\(\frac{x}{x-3}-\frac{2}{x+3}+\frac{x\left(1-x\right)}{x^2-9}\)

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

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

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

\(=\frac{x\left(7-x\right)}{\left(x-3\right)\left(x+3\right)}=\frac{7x-x^2}{\left(x-3\right)\left(x+3\right)}\)

cần câu c thôi giúp vs

Nhân hết ra rút gọn rồi tìm x 

\(\left(x+4\right)\left(x-4\right)+x\left(6-x\right)=0\)

\(x^2-16+6x-x^2=0\)

\(-16+6x=0\)

\(6x=16\)

\(x=\frac{8}{3}\)

A B C E D

Áp dụng định lí Pytago cho các tam giác vuông ta có :

\(CD^2=AC^2+DA^2\)

\(BC^2=AB^2+AC^2\)

\(\Rightarrow CD^2-BC^2=\left(AC^2+AD^2\right)-\left(AB^2+AC^2\right)=AD^2-AB^2\left(1\right)\)

------------

\(ED^2=DA^2+AE^2\)

\(BE^2=AE^2+AB^2\)

\(\Rightarrow ED^2-BE^2=\left(DA^2+AE^2\right)-\left(AE^2+AB^2\right)=AD^2-AB^2\left(2\right)\)

Từ ( 1 ) và ( 2 ) \(\Rightarrow CD^2-BC^2=ED^2-BE^2\left(đpcm\right)\)

Chúc bạn học tốt !!!

\(a^3+b^3+c^3-3abc\)

\(=a^3+3a^2b+3ab^2+b^3+c^3-3a^2b-3ab^2-3abc\)

\(=\left(a+b\right)^3+c^3-\left(3a^2b+3ab^2+3abc\right)\)

\(=\left(a+b+c\right)[\left(a+b\right)^2-c\left(a+b\right)+c^2]-3ab\left(a+b+c\right)\)

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

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

\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-ab\right)\)

a3+b3+c3−3abca^3+b^3+c^3-3abca3+b3+c3−3abc

=a3+3a2b+3ab2+b3+c3−3a2b−3ab2−3abc=a^3+3a^2b+3ab^2+b^3+c^3-3a^2b-3ab^2-3abc=a3+3a2b+3ab2+b3+c3−3a2b−3ab2−3abc

=(a+b)3+c3−(3a2b+3ab2+3abc)=\left(a+b\right)^3+c^3-\left(3a^2b+3ab^2+3abc\right)=(a+b)3+c3−(3a2b+3ab2+3abc)

=(a+b+c)[(a+b)2−c(a+b)+c2]−3ab(a+b+c)=\left(a+b+c\right)[\left(a+b\right)^2-c\left(a+b\right)+c^2]-3ab\left(a+b+c\right)=(a+b+c)[(a+b)2−c(a+b)+c2]−3ab(a+b+c)

=(a+b+c)(a2+2ab+b2−ac−bc+c2)−3ab(a+b+c)=\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2\right)-3ab\left(a+b+c\right)=(a+b+c)(a2+2ab+b2−ac−bc+c2)−3ab(a+b+c)

=(a+b+c)(a2+2ab+b2−ac−bc+c2−3ab)=\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)=(a+b+c)(a2+2ab+b2−ac−bc+c2−3ab)

=(a+b+c)(a2+b2+c2−ab−ac−ab)=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-ab\right)=(a+b+c)(a2+b2+c2−ab−ac−ab)

\(x^4-6x^3+12x^2-14x+3\)

\(=\left(x^4-2x^3+3x^2\right)-\left(4x^3-8x^2+12x\right)+\left(x^2-2x+3\right)\)

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

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

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

\(=\left(x^2-2x+3\right)\left[\left(x-2\right)^2-3\right]\)

\(=\left(x^2-2x+3\right)\left(x-2-\sqrt{3}\right)\left(x-2+\sqrt{3}\right)\)