a) \(\frac{a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)}{ab^2-ac^2-b^3+bc^2}\)

\(=\frac{a^2b-a^2c+b^2c-b^2a+c^2\left(a-b\right)}{ab^2-b^3-ac^2+bc^2}\)

\(=\frac{\left(a^2b-b^2a\right)+\left(b^2c-a^2c\right)+c^2\left(a-b\right)}{b^2\left(a-b\right)-c^2\left(a-b\right)}\)

\(=\frac{ab\left(a-b\right)+c\left(b^2-a^2\right)+c^2\left(a-b\right)}{\left(b^2-c^2\right)\left(a-b\right)}\)

\(=\frac{ab\left(a-b\right)-c\left(a-b\right)\left(a+b\right)+c^2\left(a-b\right)}{\left(b-c\right)\left(b+c\right)\left(a-b\right)}\)

\(=\frac{ab-c\left(a+b\right)+c^2}{\left(b-c\right)\left(b+c\right)}\)

\(=\frac{ab-ac+c^2-bc}{\left(b-c\right)\left(b+c\right)}\)

\(=\frac{a\left(b-c\right)-c\left(b-c\right)}{\left(b-c\right)\left(b+c\right)}\)

\(=\frac{\left(b-c\right)\left(a-c\right)}{\left(b-c\right)\left(b+c\right)}\)

\(=\frac{a-b}{b+c}\)

Sửa lại: \(\frac{a-c}{b+c}\)

http://diendantoanhoc.net/topic/152549-t%C3%ADnh-fraca2a2-b2-c2-fracb2b2-c2-a2fracc2c2-b2-a2/

Ta có: \(a+b+c=0\)

\(\Rightarrow1\hept{\begin{cases}a+b=-c\\b+c=-a\\c+a=-b\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}a^2+b^2=-2ab+c^2\\b^2+c^2=-2bc+a^2\\c^2+a^2=-2ac+b^2\end{cases}}\)

\(\Rightarrow1A=\frac{a^2}{a^2+2bc-a^2}+\frac{b^2}{b^2+2ac-b^2}+\frac{c^2}{c^2+2ab-c^2}\)

\(=\frac{a^3+b^3+c^3}{2abc}=\frac{a^3+b^3+c^3-3abc+3abc}{2abc}\)

\(=\frac{\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)+3abc}{2abc}\)

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

√(3 )(dm^2)

dien h tam giac la:

     2+2+2=6(dm)

         dap so:6dm

tk cho mk nhe

mk tk lai

chuc cau hoc gioi va thanh cong

http://diendantoanhoc.net/topic/152549-t%C3%ADnh-fraca2a2-b2-c2-fracb2b2-c2-a2fracc2c2-b2-a2/

Đặt: \(\frac{x-y}{z}+\frac{y-z}{x}+\frac{z-x}{y}=M\)

Ta có: 

\(M\cdot\frac{z}{x-y}=1+\frac{z}{x-y}\cdot\left(\frac{y-z}{x}+\frac{z-x}{y}\right)=1+\frac{z}{x-y}\cdot\frac{y^2-yz+xz-x^2}{xy}\)

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

Tương tự ta cũng có:

\(M\cdot\frac{x}{y-z}=1+\frac{2x^3}{xyz}\)              (2)

\(M\cdot\frac{y}{z-x}=1+\frac{2y^3}{xyz}\)            (3)

Từ (1);(2);(3) suy ra

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

Mà \(x+y+z=0\Rightarrow x^3+y^3+z^3=3xyz\)

Nên:

\(M\cdot\left(\frac{z}{x-y}+\frac{x}{y-z}+\frac{y}{z-x}\right)=3+\frac{2\cdot3xyz}{xyz}=9\)

=>đpcm

\(\left(x-1\right)\left(x+1\right)+3x^2=4x^2-2x-3\)

\(=>x^2-1+3x^2-4x^2+2x+3=0\)

\(=>2x+2=0\)

\(=>2x=-2\)

\(=>x=-1\)