a)a³+2a²-13a+10

Ta thấy a=1;a=2 là nghiệm của đa thức nên:

=(a-2)(a-1)(a+5)

b)(a²+4b²-5)²-16(ab+1)²

=(a²+4b²-5+4ab+4)(a²+4b²-5-4ab-4)

=[(a+2b)²-1][(a-2b)²-9]

=(a+2b+1)(a+2b-1)(a-2b+3)(a-2b-3)

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

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

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

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

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

\(\left(a^2+b^2-5\right)^2-4\left(ab+2\right)^2\)

\(=\left(a^2+b^2-5-2ab-4\right)\left(a^2+b^2-5+2ab+4\right)\)

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

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

Tham khảo nhé~

1) \(a^3+2a^2-13a+10=a^3-a^2+3a^2-3a-10a+10=\)

\(=a^2\left(a-1\right)+3a\left(a-1\right)-10\left(a-1\right)=\left(a-1\right)\left(a^2+3a-10\right)\)

\(=\left(a-1\right)\left(a^2-2a+5a-10\right)=\left(a-1\right)\left[a\left(a-2\right)+5\left(a-2\right)\right]=\)

\(=\left(a-1\right)\left(a-2\right)\left(a+5\right)\)

b) \(\left(a^2+4b^2-5\right)^2-16\left(ab+1\right)^2=\left(a^2+4b^2-5+4ab+4\right)\left(a^2+4b^2-5-4ab-4\right)\)

\(=\left(a^2+4ab+4b^2-1\right)\left(a^2-4ab+4b^2-9\right)=\left[\left(a+2b\right)^2-1\right]\left[\left(a-2b\right)^2-9\right]=\)

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

2) \(6a-5b=1\Rightarrow5b=6a-1\Rightarrow25b^2=36a^2-12a+1\)

\(\Rightarrow4a^2+25b^2=40a^2-12a+1=40\left(a^2-2\cdot a\cdot\frac{3}{20}+\left(\frac{3}{20}\right)^2\right)+1-\frac{9}{10}\)

\(=40\left(a-\frac{3}{20}\right)^2+\frac{1}{10}\)

Vậy GTNN của \(4a^2+25b^2\)= 1/10. Xảy ra khi a = 3/20 và b = -1/50.

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

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

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

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

b) \(\left(ax+by\right)^2-\left(ay+bx\right)^2\)

\(=\left(ax+by+ay+bx\right)\left(ax+by-ay-bx\right)\)

\(=\left(a+b\right)\left(x+y\right)\left(a-b\right)\left(x-y\right)\)

c) \(\left(a^2+b^2-5\right)^2-4\left(ab+2\right)^2\)

\(=\left(a^2+b^2-5+2ab+4\right)\left(a^2+b^2-5-2ab-4\right)\)

\(=\left[\left(a+b\right)^2-1\right]\left[\left(a-b\right)^2-9\right]\)

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

d) \(\left(4x^2-3x-18\right)^2-\left(4x^2+3x\right)^2\)

\(=\left(4x^2-3x-18+4x^2+3x\right)\left(4x^2-3x-18-4x^2-3x\right)\)

\(=\left(8x^2-18\right)\left(-6x-18\right)\)

\(=\left[2\left(4x^2-9\right)\right]\left[-6\left(x+3\right)\right]\)

\(=12\left(2x+3\right)\left(2x-3\right)\left(x+3\right)\)

(x - 5)² - 4(x - 3)² + 2(2x - 1)(x - 5) + (2x - 1)²

= [(x - 5)² + 2(2x - 1)(x - 5) + (2x - 1)²) - [2(x - 3)]²

= (x - 5 + 2x - 1)² - (2x - 6)²

= (3x - 6)² - (2x - 6)²

= (3x - 6 - 2x + 6)(3x - 6 + 2x - 6) = x(5x - 12)

( x - 5 )² - 4( x - 3 )² + 2( 2x - 1 )( x - 5 ) + ( 2x - 1 )²

= [ ( x - 5 )² + 2( 2x - 1 )( x - 5 ) + ( 2x - 1 )² ] - 2²( x - 3 )²

= ( x - 5 + 2x - 1 )² - ( 2x - 6 )²

= ( 3x - 6 )² - ( 2x - 6 )²

= ( 3x - 6 - 2x + 6 )( 3x - 6 + 2x - 6 )

= x( 5x - 12 )

\(\left(x-5\right)^2-4\left(x-3\right)^2+2\left(2x-1\right)\left(x-5\right)+\left(2x-1\right)^2\)

\(=\left(x-5\right)^2+2\left(2x-1\right)\left(x-5\right)+\left(2x-1\right)^2-4\left(x-3\right)^2\)

\(=\left(x-5+2x-1\right)^2-\left(2x-6\right)^2\)

\(=\left(3x-6\right)^2-\left(2x-6\right)^2\)

\(=\left[\left(3x-6\right)-\left(2x-6\right)\right].\left[\left(3x-6\right)+\left(2x-6\right)\right]\)

\(=\left(3x-6-2x+6\right)\left(3x-6+2x-6\right)\)

\(=\left(5x-12\right)x\)

a) 

(x-y+5)²-2.(x-y+5)+1

=(x-y+5-1)²

=(x-y+4)²

b)

(x²+4y²-5)²-16.(x².y²+2xy+1)

=(x²+4y²-5)²-[4.(xy+1)]²

=(x²+4y²-5-4xy-4)(x²+4y²-5+4xy+4)

=(x²+4y²-4xy-9)(x²+4y²+4xy-1)

=[(x-2y)²-9][(x+2y)²-1]

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

=(x²+x-3x-3)((x-2y+3)(x+2y-1)(x+1)²

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

=(x+1)(x-3)(x-2y+3)(x+2y-1)(x+1)²

a) \(100x^2-\left(x^2+25\right)^2=\left(10x\right)^2-\left(x^2+25\right)^2=\left(10x-x^2-25\right)\left(10x+x^2+25\right)\)

\(=-\left(x-5\right)^2\left(x+5\right)^2\)

b) \(\left(x-y+5\right)^2-2\left(x-y+5\right)+1=\left(x-y+5-1\right)^2=\left(x-y+4\right)^2\)

c) \(\left(x^2+4y^2-5\right)^2-16\left(x^2+y^2+2xy+1\right)\)

Có lẽ bạn ghi sai đề rồi nha. 

\(\left(ax+by\right)^2-\left(ay+bx\right)^2\)

\(=\left(ax+by+ay+bx\right)\left(ax+by-ay-bx\right)\)

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

\(=\left(a+b\right)\left(a-b\right)\left(x+y\right)\left(x-y\right)\)

\(\left(a^2+b^2-5\right)^2-4\left(ab+2\right)^2\)

\(=\left[\left(a^2+b^2-5\right)+2\left(ab+2\right)\right]\left[\left(a^2+b^2-5\right)-2\left(ab+2\right)\right]\)

\(=\left[a^2+b^2-5+2ab+4\right]\left[a^2+b^2-5-2ab-4\right]\)

\(=\left[\left(a+b\right)^2-1\right]\left[\left(a-b\right)^2-9\right]\)

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

a)

(ax+by)² - (ay+bx)²

=(ax+by-ay-bx)(ax+by+ay+bx)

=[ a(x-y) -b(x-y)][ a(x+y) + b(x+y)]

=(a-b)(x-y)(a+b)(x+y)

b)(a²+b²-5)² - 4(ab+2)²

=(a²+b²-5-2ab-4)(a²+b²-5+2ab+4)

=[ (a-b)² -9][ (a+b)² -1]

=(a-b-3)(a-b+3)(a+b-1)(a+b+1)

a, \(\left(ax+by\right)^2-\left(ay+bx\right)^2\)

\(=\left(ax+ay+bx+by\right)\left(ax-ay+bx-by\right)\)

\(=\left(a+b\right)\left(x+y\right)\left(a-b\right)\left(x-y\right)\)

b, \(\left(a^2+b^2-5\right)^2-4\left(ab+2\right)^2\)

\(=\left(a^2+b^2-5-2ab-4\right)\left(a^2+b^2-5+2ab+4\right)\)

\(=\left[\left(a-b\right)^2-9\right]\left[\left(a+b\right)^2-1\right]\)

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