( a+ b+ c)^2 = a^2 + b^2 + c^2 + 2ab+ 2bc+ 2ac

a, ( x + y - 2)^2 = x^2 + y^2 + 4 + 2xy - 4x - 4y 

bl, ( 2x + 3y+ 5)^2 = 4x^2 + 9y^2 + 25 + 12xy + 20x + 30y

c < ( 3x - y + 2)^2 = 9x^2 + y^2 + 4 -6xy + 12x - 4y

1. \(4x^2-9y^2=\left(2x\right)^2-\left(3y\right)^2=\left(2x-3y\right)\left(2x+3y\right)\)  

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

3. \(x^2-1=x^2-1^2=\left(x-1\right)\left(x+1\right)\)  

4.  \(8+x^3=2^3+x^3=2^3+3.2^2.x+3.2.x^2+x^3\) 

                                   \(=8+12x+6x^2+x^3\)  

5. \(8x^3+27=\left(2x\right)^3+3^3\)  

                    \(=\left(2x\right)^3+3.\left(2x\right)^2.3+3.2x.3^2+3^3\)  

                    \(=8x^3+36x^2+54x+27\)  

6. \(\left(a+b+c\right)^2=a^2+b^2+c^2+2ab+2bc+2ac\) 

7. \(\left(a-b+c\right)^2=a^2+b^2+c^2-2ab-2bc+2ac\) 

8. \(\left(a-b-c\right)^2=a^2+b^2+c^2-2ab-2ac+2bc\) 

a: \(=\dfrac{\left(x^4-y^4\right)^2}{x^2+y^2}=\left(x^2-y^2\right)^2\cdot\left(x^2+y^2\right)\)

b: \(=\dfrac{\left(4x+3\right)\left(16x^2-12x+9\right)}{16x^2-12x+9}=4x+3\)

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

b) \(=\left(3a-1\right)^2+2\left(3a-1\right)\left(3a+1\right)+\left(3a+1\right)^2=\left(3a-1+3a+1\right)^2=\left(6a\right)^2=36a^2\)

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

nhớ LI KE

jjj

1/ Chứng minh các hằng đẳng thức:

^{\(x^4 + y^4 +(x+y)^4 = x^4 + y^4 + x^4 + 4x^3y + 6x^2y^2 +4xy^3 + y^4 \\\ = 2x^4 +2y^4 +4x^2y^2+4x^3y+4xy^3+2x^2y^2\)}

\(= 2(x^4 +y^4 +2x^2y^2)+4xy(x^2+y^2) + 2x^2y^2 \\\ = 2(x^2 + y^2)2 + 4xy(x^2 + y^2) +2x^2y^2\)

\(=2(x^2 +y^2) +2xy(x^2+ y^2) +x^2y^2) = 2(x^2 + y^2 + xy)^2 \\\ ⇒ đpcm\)

2/

Ta có : \([(5a - 3b) + 8c][(5a - 3b) - 8c] \)
\(= (5a - 3b)^2 - 64c^2\) (theo hiệu hai bình phương)
\(= 25a^2 - 30ab + 9b^2 - 64c^2\) (theo bình phương của hiệu)
\(= 25a^2 - 30ab + 9b^2 - 16(a^2 - b^2)\) (vì \(4c^2 = a^2 - b^2\))
\(= 9a^2 - 30ab + 25b^2 \)
\(= (3a - 5b)^2\) (theo bình phương của hiệu).