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2 nha bn

time in the past; present; future

Câu 45: 

Đặt \(g\left(x\right)=\frac{x}{x^2+x+1}-\frac{m}{3}\)

\(g'\left(x\right)=\left(\frac{x}{x^2+x+1}\right)'=\frac{-x^2+1}{\left(x^2+x+1\right)^2}\)

\(g'\left(x\right)=0\Rightarrow x=\pm1\), \(g'\left(x\right)\)xác định với mọi \(x\inℝ\).

Suy ra để hàm số \(f\left(x\right)=\left|g\left(x\right)\right|\)có \(4\)điểm cực trị thì phương trình \(g\left(x\right)=0\)có hai nghiệm phân biệt khác \(\pm1\). 

\(g\left(x\right)=0\Leftrightarrow\frac{x}{x^2+x+1}=\frac{m}{3}\)

\(lim_{x\rightarrow-\infty}\frac{x}{x^2+x+1}=0,lim_{x\rightarrow+\infty}\frac{x}{x^2+x+1}=0\)

\(g\left(-1\right)=-1,g\left(1\right)=\frac{1}{3}\)

Suy ra BBT của hàm \(\frac{x}{x^2+x+1}\).

Từ đó suy ra để phương trình \(\frac{x}{x^2+x+1}\)có hai nghiệm phân biệt thì 

\(\orbr{\begin{cases}0< \frac{m}{3}< \frac{1}{3}\\-1< \frac{m}{3}< 0\end{cases}}\Leftrightarrow m\in\left\{-2,-1\right\}\)(vì \(m\)nguyên) 

Chọn A. 

Kí cần bt đâu bn mà tôi cũng ko bt vì mới lớp sáu 

ko bt

mong anh thông cảm em mới có lớp 5

1 + 1 = 2 

2 + 2 = 4 

Nếu bạn bảo sai thì đây là đố mẹo 

2 và 4 nếu ko thì là đố mẹo

1+1=5                         2+2=22

tieng dog 

mk ko co y xuc pham

:>