Question

Cm ta có tỉ lệ thức \(\dfrac{a}{b}=\dfrac{c}{d}\) nếu có đẳng thức:

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

Answer 1

Đặt a/b=c/d=k

=>a=bk; c=dk

\(\left(a+b+c+d\right)\left(a-b-c+d\right)\)

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

\(=\left(bk+d\right)^2-\left(b+dk\right)^2\)

\(=b^2k^2+2bkd+d^2-b^2-2bkd-d^2k^2\)

\(=b^2\left(k^2-1\right)+d^2\left(1-k^2\right)\)

\(=\left(k^2-1\right)\left(b^2-d^2\right)\)(1)

\(\left(a-b+c-d\right)\left(a+b-c-d\right)\)

\(=\left(a-d\right)^2-\left(b-c\right)^2\)

\(=\left(bk-d\right)^2-\left(b-dk\right)^2\)

\(=b^2k^2-2bkd+d^2-b^2+2dk-d^2k^2\)

\(=k^2\left(b^2-d^2\right)-\left(b^2-d^2\right)=\left(b^2-d^2\right)\left(k^2-1\right)\)(2)

Từ (1) và (2) suy ra ĐPCM