dễ vãi lol ra bn ngu vãi đái

a, \(\left(x-y+4\right)^2-\left(2x+3y-1\right)^2=\left(x-y+4+2x+3y-1\right)\left(x-y+4-2x-3y+1\right)\)

\(=\left(3x+2y+3\right)\left(-x-4y+5\right)\)

b, \(x^6+y^6=\left(x^2\right)^3+\left(y^2\right)^3=\left(x^2+y^2\right)\left(x^4-x^2y^2+y^4\right)\)

c, \(x^{16}-1=\left(x^2\right)^8-1=\left[\left(x^2\right)^4\right]^2-1=\left(x^8-1\right)\left(x^8+1\right)\)

Phân tích à :)

a) ( x - y + 4 )² - ( 2x + 3y - 1 )²

= [ ( x - y + 4 ) - ( 2x + 3y - 1 ) ][ ( x - y + 4 ) + ( 2x + 3y - 1 ) ]

= ( x - y + 4 - 2x - 3y + 1 )( x - y + 4 + 2x + 3y - 1 )

= ( -x - 4y + 5 )( 3x + 2y + 3 )

b) x⁶ + y⁶ = ( x² )³ + ( y² )³ = ( x² + y² )( x⁴ - x²y² + y⁴ )

c) x¹⁶ - 1 = ( x⁸ )² - 1² = ( x⁸ - 1 )( x⁸ + 1 )

= ( x⁴ - 1 )( x⁴ + 1 )( x⁸ + 1 )

= ( x² - 1 )( x² + 1 )( x⁴ + 1 )( x⁸ + 1 )

= ( x - 1 )( x + 1 )( x² + 1 )( x⁴ + 1 )( x⁸ + 1 )

d) x³⁶ - 64 = ( x¹⁸ )² - 8²

= ( x¹⁸ - 8 )( x¹⁸ + 8 )

= ( x⁶ - 2 )( x¹² + 2x + 4 )( x⁶ + 2 )( x¹² - 2x + 4 )

e) 9x² + 90x + 225 - ( x - 7 )²

= ( 3x + 15 )² - ( x - 7 )²

= ( 3x + 15 - x + 7 )( 3x + 15 + x - 7 )

= ( 2x + 22 )( 2x - 8 )

= 2( x + 11 ).2( x - 4 )

= 4( x + 11 )( x - 4 )

f) 49( y - 4 )² - 9y² - 36y - 36

= 7²( y - 4 )² - ( 9y² + 36y + 36 )

= ( 7y - 28 )² - ( 3y + 6 )² 

= ( 7y - 28 - 3y - 6 )( 7y - 28 + 3y + 6 )

= ( 4y - 34 )( 10y - 22 )

= 2( 2y - 17 ).2( 5y - 11 )

= 4( 2y - 17 )( 5y - 11 )

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

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

\(=\left(3x+2y+3\right)\left(-x-4y+5\right)\)

\(49\left(y-4\right)^2-9y^2-36y-36\)

\(=49\left(y-4\right)^2-\left(9y^2+36y+36\right)\)

\(=49\left(y-4\right)^2-\left(3y+6\right)^2\)

\(=[7\left(y-4\right)]^2-\left(3y+6\right)^2\)

\(=\left(7y-28\right)^2-\left(3y+6\right)^2\)

\(=\left(7y-28+3y+6\right)\left(7y-28-3y-6\right)\)

\(=\left(10y-22\right)\left(4y-34\right)\)

\(9x^2+90x+225-\left(x-7\right)^2\)

\(=\left(3x+15\right)^2-\left(x-7\right)^2\)

\(=\left(3x+15+x-7\right)\left(3x+15-x+7\right)\)

\(=\left(4x+8\right)\left(2x+22\right)\)

a) =( x²+y²-5)²-[2(xy-2)]²

⁼( x²+y²-5)²- (2xy-4)²

=(x²+y²-5+2xy-4)(x²+y²-5-2xy+4)

=[(x+y)²-9][(x-y)²-1]

phân tích tiếp HĐT 2 ở 2 thừa số

B1:

a) \(9x^2+90x+225-\left(x-7\right)^2\)

= \(9x^2+90x+225-x^2+14x-49\)

= \(8x^2+104x+176\)

= \(\left(x+2\right)\left(x+11\right)\)

b) \(49\left(y-4\right)^2-9y^2-36y+36\)

= \(49\left(y^2-8y+16\right)-9y^2-36y+36\)

= \(49y^2-392y+784-9y^2-36y+36\)

= \(40y^2-428y+820\)

= \(\left(5y-41\right)\left(8y-20\right)\)

B2:

a) A = \(xy-4y-5y+20=xy-9y+20\)

A = \(y\left(x-9\right)+20\)

Với x = 14, y = \(\dfrac{11}{2}\)

A = \(\dfrac{11}{2}\left(14-9\right)+20=47,5\)

b) B = \(x^2+xy-5x-5y\)

B = \(x\left(x+y\right)-5\left(x+y\right)=\left(x+y\right)\left(x-5\right)\)

Với x = -5, y = -8

B = \(\left(-5-8\right)\left(-5-5\right)=130\)

B3:

a) \(4x^2-25-\left(2x-5\right)\left(2x+7\right)=0\)

\(\left(2x-5\right)\left(2x+5\right)-\left(2x-5\right)\left(2x+7\right)=0\)

\(\left(2x-5\right)\left(2x+5-2x-7\right)=0\)

\(\left(2x-5\right)\left(-2\right)=0\)

\(x=\dfrac{5}{2}\)

b) \(\left(x^3+27\right)+\left(x+3\right)\left(x-9\right)=0\)

\(\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)=0\)

\(\left(x+3\right)\left(x^2-3x+9+x-9\right)=0\)

\(\left(x+3\right)x\left(x-2\right)=0\)

\(\left[{}\begin{matrix}x=-3\\x=0\\x=2\end{matrix}\right.\)

c) \(\left(2x^3+2x^2\right)+\left(3x^2+3\right)=0\)

\(2x^3+5x^2+3=0\)

\(\Rightarrow\) Đề sai rồi, nghiệm khủng bố lắm.

Câu 2 nha

\(a,x^4+2x^3+x^2\)

\(=x^2\left(x^2+2x+1\right)\)

\(=x^2\left(x+1\right)^2\)

\(c,x^2-x+3x^2y+3xy^2+y^3-y\)

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

\(=\left(x+y\right)^3-\left(x+y\right)\)

\(=\left(x+y\right)\left(x^2+2xy+y^2-1\right)\)

\(d,5x^2-10xy+5y^2-20z^2\)

\(=5\left(x^2-2xy+y^2-4z^2\right)\)

\(=5\left[\left(x-y\right)^2-4z^2\right]\)

\(=5\left[\left(x-y+2z\right)\left(x-y-2z\right)\right]\)

chiều lm tiếp cho 

Vụ này khoai à nha !

\(b,9x^2+90x+225-\left(x-y\right)^2\)

\(=\left(3x+15\right)^2-\left(x-y\right)^2\)

\(=\left(3x+15-x+y\right)\left(3x+15+x-y\right)\)

\(=\left(2x+y+15\right)\left(4x-y+15\right)\)

\(c,49\left(y-4\right)^2-9y^2-36y-36\)

\(=49\left(y-4\right)^2-\left(9y^2+36y+36\right)\)

\(=49\left(y-4\right)^2-\left(3y+6\right)^2\)

\(=\left[7\left(y-4\right)-3y-6\right]\left[7\left(y-4\right)+3y+6\right]\)

\(=\left(7y-28-3y-6\right)\left(7y-28+3y+6\right)\)

\(=\left(4y-34\right)\left(10y-22\right)\)

Dễ nhưng mà dài chết người 

Bài 1:

Ta có: \(x^2-2x+2=x^2-2x+1+1\)

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

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

Ta thấy rằng: \(\left(x-1\right)^2\ge0\) ( Với mọi \(x\in Z\) )

mà 1 > 0

=> \(\left(x-1\right)^2+1\ge0\)

<=> \(x^2-2x+1\ge0\)

Bài 3:

a) 53^2 + 47^2 + 94.53

= 53^2 + 47^2 + 2.47.53

= ( 53 + 47 )^2

= 100^2

= 10000

b) 50^2 - 49^2 + 48^2 - 47^2 + 2^2 - 1^2

= ( 50^2 - 49^2 ) + ( 48^2 - 47^2 ) + ( 2^2 - 1^2 )

= (50+49).(50-49) + (48+47).(48-47) + (2+1).(2-1)

= 50 + 49 + 48 + 47 + 2 + 1

= (49 + 1) + (48 + 2) + 50 + 47

= 50 + 50 + 50 + 47

= 197

Bài 1 : 

x²-2x+2>0 với mọi x

=x²-2.x.1/4+1/16+31/16

=(x-1/4)² + 31/16

Vì (x-1/4)² \(\ge\) 0 nên (x-1/4)² + 31/16 \(\ge\) 0 với mọi x (đfcm)

a/Dùng hằng đẳng thức A²-B²=(A+B)(A-B) phân tích được ngay

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

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

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

b/Chắc chỉ phân tích hằng đẳng thức (A-B)²=A²-2ab+B²

\(49\left(y-4\right)^2-9y^2-3y-36=49y^2-392y+784-9y^2-3y-36\)

\(=40y^2-395y+748\)

Mình dùng biệt thức cho ra nghiệm vô tỉ, không biết cho phải tại mình tính sai hay đề thiếu nữa

c/Khai triển biểu thức ban đầu ta được

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