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\(\dfrac{x^2}{y+z}+\dfrac{y^2}{x+z}+\dfrac{z^2}{x+y}\)
\(=x.\left(\dfrac{x}{y+z}+1-1\right)+y.\left(\dfrac{y}{x+z}+1-1\right)+z.\left(\dfrac{z}{x+y}+1-1\right)\)
\(=x.\left(\dfrac{x+y+z}{y+z}\right)+y.\left(\dfrac{x+y+z}{x+z}\right)+z.\left(\dfrac{x+y+z}{x+y}\right)-\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left(\dfrac{x}{y+z}+\dfrac{y}{z+x}+\dfrac{z}{x+y}\right)-\left(x+y+z\right)=\left(x+y+z\right)-\left(x+y+z\right)=0\)
1) \(\dfrac{3}{x-3}-\dfrac{6x}{9-x^2}+\dfrac{x}{x+3}=0\)
\(\Leftrightarrow\dfrac{3}{x-3}+\dfrac{6x}{x^2-9}+\dfrac{x}{x+3}=0\)
\(\Leftrightarrow\dfrac{3}{x-3}+\dfrac{6x}{\left(x-3\right)\left(x+3\right)}+\dfrac{x}{x+3}=0\)
\(\Leftrightarrow\dfrac{3\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}+\dfrac{6x}{\left(x-3\right)\left(x+3\right)}+\dfrac{x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=0\)
\(\Leftrightarrow\dfrac{3\left(x+3\right)+6x+x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=0\)
\(\Leftrightarrow\dfrac{3x+9+6x+x^2-3x}{\left(x-3\right)\left(x+3\right)}=0\)
\(\Leftrightarrow\dfrac{x^2+6x+9}{\left(x-3\right)\left(x+3\right)}=0\)
\(\Leftrightarrow\dfrac{x^2+2.x.3+3^2}{\left(x-3\right)\left(x+3\right)}=0\)
\(\Leftrightarrow\dfrac{\left(x+3\right)^2}{\left(x-3\right)\left(x+3\right)}=0\)
\(\Leftrightarrow\dfrac{x+3}{x-3}=0\)
\(\Leftrightarrow x+3=0\)
\(\Leftrightarrow x=-3\)
Vậy x=-3
bạn ơi x ko thể bằng -3 đc vì
\(\dfrac{x}{x+3}=\dfrac{-3}{-3+3}=\dfrac{-3}{0}\) là sai
Câu hỏi của Hoàng Liên - Toán lớp 9 - Học toán với OnlineMath Em tham khảo tại link này nhé !
Ta có:
\(\left(x+y+z\right)\left(\dfrac{x}{y+z}+\dfrac{y}{z+x}+\dfrac{z}{x+y}\right)=\dfrac{x^2}{y+z}+\dfrac{y^2}{z+x}+\dfrac{z^2}{x+y}+x+y+z\)
\(\Leftrightarrow x+y+z=\dfrac{x^2}{y+z}+\dfrac{y^2}{z+x}+\dfrac{z^2}{x+y}+x+y+z\)
\(\Leftrightarrow\dfrac{x^2}{y+z}+\dfrac{y^2}{z+x}+\dfrac{z^2}{x+y}=0\)
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