\(2x^2-7x+3\)

\(=2\left(x^2-\frac{7}{2}x+\frac{3}{2}\right)\)

Vậy thôi đâu cần dùng HĐT

2x²-7x+3

=2x²-x-6x+3

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

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

áp dụng hằng đẳng thức 7 và 3 nha dễ dàng mà . bn tách hết cho về mũ 3 hoặc mũ 2 nha 

somebody help me 

\(1,2x^2-3x-2\) 

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

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

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

\(2,4x^2-7x-2\)

\(=4x^2-8x+x-2\) 

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

\(\left(4x+1\right)\left(x-2\right)\)

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

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

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

=a(a-4b)

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

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

d: \(\left(a+3b\right)^2-9b^2\)

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

=(a+3b-3b)(a+3b+3b)=a(a+6b)

e: \(\left(x-3\right)^3-27\)

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

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

f: \(\left(x+1\right)^3-125\)

\(=\left(x+1-5\right)\left\lbrack\left(x+1\right)^2+5\left(x+1\right)+25\right\rbrack\)

\(=\left(x-4\right)\left(x^2+2x+1+5x+5+25\right)=\left(x-4\right)\left(x^2+7x+31\right)\)

dấu ^ là mũ nha mn

1: =3(y^2+x^2+2xy-z^2)

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

2: =2(8x^3+27y^3)

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

3: =(x-y)^2-25

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

\(\left(6x-1\right)^2-\left(3x+2\right)\)

\(=36x^2-12x+1-3x-2\)

\(=36x^2-15x-1\)

bn ktra lại đề nhé

hk tốt

16x³+54y³

= 16.(x³+4y³)

\(16x^3+54y^3\)

\(=2\left(8x^3-27y^3\right)\)

\(=2.\left(2x-3y\right)\left(4x^2-6xy+9y^2\right)\)