g) (x+2)(x+3)(x+4)(x+5)-24 = \(\left[\left(x+2\right)\left(x+5\right)\right]\left[\left(x+3\right)\left(x+4\right)\right]-24\)

                                       =\(\left[x^2+7x+10\right]\left[x^2+7x+12\right]\)

đặt \(x^2+7x+10=a\)

ta có \(a\left(a+2\right)-24=a^2+2a-24\)

                                      \(=a^2+2a+1-25\)

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

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

                                     \(=\left(a-4\right)\left(a+6\right)\)

\(\Rightarrow\) \(\left(x^2+7x+10-4\right)\left(x^2+7x+10+6\right)=\left(x^2+7x+6\right)\left(x^2+7x+16\right)\)

a) = (x +5)² - 2² = (x+5 -2)(x+5 +2) = (x+3)(x+7)

b) = x(x² -1) -6(x-1)= x(x+1)(x-1) -6(x-1) = (x-1)(x(x+1)-6)

a) \(x^2+10x+21=x^2+3x+7x+21\)

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

từ ý b đến ý e dùng chức lăng EQN của máy tính

bạn mở mt bấm mode rồi ấn 5 rồi tìm cái có ax³+bx²+cx+d

b) nhập 1 = ; 0 =;-7=;6= rồi ấn = sao cho hết nghiệm

có \(X_1=2;X_2=1;X_3=-3\)

\(1,=\left(x-3\right)\left(x+3\right)\\ 2,=\left(x-y\right)\left(5+a\right)\\ 3,=\left(x+3\right)^2\\ 4,=\left(x-y\right)\left(10x+7y\right)\\ 5,=5\left(x-3y\right)\\ 6,=\left(x-y\right)^2-z^2=\left(x-y-z\right)\left(x-y+z\right)\)

bạn gõ lại công thức cho rõ đi, khó đọc quá

a)\(x^2+10x+25-y^2\)

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

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

b)\(5x^3-7x^2+10x-14\)

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

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

c)\(-5y^2+30y-45\)

\(=-5\left(y^2-6y+9\right)\)

\(=-5\left(y-3\right)^2\)

e)\(4xy^2-8xyz+4xz^2\)

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

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

f)\(x^2+7x+10\)

\(=x^2+5x+2x+10\)

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

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

k)\(2x^7+6x^6+6x^5-2x^4\)

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

a)\(x^2+10x+25-y^2\)

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

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

b)\(5x^3-7x^2+10x-14\)

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

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

c)\(-5y^2+30y-45\)

\(=-5\left(y^2-6y+9\right)\)

\(=-5\left(y-3\right)^2\)

e)\(4xy^2-8xyz+4xz^2\)

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

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

f)\(x^2+7x+10\)

\(=x^2+5x+2x+10\)

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

k)\(2x^7+6x^6+6x^5-2x^4\)

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

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

làm lại

 = (x^4-4x^3)+(3x^3-12x^2)+(2x^2-8x)-(2x-8)

 = x^3.(x-4)+3x^2.(x-4)+2x.(x-4)-2.(x-4)

 = (x-4).(x^3+3x^2+2x-2)

Tk mk nha

Đặt x²+10x+5=a.Ta có biểu thức là : a(a+8)+16=a²+8a+16=(a+4)²=(x²+10x+9)²

a) xy+3x-7y-21

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

=(x-7)(y+3)

b)2xy-15-6x-5y

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

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

c)2x^2y+2xy^2-2x-2y

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

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

x(x+3)-5x(x-5)-5(x+3)

=(x-5)(x+3)-5x(x-5)

=(x-5)(x+3-5x)

Câu cuối mình bị nhầm dòng cuối phải là (x-5)(x+3+x-5)=(x-5)(2x-2)nha bạn

a) xy+3x-7y-21=(xy+3x)-(7y+21)= x(y+3)-7(y+3)=(y+3)(x-7)

b)2xy-15-6x+5y=(2xy-6x)+(5y-15)=2x(y-3)+5(y-3)=(y-3)(2x+5)

c)2x^2y+2xy^2-2x-2y=2xy(x+y)-2(x+y)=2(x+y)(xy-1)

d) x(x+3)-5x(x-5)-5(x+3)=[x(x+3)-5(x+3)]-5x(x-5)=(x+3)(x-5)-5x(x-5)=(x-5)(x+3-5x)=(x-5)(3-4x)

a,2x²-7x+6=(2x²-4x)-(3x-6)
=2x(x-3)-3(x-2)=(x-2)(2x-3)
b,x²+x-6=(x²+3x)-(2x+6)
=x(x-3)-2(x-3)=(x-3)(x-2)
c,x³+3x²+6x+4=x³+x²+2x²+2x+4x+4
=(x+1)(x²+2x+4)
d,x¹⁰+x⁵+1=(x¹⁰-x)+(x⁵-x²)+(x²+x+1)
=x((x³)³-1)+x²(x³-1)+(x²+x+1)
=x(x³-1)(x⁶+x³+1)+x²(x-1)(x²+x+1)+(x²+x+1)
=x(x-1)(x²+x+1)+x²(x-1)(x²+x+1)+(x²+x+1)
(x²+x+1)(x²-x+x³-x²+1)
e,(12x²-12xy+3y²)-10x(2x-y)=3(4x²-4xy+y²)-10x(2x-y)
=3(2x-y)²-10x(2x-y)=(2x-y)(6x-3y-10x)=(2x-y)(-4x-3y)

phân tích đa thức thành nhân tử

a,2x^2-7x+6
b,x^2+x-6
c,x^3+3x^2+6x+4
d,x^10+x^5+1
e,(12x^2-12xy+3y^2)-10x(2x-y)

\(2x^2-7x+6\)

\(=2x^2-3x-4x+6\)

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

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

a)  \(x^2\)\(+\)\(6x\)\(+\)\(9\)

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

b)  \(x^3\)\(+\)\(3x^2\)\(+\)\(3x\)\(+\)\(1\)

\(=\left(x+1\right)^3\)

c)  \(8x^3\)\(-\)\(\frac{1}{8}\)

\(=\left(2x-\frac{1}{2}\right)\left(4x^2+x+\frac{1}{4}\right)\)

d)  \(10x\)\(-\)\(25\)\(-\)\(x^2\)

\(=\)\(-x^2\)\(+\)\(10\)\(-\)\(25\)

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

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

e)  \(\frac{1}{25}x^2\)\(-\)\(64y^2\)

=\(\left(\frac{1}{25}x-8y\right)\left(\frac{1}{5}x+8y\right)\)