b: \(4m^2n-8m^2n-12n\)

\(=-4m^2n-12n\)

\(=-4n\left(m^2+3\right)\)

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

d: \(16b^4-9c^2\)

\(=\left(4b^2-3c\right)\left(4b^2+3c\right)\)

e: \(5a\left(x+y\right)+x+y=\left(x+y\right)\left(5a+1\right)\)

\(\left(2x+5\right)^2-6x-15=\left(2x+5\right)^2-3\left(2x-5\right)=\left(2x-5\right)^2-3\left(2x-5\right)\)

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

(2x+5)^2-6x-15

=4x²+20x+25-6x-15

=4x²+14x+10

=(2x+1)²+9        (đề bài có nhầm không)

ღ๖ۣۜLinh's ๖ۣۜLinh'sღ] ★we are one★ hình như bạn làm sai rồi thì phải

đề là phân tích thành nhân tử kia mà

x^2 + 7x -15

= x^2 + 7x +12,25  -27,25

= (x+3,5)^2 - 27, 25

= ( x+3,5 - \(\sqrt{27,25}\))(x+3,5+\(\sqrt{27,25}\))

\(3x-7\sqrt{x}+4=3x-4\sqrt{x}-3\sqrt{x}+4=\sqrt{x}\left(3\sqrt{x}-4\right)-\left(3\sqrt{x}-4\right)=\left(3\sqrt{x}-4\right)\left(\sqrt{x}-1\right)\)

\(3x-7\sqrt{x}+4=3x-3\sqrt{x}-4\sqrt{x}+4\)

\(=3\sqrt{x}\left(\sqrt{x}-1\right)-4\left(\sqrt{x}-1\right)\)

\(=\left(3\sqrt{x}-4\right)\cdot\left(\sqrt{x}-1\right)\)

\(3x-7\sqrt{x}+4=3x-3\sqrt{x}-4\sqrt{x}+4=3.\left(\sqrt{x}\right)^2-3\sqrt{x}-4\sqrt{x}+4=3\sqrt{x}\left(\sqrt{x}-1\right)-4\left(\sqrt{x}-1\right)=\left(3\sqrt{x}-4\right)\left(\sqrt{x}-1\right)\)

\(x^2-4x+4-y^2\)

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

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

(x - 2)² - y² = (x - 2 - y)(x - 2 + y)

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

\(x^2-4x+4-y^2\)

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

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

(x - 2)² - y² = (x - 2 - y)(x - 2 + y)

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

(x² + 2.x.3 + 3² - 1).(x² + 2.x.4 + 16 - 1) - 24

=[(x+3)² - 1]. [(x+4)²-1] -24

=(x+3+1)(x+3-1)(x+4+1)(x+4-1) - 24

=(x+4)(x+2)(x+5)(x-3) - 24

(x²+6x+8)(x²+8x+15)-24

<=>(x²+4x+2x+8)(x²+5x+3x+15)-24

<=> [x(x+4)+2(x+4)][x(x+5)+3(x+5)]-24

<=> (x+4)(x+2)(x+5)(x+3)-24

<=> (x+4)(x+3)(x+2)(x+5)-24

<=>(x²+7x+12)(x²+7x+10)

đặt t=x²+7x+11 ta có:

(t-1)(t+1)-24

<=> t²-1-24

<=>t²-25

<=>(t-5)(t+5)

thay t=x²+7x+11 vào ta có:

(x²+7x+11-5)(x²+7x+11+5)

<=>(x²+7x+6)(x²+7x+16)