\(\dfrac{bz-cy}{a}=\dfrac{cx-az}{b}=\dfrac{ay-bx}{c}\)

\(\Rightarrow\dfrac{abz-acy}{a^2}=\dfrac{bcx-abz}{b^2}=\dfrac{acy-bcx}{c^2}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{abz-acy}{a^2}=\dfrac{bcx-abz}{b^2}=\dfrac{acy-bcx}{c^2}=\dfrac{abz-acy+bcx-abz+acy-bcx}{a^2+b^2+c^2}=0\)

Suy ra \(\left\{{}\begin{matrix}bz=cy\Leftrightarrow\dfrac{y}{b}=\dfrac{z}{c}\\cx=az\Leftrightarrow\dfrac{x}{a}=\dfrac{z}{c}\\ay=bx\Leftrightarrow\dfrac{x}{a}=\dfrac{y}{b}\end{matrix}\right.\Leftrightarrow\dfrac{x}{a}=\dfrac{y}{b}=\dfrac{z}{c}\left(đpcm\right)\)

p/s: đã sửa đề

\(=\frac{bzx-cxy}{ax}=\frac{cxy-ayz}{by}=\frac{ayz-bzx}{cz}=\frac{bzx-cxy+cxy-ayz+ayz-bzx}{ax+by+cz}=0\)

=>bz-cy=0;cx-az=0;ay-bx=0

\(\Rightarrow\frac{x}{a}=\frac{y}{b}=\frac{z}{c}\left(đpcm\right)\)

2.

Vì \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{a+b+c}{b+c+d}=\dfrac{\left(a+b+c\right)^3}{\left(b+c+d\right)^3}\left(1\right)\)

Vì \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}\Rightarrow\dfrac{a}{b}.\dfrac{a}{b}.\dfrac{a}{b}=\dfrac{a}{b}.\dfrac{b}{c}.\dfrac{c}{d}=\dfrac{a}{d}\left(2\right)\)

Từ \(\left(1\right);\left(2\right)\Rightarrow\left(\dfrac{a+b+c}{b+c+d}\right)^3=\dfrac{a}{d}\left(dpcm\right)\)

a) x^3 + x^2 - x - 1

=(x³+x²)+(-x-1)

=x².(x+1)-(x+1)

=(x+1)(x²-1)

=(x+1)(x-1)(x+1)

=(x+1)²(x-1)

b) a^3 + a^2.b - a^2.c - a.b.c

=(a³+a²b)+(-a²c-abc)

=a².(a+b)-ab.(a+b)

=(a+b)(a²-ab)

=a.(a+b)(a-b)

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