2:
ĐKXĐ: x≠-1
Ta có: \(x\left(\frac{5-x}{x+1}\right)\left(x+\frac{5-x}{x+1}\right)=6\)
\(\Leftrightarrow\frac{5x-x^2}{x+1}\cdot x+\frac{5-x}{x+1}\cdot\frac{5x-x^2}{x+1}-6=0\)
\(\Leftrightarrow\frac{5x^2-x^3}{x+1}+\frac{x^3-10x^2+25x}{\left(x+1\right)^2}-\frac{6\left(x+1\right)^2}{\left(x+1\right)^2}=0\)
\(\Leftrightarrow\frac{\left(5x^2-x^3\right)\left(x+1\right)}{\left(x+1\right)^2}+\frac{x^3-10x^2+25x}{\left(x+1\right)^2}-\frac{6\left(x^2+2x+1\right)}{\left(x+1\right)^2}=0\)
Suy ra: \(-x^4+4x^3+5x^2+x^3-10x^2+25x-6x^2-12x-6=0\)
\(\Leftrightarrow-x^4+5x^3-11x^2+13x-6=0\)
\(\Leftrightarrow-x^4+x^3+4x^3-4x^2-7x^2+7x+6x-6=0\)
\(\Leftrightarrow-x^3\left(x-1\right)+4x^2\left(x-1\right)-7x\left(x-1\right)+6\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(-x^3+4x^2-7x+6\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(-x^3+2x^2+2x^2-4x-3x+6\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left[-x^2\left(x-2\right)+2x\left(x-2\right)-3\left(x-2\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)\left(-x^2+2x-3\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)\left(x^2-2x+3\right)=0\)
mà \(x^2-2x+3=\left(x-1\right)^2+2>0\forall x\)
nên \(\left[{}\begin{matrix}x-1=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\left(tm\right)\\x=2\left(tm\right)\end{matrix}\right.\)
Vậy: S={1;2}
a/ Đề bài sai, ví dụ tam giác điển hình \(a=3;b=4;c=5\)
\(\Rightarrow\left(3+4+5\right)^2\le9.3.4\Rightarrow144\le108\) (vô lý)
b/ Bạn tham khảo:
Câu hỏi của Vo Thi Minh Dao - Toán lớp 9 | Học trực tuyến
