Question

[chim cánh cụt]

CMR: Nếu

\(\left(a+b+c+d\right)\left(1-b-c-d\right)=\left(a-b+c-d\right)\left(a+b-c-d\right)\)

Thì:

\(\frac{a}{b}=\frac{c}{d}\)

Answer 1

[tth_new]

Xin phép sửa đề nhé: " Nếu \(\left(a+b+c+d\right)\left(1-b-c-d\right)=\left(a-b-c-d\right)\left(a+b+c+d\right)\)thì \(\frac{a}{b}=\frac{c}{d}\)"

                        Giải

Từ giả thiết suy ra a = b = c = d

Ta có:\(\left(a+b+c+d\right)\left(1-b-c-d\right)=\left(a-b-c-d\right)\left(a+b+c+d\right)\)

Suy ra: \(\frac{a+b+c+d}{a+b+c+d}=\frac{a-b-c-d}{1-b-c-d}\)

Do a = b =c =d nên \(\frac{a+b+c+d}{a+b+c+d}=\frac{a-b-c-d}{1-b-c-d}\Leftrightarrow\frac{4a}{4a}=\frac{4b}{4b}=\frac{4c}{4c}=\frac{4d}{4d}\)

Theo tỉ lệ thức ta có thể suy ra \(\frac{4a}{4b}=\frac{4c}{4d}\Leftrightarrow\frac{a}{b}=\frac{c}{d}^{\left(đpcm\right)}\)

Answer 2

[kudo shinichi]

Mạo phép sửa đề:

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

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