a)(a+2b-3c-d)(a+2b+3c+d)=[(a+2b)-(3c+d)][(a+2b)+(3c-d)]

​=(a+2b)²​-(3c-d)²=a²​+4ab+4b²​-9c²​+6cd-d²

^​câu b tương tự

\(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{2a}{2b}=\dfrac{3c}{3d}=\dfrac{2a+3c}{2b+3d}=\dfrac{2a-3c}{2b-3d}\)

\(\Rightarrow\dfrac{2a+3c}{2a-3c}=\dfrac{2b+3d}{2b-3d}\)

\(\Rightarrow dpcm\)

A =(a+b-2c) -(-a+b+c) -(2a-b-c)

   = a+b-2c+a-b-c-2a+b+c

   = b-2c

B=-(2a-b+c) + (b-2c-3a) -(-5a-3c+b)

  = -2a+b-c+b-2c-3a+5a+3c-b

  = b-c

C=(3a-b-2c)-( 2b+3c-a) +(2a-3b)

  = a-b-2c-2b-3c+a+2a-3b

  = -6b-5c

D=(5a-3b+c) +( 2a-3b+5) -( b-c+a)

   = 5a-3b+c+2a-3b+5-b+c-a

   = 6a-7b+2c

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

\(=a+b-2c+a-b-c-2a+b+c=b-2c\)

\(B=-\left(2a-b+c\right)+\left(b-2c-3a\right)-\left(-5a-3c+b\right)\)

\(=-2a+b-c+b-2c-3a+5a+3c-b=b\)

\(C=\left(3a-b-2c\right)-\left(2b+3c-a\right)+\left(2a-3b\right)\)

\(=3a-b-2c-2b-3c+a+2a-3b=6a-6b-5c\)

\(D=\left(5a-3b+c\right)+\left(2a-3b+5\right)-\left(b-c+a\right)\)

\(=5a-3b+c+2a-3b+5-b+c-a=6a-7b+2c\)

Ta có: \(\left(a+2b-3c-d\right)\left(a+2b+3c+d\right)\)

\(=\left[\left(a+2b\right)-\left(3c+d\right)\right]\cdot\left[\left(a+2b\right)+\left(3c+d\right)\right]\)

\(=\left(a+2b\right)^2-\left(3c+d\right)^2\)

\(=a^2+4ab+4b^2-9c^2-6cd-d^2\)

( a + 2b - 3c - d )( a + 2b + 3c + d )

= [ ( a + 2b ) - ( 3c + d ) ][ ( a + 2b ) + ( 3c + d ) ]

= ( a + 2b )² - ( 3c + d )²

= a² + 4ab + 4b² - ( 9c² + 6cd + d² )

= a² + 4ab + 4b² - 9c² - 6cd - d²

(a + 2b − 3c − d) (a + 2b + 3c + d) 

= [(a + 2b) − (3c + d) ] · [(a + 2b) + (3c + d)] 

= (a + 2b)² − (3c + d)² 

= a² + 4ab + 4b² − 9c² − 6cd − d²

Hok tốt

đề bài của bài 1 là rút gọn

\(\dfrac{2a+3c}{2b+3d}=\dfrac{2a-3c}{2b-3d}=\dfrac{2a+3c+2a-3c}{2b+3d+2b-3d}=\dfrac{a}{b}\)

\(\dfrac{2a+3c}{2b+3d}=\dfrac{2a-3c}{2b-3d}=\dfrac{2a+3c-\left(2a-3c\right)}{2b+3d-\left(2b-3d\right)}=\dfrac{c}{d}\)

Suy ra \(\dfrac{a}{b}=\dfrac{c}{d}\)