Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) \(x^2+5x-6\)
\(=x^2-2x+3x-6\\ =\left(x^2-2x\right)+\left(3x-6\right)\\ =x\left(x-2\right)+3\left(x-2\right)\\ =\left(x-2\right)\left(x+3\right)\)
b) \(5x^2+5xy-x-y\)
\(=\left(5x^2+5xy\right)-\left(x+y\right)\\ =5x\left(x+y\right)-\left(x+y\right)\\ =\left(x+y\right)\left(5x+1\right)\)
c)\(7x-6x^2-2\)
\(=3x+4x-6x^2-2\\ =\left(3x-6x^2\right)+\left(4x-2\right)\\ =3x\left(1-2x\right)+2\left(2x-1\right)\\ =3x\left(1-2x\right)-2\left(1-2x\right)\\ =\left(1-2x\right)\left(3x-2\right)\)
a ) \(x^2+5x+6\)
\(=x^2+5x+\frac{25}{4}-\frac{1}{4}\)
\(=\left(x+\frac{5}{2}\right)^2-\frac{1}{4}\)
b ) \(x^2\left(1-x^2\right)-4+4x^2\)
\(=x^2\left(1-x^2\right)-4\left(1-x^2\right)\)
\(=\left(x^2-4\right)\left(1-x^2\right)\)
\(=\left(x-2\right)\left(x+2\right)\left(1-x\right)\left(1+x\right)\)
a) \(x^2+5x+6\\ =x^2+5x+\frac{25}{4}-\frac{1}{4}\\ =\left(x+\frac{5}{2}\right)^2-\frac{1}{4}\\ \)
b) \(x^2\left(1-x^2\right)-4+4x^2\\ =x^2\left(1-x^2\right)-4\left(1-x^2\right)\\ =\left(x^2-4\right)\left(1-x^2\right)\\ =\left(x-2\right)\left(x+2\right)\left(1-x\right)\left(1+x\right)\)
a/ \(x^2+5x+6\)
\(=x^2+5x+\frac{25}{4}-\frac{1}{4}\)
\(=\left(x+\frac{5}{2}\right)^2-\frac{1}{4}\)
\(=\left(x+3\right)\left(x+2\right)\)
b/ \(x^2\left(1-x^2\right)-4+4x^2\)
\(=x^2\left(1-x^2\right)-4\left(1-x^2\right)\)
\(=\left(x^2-4\right)\left(1-x^2\right)\)
\(=\left(x-2\right)\left(x+2\right)\left(1-x\right)\left(1-x\right)\)
a ) \(x^3-3x^2-4x+12\)
\(=x^2\left(x-3\right)-4\left(x-3\right)\)
\(=\left(x^2-4\right)\left(x-3\right)\)
\(=\left(x-3\right)\left(x-2\right)\left(x+2\right)\)
b ) \(x^4-5x^2+4\)
\(=x^4-4x^2-x^2+4\)
\(=x^2\left(x^2-4\right)-\left(x^2-4\right)\)
\(\Leftrightarrow\left(x^2-1\right)\left(x^2-4\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(x-2\right)\left(x+2\right)\)
\(=\left(x-2\right)\left(x-1\right)\left(x+1\right)\left(x+2\right)\)
a) x2 – 4x + 3 = x2 – x - 3x + 3
= x(x - 1) - 3(x - 1) = (x -1)(x - 3)
b) x2 + 5x + 4 = x2 + 4x + x + 4
= x(x + 4) + (x + 4)
= (x + 4)(x + 1)
c) x2 – x – 6 = x2 +2x – 3x – 6
= x(x + 2) - 3(x + 2)
= (x + 2)(x - 3)
d) x4+ 4 = x4 + 4x2 + 4 – 4x2
= (x2 + 2)2 – (2x)2
= (x2 + 2 – 2x)(x2 + 2 + 2x)
Bài giải:
a) x2 – 4x + 3 = x2 – x - 3x + 3
= x(x - 1) - 3(x - 1) = (x -1)(x - 3)
b) x2 + 5x + 4 = x2 + 4x + x + 4
= x(x + 4) + (x + 4)
= (x + 4)(x + 1)
c) x2 – x – 6 = x2 +2x – 3x – 6
= x(x + 2) - 3(x + 2)
= (x + 2)(x - 3)
d) x4+ 4 = x4 + 4x2 + 4 – 4x2
= (x2 + 2)2 – (2x)2
= (x2 + 2 – 2x)(x2 + 2 + 2x)
a,
\(x^2+5x+6=x^2+2x+3x+6=x\left(x+2\right)+3\left(x+2\right)=\left(x+2\right)\left(x+3\right)\)
b,
\(3x^2-7x+2=3x^2-x-6x+2=x\left(3x-1\right)-2\left(3x-1\right)=\left(3x-1\right)\left(x-2\right)\)
c,
\(a^3+b^3+c^3-3abc=\left(a+b\right)^3-3ab\left(a+b+c\right)+c^3\)
\(=\left(a+b+c\right)\left(\left(a+b\right)^2-\left(a+b\right)c+c^2\right)-3ab\left(a+b+c\right)\)
=)
a) \(x^2+5x+6\)
\(=x^2+2x+3x+6\)
\(=x\left(x+2\right)+3\left(x+2\right)\)
\(=\left(x+3\right)\left(x+2\right)\)
b) \(3x^2-7x+2\)
\(=3x^2-x-6x+2\)
\(=x\left(3x-1\right)-2\left(3x-1\right)\)
\(=\left(x-2\right)\left(3x-1\right)\)
c) Phân tích thành nhân tử $a^3 + b^3 + c^3 - 3abc$ - Đại số - Diễn đàn Toán học
\(x^2+5x-6=x^2-6x+x-6=x\left(x-6\right)+\left(x-6\right)=\left(x+1\right)\left(x-6\right)\)
\(5x^2+5xy-x-y=5x\left(x+y\right)-\left(x+y\right)=\left(5x-1\right)\left(x+y\right)\)
\(7x-6x^2-2=-6x^2+7x-2=-6\left(x^2-\frac{7}{6}x+\frac{1}{3}\right)=-6\left(x^2-\frac{7}{6}x+\frac{49}{144}-\frac{1}{144}\right)=-6\left[\left(x-\frac{7}{12}\right)^2-\frac{1}{144}\right]\)
1) x2-x+6x-6 = x(x-1)+6(x-1)=(x+6)(x-1)
2) 5x(x+y)-(x+y) =(5x-1)(x+y)
3) -6x2+7x-2 = -6x2+3x+4x-2= -6x(x-\(\frac{1}{2}\)) +4(x-\(\frac{1}{2}\)) =(-3x+2)(2x-1)
Ta có : \(4x^2-3x-1\)
\(=4x^2-4x+x-1\)
\(=4x\left(x-1\right)+\left(x-1\right)\)
\(=\left(x-1\right)\left(4x+1\right)\)
Ta có : \(x^2-7x+12\)
\(=x^2-3x-4x+12\)
\(=x\left(x-3\right)-\left(4x-12\right)\)
\(=x\left(x-3\right)-4\left(x-3\right)\)
\(=\left(x-4\right)\left(x-3\right)\)
a) \(x^2+5x+6=x^2+2x+3x+6=x\left(x+2\right)+3\left(x+2\right)=\left(x+3\right)\left(x+2\right)\)
b)