left(x\right)=\left(x-1\right)\left(x-3\right)\left(x-5\right)\left(x-a\right)\)

\(\Rightarrow P\left(x\right)=H\left(x\right)+x^2+2=\left(x-1\right)\left(x-3\right)\left(x-5\right)\left(x-a\right)+x^2+2\)

\(\Rightarrow P\left(-2\right)+7P\left(6\right)=-105\left(-2-a\right)+4+2+7\left[15\left(6-a\right)+36+2\right]=1112\)

\(-\frac{17}{21}:\left(\frac{5}{4}-\frac{2}{5}\right)< x+\frac{4}{7}< 1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}\)

\(\Leftrightarrow-\frac{17}{21}:\frac{17}{20}< x+\frac{4}{7}< \frac{12}{12}-\frac{6}{12}+\frac{4}{12}-\frac{3}{12}\)

\(\Leftrightarrow-\frac{17}{21}.\frac{20}{17}< x+\frac{4}{7}< \frac{7}{12}\)

\(\Leftrightarrow-\frac{20}{21}< x+\frac{4}{7}< \frac{7}{12}\)

\(\Leftrightarrow-\frac{20}{21}< x< \frac{1}{84}\)

\(\Leftrightarrow-\frac{80}{84}< x< \frac{1}{84}\)

\(\Leftrightarrow-80< x< 1\Leftrightarrow x\in\left\{-79;-78;...;0\right\}\)

mà để Giá trị nguyên lớn nhất của x

\(\Rightarrow x=-1\)

\(\left(x-\frac{3}{5}\right).\left(x+\frac{2}{7}\right)< 0\)

\(\Rightarrow\hept{\begin{cases}x-\frac{3}{5}< 0\\x+\frac{2}{7}>0\end{cases}\text{hoặc}\hept{\begin{cases}x-\frac{3}{5}>0\\x+\frac{2}{7}< 0\end{cases}}}\)

\(\Rightarrow\hept{\begin{cases}x< \frac{3}{5}\\x>-\frac{2}{7}\end{cases}\text{hoặc}\hept{\begin{cases}x>\frac{3}{5}\\x< -\frac{2}{7}\end{cases}}}\)

\(\Rightarrow\orbr{\begin{cases}-\frac{2}{7}< x< \frac{3}{5}\\x\in\varnothing\end{cases}}\)

\(\Rightarrow-\frac{2}{7}< x< \frac{3}{5}\)

\(\Rightarrow x=0\)

Vậy x = 0

\(\left(x-\frac{3}{5}\right)\cdot\left(x+\frac{2}{7}\right)< 0\)

TH1 : \(\Rightarrow\hept{\begin{cases}x-\frac{3}{5}< 0\\x+\frac{2}{7}>0\end{cases}}\)                 \(\Rightarrow\hept{\begin{cases}x< \frac{3}{5}\\x>-\frac{2}{7}\end{cases}}\)              \(\Rightarrow\text{ }-\frac{2}{7}< x< \frac{3}{5}\)

TH2 : \(\Rightarrow\hept{\begin{cases}x-\frac{3}{5}>0\\x+\frac{2}{7}< 0\end{cases}}\)                 \(\Rightarrow\hept{\begin{cases}x>\frac{3}{5}\\x< -\frac{2}{7}\end{cases}}\)             \(\Rightarrow\text{ Không xảy ra}\)

                            Vì \(x\in Z\text{ }\Rightarrow\text{ }x=0\)

\(\left(x-\frac{3}{5}\right)\cdot\left(x+\frac{2}{7}\right)< 0\) 

TH1 : \(\Rightarrow\hept{\begin{cases}x-\frac{3}{5}< 0\\x+\frac{2}{7}>0\end{cases}}\)                 \(\Rightarrow\hept{\begin{cases}x< \frac{3}{5}\\x>-\frac{2}{7}\end{cases}}\)              \(\Rightarrow\text{ }-\frac{2}{7}< x< \frac{3}{5}\)

TH2 : \(\Rightarrow\hept{\begin{cases}x-\frac{3}{5}>0\\x+\frac{2}{7}< 0\end{cases}}\)                 \(\Rightarrow\hept{\begin{cases}x>\frac{3}{5}\\x< -\frac{2}{7}\end{cases}}\)             \(\Rightarrow\text{ Không xảy ra}\)

                            Vì \(x\in Z\text{ }\Rightarrow\text{ }x=0\)

Trả lời giùm mk vs các bn ạ

Thi vòng 12 à bạn!!! Để mk chép đề mà làm