a) ( a - b ) + ( c - d ) - ( a + c ) 

= a - b + c - d - a - c

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

= 0 + 0 - ( b + d )

= - ( b + d )

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

= a - b - c + d + b + c

= a + ( -b + b ) + ( -c +  c ) + d

= a + 0 + 0 + d

= a + d

~ HỌC TỐT ~

A + B = (a + b - 5) + (b - c - 9) = a + 2b - c - 14

C + D = (b - c - 4) + (-b + a) = a - b - c - 4

Ta thấy A + B = C + D = a + 2b - c - 14 = a - b - c - 4

Vậy A+B = C+D(điều phải chứng minh)

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

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

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

Từ ( 1 ) và ( 2 ) suy ra 

\(b^2+2bc+c^2-a^2-2ad-d^2=a^2-2ad+d^2-b^2+2bc-c^2\)

\(4ad=4ac\Rightarrow ad=bc\)

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

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

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

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

\(\Leftrightarrow2d\cdot2a=2c\cdot2b\)

\(\Leftrightarrow ad=bc\)

hay \(\dfrac{a}{c}=\dfrac{b}{d}\)

a) Áp dụng tính chất của dãy tỉ số bằng nhau, ta có :

\(\frac{a}{b}\)=\(\frac{c}{d}\)=\(\frac{a+c}{b+d}\)=\(\frac{a-c}{c-d}\)(đpcm)

Vậy .......