Đáp án:
$B=1$
Giải thích các bước giải:
$B=\frac{yz}{(x-y)(x-z)}+$ $\frac{xz}{(y-x)(y-z)}+$ $\frac{xy}{(z-x)(z-y)}$
$B=\frac{yz(z-y)+zx(x-z)+xy(y-x)}{(x-y)(y-z)(z-x)}$
$B=\frac{yz(z-y)+zx(x-z)-xy[(z-y)+(x-z)]}{(x-y)(y-z)(z-x)}$
$B=\frac{y(z-y)(z-x)+x(x-z)(z-y)}{(x-y)(y-z)(z-x)}$
$B=\frac{(x-y)(y-z)(z-x)}{(x-y)(y-z)(z-x)}$
$B=\dfrac{yz}{(x-y)(x-z)}+\dfrac{xz}{(y-x)(y-z)}+\dfrac{xy}{(z-x)(z-y)}$
$⇔ B=\dfrac{-yz(y-z)-xz(z-x)-xy(x-y)}{(x-y)(y-z)(z-x)}$
$⇔ B=\dfrac{-y^2z+yz^2-xz^2+x^2z-x^2y+xy^2}{(x-y)(y-z)(z-x)}$
$⇔ B=\dfrac{-yz(y-z)-x^2(y-z)+x(y^2-z^2)}{(x-y)(y-z)(z-x)}$
$⇔ B=\dfrac{(y-z)(-yz-x^2+xy+xz)}{(x-y)(y-z)(z-x)}$
$⇔ B=\dfrac{(y-z)[-x(x-y)+z(x-y)}{(x-y)(y-z)(z-x)}$
$⇔ B=\dfrac{(y-z)(x-y)(z-x)}{(x-y)(y-z)(z-x)}$
$⇔ B=1$
Chúc bạn học tốt !!!
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