\(a,\)\(đkxđ\Leftrightarrow\hept{\begin{cases}x\ne0\\x\ne-2\end{cases}}\)

\(A=\frac{3m^3+6m^2}{m^3+2m^2+m+2}=\frac{3m^2\left(m+2\right)}{m^2\left(m+2\right)+m+2}.\)

\(=\frac{3m^2\left(m+2\right)}{\left(m+2\right)\left(m^2+1\right)}=\frac{3m^2}{m^2+1}\)

Để \(A=3\Rightarrow\frac{3m^2}{m^2+1}=3\)

\(\Rightarrow3m^2=3\left(m^2+1\right)\)

\(\Rightarrow m^2=m^2+1\)

\(\Rightarrow0=1\)(vô lí )

Vậy không có giá trị nào của m để A = 3

a) A xác định khi \(m^3+2m^2+m+2\ne0\)

\(\Leftrightarrow m^2\left(m+2\right)+\left(m+2\right)\ne0\)\(\Leftrightarrow\left(m^2+1\right)\left(m+2\right)\ne0\)

\(\Rightarrow m+2\ne0\)\(\Rightarrow m\ne-2\)\(\RightarrowĐKXĐ:x\ne-2\)

b) \(A=\frac{3m^3+6m^2}{m^3+2m^2+m+2}=\frac{3m^2\left(m+2\right)}{\left(m^2+1\right)\left(m+2\right)}=\frac{3m^2}{m^2+1}\)

c) \(A=3\)\(\Leftrightarrow\frac{3m^2}{m^2+1}=3\)\(\Leftrightarrow3m^2=3\left(m^2+1\right)\)

\(\Leftrightarrow3m^2=3m^2+3\)\(\Leftrightarrow3m^2-3m^2=3\)\(\Leftrightarrow0=3\)(vô lý)

Vậy không có giá trị m thoả mãn A=3

Ta có : \(x+y+z=0\Rightarrow x+y=-z\)

\(\Rightarrow\left(x+y\right)^2=z^2\Rightarrow x^2+y^2+2xy=z^2\)

\(\Rightarrow x^2+y^2=z^2-2xy\)

Tương tự ta có : \(y^2+z^2=x^2-2yz\)

\(x^2+z^2=y^2-2xz\)

Thay vào biểu thức ta có :

\(A=\frac{x^2}{y^2+z^2-x^2}+\frac{y^2}{x^2+z^2-y^2}+\frac{z^2}{x^2+y^2-z^2}\)

\(=\frac{x^2}{x^2-2yz-x^2}+\frac{y^2}{y^2-2xz-y}+\frac{z^2}{z^2-2xy-z^2}\)

\(=-\frac{x^2}{2yz}-\frac{y^2}{2xz}-\frac{z^2}{2xy}\)

\(=\frac{-x^3-y^3-z^3}{2xyz}=-\frac{x^3+y^3+z^3}{2xyz}\)

\(=\frac{3xyz}{2xyz}=-\frac{3}{2}\)

Chỗ \(x^3+y^3+z^3=3xyz\)là do \(x+y+z=0\)nhé, bạn cần chứng minh không ?

làm sao nhỉ

\(\Leftrightarrow\frac{3\left(x-3\right)+3\left(x+3\right)}{\left(x+3\right)\left(x-3\right)}.\frac{\left(x-3\right)^2}{9}\)

\(\Leftrightarrow\frac{3x-9+3x+9}{x+3}.\frac{x-3}{9}\)

\(\Leftrightarrow\frac{6x\left(x-3\right)}{9\left(x+3\right)}\)

\(\Leftrightarrow\frac{2x\left(x-3\right)}{3\left(x+3\right)}\)

cho mik hỏi bạn cần đề toán hình hay toán đại

hình nhé bạn

\(\frac{2x^2y+2xy+2y}{x^5+x+1}=\frac{2y\left(x^2+x+1\right)}{x^3.x^2+x+1}=\frac{2y}{x^3}\)

#Học tốt!!!

\(\frac{2x^2y+2xy+2y}{x^5+x+1}=\frac{2y\left(x^2+x+1\right)}{x^5-x^2+x^2+x+1}.\)

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

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

\(=\frac{2y\left(x^2+x+1\right)}{\left(x^2+x+1\right)\left[x^2\left(x-1\right)+1\right]}\)

\(=\frac{2y}{x^3-x^2+1}\)