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

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

= a.(d - b) + (bc + bd - db - dc) = a.(d - b) + c.(b - d) = a.(d - b) - c.(d - b) = (a - c).(d - b)  = Vế phải

Vậy....

b) làm  tương tự:

a) (a+b) (c+d) - (a+d) (b+c) = (ac + ad + bc + bd) - (ab + ac +bd + cd) = ac + ad + bc + bd - ab -ac - bd - cd

 và bằng ad + bc - ab - cd = a( d-b ) + c( b-d ) = a (d-b) - c (d-b) = (a-c)(d-b) (dpcm)

p/s: ý B chứng minh tương tự.

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

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

\(=\)\(a-b+2c\)( đpcm )

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

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

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

\(=-a\left(c+d\right)\)( đpcm )

học tốt

Cảm ơn Thanh Ngân

a. VT:(x-y)-(x-z)

= x-y-x+z

= z-y

VP:(z+x)-(y+x)

=z+x-y-x

=z-y

=> VT=VP => đpcm.

b. VT:(x-y+z)-(y+z-x)-(x-y)

= x-y+z-y-z+x-x+y

= x-y

VP:(z-y)-(z-x)

= z-y-z+x

= x-y

=> VT=VP => đpcm.

c. VT: a(b+c)-b(a-c)

=ab+ac-ab+bc

= ac+bc

VP: (a+b)c

= ac+bc

=> VT=VP => đpcm.

d. VT: a(b-c)-a(b+d)

= ab-ac-ab-ad

= -ac-ad

VP: -a(c+d)

= -ac-ad 

=> VT=VP => đpcm

tương tự...

b) Ta có :

\(VT=\left(4x-3y+2\right)-\left(3x-4y+2\right)\)

\(=4x-3y+2-3x+4y-2\)

\(=\left(4x-3x\right)-\left(3y-4y\right)+\left(2-2\right)\)

\(=x+y\)

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

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

\(=x+y\)

\(\Rightarrow VT=VP\)

\(\Rightarrow\)đpcm

Sửa đề: a,b,c,d>0

C/m: \(\left(\frac{a+b}{2}+\frac{c+d}{2}\right)^2\ge\left(a+c\right)\left(c+d\right)\)

Áp dụng BĐT AM-GM ta có:

\(\left(\frac{a+b}{2}+\frac{c+d}{2}\right)^2=\left[\frac{\left(a+c\right)+\left(b+d\right)}{2}\right]^2\ge\left[\frac{2.\sqrt{\left(a+c\right)\left(b+d\right)}}{2}\right]^2=\left(a+c\right)\left(b+d\right)\)

Dấu " = " xảy ra <=> a+c=b+d

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

\(\Leftrightarrow ab+2ad-2bc-4cd=ab+2bc-2ad-4cd\)

\(\Leftrightarrow2ad+2ad=2bc+2bc\Leftrightarrow4ab=4bc\)

\(\Leftrightarrow ad=bc\Rightarrow\dfrac{a}{b}=\dfrac{c}{d},\left(a,b,c,d\ne0\right)\)