Toán Lớp 8: Giúp mình với!!!!!!!!!
Chứng minh đẳng thức sau:
$\frac{1}{x(x-y)(x-z)}+$ $\frac{1}{y(y-z)(y-x)}+$ $\frac{1}{z(z-x)(z-y)}=$ $\frac{1}{xyz}$
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Toán Lớp 8: Giúp mình với!!!!!!!!!
Chứng minh đẳng thức sau:
$\frac{1}{x(x-y)(x-z)}+$ $\frac{1}{y(y-z)(y-x)}+$ $\frac{1}{z(z-x)(z-y)}=$ $\frac{1}{xyz}$
Comments ( 1 )
Giải đáp:
1/(x(x-y)(x-z)) + 1/(y(y-z)(y-x)) + 1/(z(z-x)(z-y)) = 1/(xyz)
Lời giải và giải thích chi tiết:
Ta có:
1/(x(x-y)(x-z)) + 1/(y(y-z)(y – x) ) + 1/(z(z-x)(z-y))
= 1/(x(x-y)(x-z)) – 1/(y(y-z)(x-y) ) + 1/(z(x-z)(y-z))
= ( yz(y – z) – xz(x – z) + xy(x – y) )/( xyz(x – y)(x – z)(y – z) )
= ( y^2z – yz^2 – x^2z + xz^2 + x^2y – xy^2)/( xyz(x – y)(x – z)(y – z) )
= ( x(z^2 – y^2) – yz(z – y) + x^2(y – z) )/( xyz(x – y)(x – z)(y – z) )
= ( x(z – y)(z + y) – yz(z – y) – x^2(z – y) )/( xyz(x – y)(x – z)(y – z ))
= ( (z – y)[ x(z + y) – yz – x^2] )/( xyz(x – y)(x – z)(y – z) )
= – ( xz + xy – yz – x^2)/( xyz(x – y)(x – z) )
= – ( z(x – y) – x(x – y) )/( xyz(x – y)(x – z) )
= -( (z – x)(x – y))/( xyz(x – y)(x – z) )
= ( (x – z)(x – y))/(xyz(x – y)(x-z))
= 1/(xyz)