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

\(=ab+ac-\left(ab-bc\right)\)

\(=ab+ac-ab+bc\)

\(=ac+bc\)

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

b,\(\left(a+b\right)\left(a-b\right)\)

\(=\left(aa+ab\right)-\left(ab+bb\right)\)

\(=aa+ab-ab-bb\)

\(=aa-bb\)

\(=a^2-b^2\)

\(x\left(y+z\right)-y\left(x-z\right)=xy+xz-yx+yz\)

\(=xy-xy+\left(zx+zy\right)\)

\(=\left(x+y\right)z\)

b, \(\left(m-n\right)\left(m+n\right)=m^2+mn-nm-n^2\)

\(=m^2-n^2\)

VT = a + b 2 = a + b . a + b    = a ( a + b ) + b ( a + b ) = a 2 + ab + ba + b 2    = a 2 + 2 ab + b 2 = VP   ( dpcm )

VT = a − b . a + b    = a ( a + b ) − b ( a + b ) = a 2 + ab − ba − b 2    = a 2 − b 2 = VP   ( dpcm )

VT = ( a + b ) ( a − b ) = a 2 − a . b + b . a − b 2 = a 2 − b 2 + ab − ab = a 2 − b 2 + 0 = a 2 − b 2 = VP . Vậy   ( a + b ) ( a − b ) = a 2 − b 2

c) Ta có: \(C=4x^2+y^2-4xy+8x-4y+4\)

\(=\left(2x-y\right)^2+2\cdot\left(2x-y\right)\cdot2+2^2\)

\(=\left(2x-y+2\right)^2\)

Vế trái: x2; Vế phải: 6x –5

Vế trái của đẳng thức là: 2.(b+1)

Vế phải của đẳng thức là: 2b+2

VT = a ( b + c ) − b ( a − c ) = ab + ac − ba + bc = ( ab − ab ) + ( ac + bc ) = 0 + a + b . c = VP Vậy   a ( b + c ) − b ( a − c ) = ( a + b ) . c

VT = a ( b + c ) − b ( a − c ) = ab + ac − ba − bc    = ab + ac − ba + bc = ac + bc = c ( a + b ) = VP   ( dpcm )