a ) \(\left(2a+b\right)^2-\left(2a+b\right)\left(2a-b\right)-2a\left(b-a\right)\)

\(=4a^2+4ab+b^2-\left(4a^2-b^2\right)-2ab+2a^2\)

\(=4a^2+4ab+b^2-4a^2+b^2-2ab+2a^2\)

\(=2a^2+2ab+2b^2\)

\(=\left(a^2+2ab+b^2\right)+a^2+b^2\)

\(=\left(a+b\right)^2+a^2+b^2\)

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

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

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

\(=c^2\)

@Khôi Bùi

\(\left(2a+b\right)^2-\left(2a+b\right)\left(2a-b\right)-2a\left(b-a\right)\)

\(=\left(2a+b\right)\left[\left(2a+b\right)-\left(2a-b\right)\right]-2a\left(b-a\right)\)

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

\(=4ab+2b^2-2ab+2a^2=2\left(a^2+ab+b^2\right)\)

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

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

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

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

a: \(=a^2-b^4\)

b: \(=\left(a^2+2a\right)^2-9\)

c: \(=a^2-\left(2a+3\right)^2\)

d: \(=a^4-\left(2a-3\right)^2\)

e: \(=\left(-a^2-2a+3\right)^2\)

g: \(=4a^2-a^4\)

a) A=(4-5x)²-(3+5x)²=(4-5x-3-5x)(4-5x+3+5x)=(-25x+1)1=-25x+1

B=(3x-1)(1+3x)-(3x+1)²=9x²-1-(3x+1)²=9x²-1-(9x²+6x+1)=9x²-1-9x²-6x-1=-6x-2=-2(3x+1)

MN K BT?

\(\left(2a+b\right)^2-\left(2a-b\right)^2\)

\(=\left(4a^2+4ab+b^2\right)-\left(4a^2-4ab+b^2\right)\)

\(=4a^2+4ab+b^2-4a^2+4ab-b^2\)

\(=8ab\)

a/ \(16a^4+48a^3+4a^2-120a-72\)

b.\(2b^2+2a^2=4a\)

a^4+a^3b+ab^3+b^4>=4

Bài 2:

a) \(\left(x+5\right)^2=x^2+10x+25\)

b) \(\left(\dfrac{5}{2}-t\right)^2=\dfrac{25}{4}-5t+t^2\)

c) \(\left(2u+3v\right)^2=4u^2+12uv+9v^2\)

d) \(\left(-\dfrac{1}{8}a+\dfrac{2}{3}bc\right)^2=\dfrac{1}{64}a^2-\dfrac{1}{6}abc+\dfrac{4}{9}b^2c^2\)

e) \(\left(\dfrac{x}{y}-\dfrac{1}{z}\right)^2=\dfrac{x^2}{y^2}-\dfrac{2x}{yz}+\dfrac{1}{z^2}\)

f) \(\left(\dfrac{mn}{4}-\dfrac{x}{6}\right)\left(\dfrac{mn}{4}+\dfrac{x}{6}\right)=\dfrac{m^2n^2}{16}-\dfrac{x^2}{36}\)

Bài 1:

$M=(2a+b)^2-(b-2a)^2=[(2a+b)-(b-2a)][(2a+b)+(b-2a)]$

$=4a.2b=8ab$

$N=(3a+1)^2+2a(1-2b)+(2b-1)^2$

$=(9a^2+6a+1)+2a-4ab+(4b^2-4b+1)$
$=9a^2+8a+4b^2-4b-4ab+2$

$A=(m-n)^2+4mn=m^2-2mn+n^2+4mn$

$=m^2+2mn+n^2=(m+n)^2$

Bài 1: 

a: Ta có: \(M=\left(2a+b\right)^2-\left(b-2a\right)^2\)

\(=4a^2+4ab+b^2-b^2+4ab-4a^2\)

\(=8ab\)

b: Ta có: \(N=\left(3a+2\right)^2+2a\left(1-2b\right)+\left(2b-1\right)^2\)

\(=\left(3a+2+1-2b\right)^2\)

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

\(=9a^2+4b^2+9-12ab+18a-12b\)

c: Ta có: \(A=\left(m-n\right)^2+4nm\)

\(=m^2-2mn+n^2+4mn\)

\(=m^2+2mn+n^2\)

\(=\left(m+n\right)^2\)

2: 

a: \(\left(x+5\right)^2=x^2+10x+25\)

b: \(\left(\dfrac{5}{2}-t\right)^2=\dfrac{25}{4}-5t+t^2\)