Phân tích đa thức thành nhân tử
A. a2 - 2x - 4b2 - 4b
B. (1 + 2x)(1 - 2x) - (x + 2)(x - 2)
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\(a,a^2-2a-4b^2-4b\)
\(=\left(a^2-4b^2\right)-\left(2a+4b\right)\)
\(=\left(a-2b\right)\left(a+2b\right)-2\left(a+2b\right)\)
\(=\left(a+2b\right)\left(a-2b-2\right)\)
\(b,x^3-2x^2+4x-8\)
\(=x^2\left(x-2\right)+4\left(x-2\right)\)
\(=\left(x-2\right)\left(x^2+4\right)\)
\(c,x^3+36x-12x^2\)
\(=x^3-6x^2-6x^2+36x\)
\(=x^2\left(x-6\right)-6x\left(x-6\right)\)
\(=\left(x-6\right)\left(x^2-6x\right)\)
\(=x\left(x-6\right)^2\)
\(d,5a^2+3\left(a+b\right)^2-5b^2\)
\(=\left(5a^2-5b^2\right)+3\left(a+b\right)^2\)
\(=5\left(a^2-b^2\right)+3\left(a+b\right)^2\)
\(=5\left(a-b\right)\left(a+b\right)+3\left(a+b\right)^2\)
\(=\left(a+b\right)\left[5\left(a-b\right)+3\left(a+b\right)\right]\)
\(=\left(a+b\right)\left(5a-5b+3a+3b\right)\)
\(=\left(a+b\right)\left(8a-2b\right)\)
\(=2\left(a+b\right)\left(4a-b\right)\)
\(e,x^3-3x^2+3x-1-y^3\)
\(=\left(x^3-3x^2+3x-1\right)-y^3\)
\(=\left(x-1\right)^3-y^3\)
\(=\left(x-1-y\right)\left[\left(x-1\right)^2+\left(x-1\right)y+y^2\right]\)
\(=\left(x-y-1\right)\left(x^2-2x+1+xy-y+y^2\right)\)
\(=\left(x-y-1\right)\left(x^2+y^2-xy-y+1\right)\)
#Urushi☕
\(c.\\ x^3+36x-12x^2\\ =x\left(x^2-12x+36\right)\\ =x.\left(x^2-2.x.6+6^2\right)\\ =x.\left(x-6\right)^2\\ ---\\ d.\\ 5a^2+3\left(a+b\right)^2-5b^2\\ =\left(5a^2-5b^2\right)+3\left(a+b\right)^2\\ =5.\left(a^2-b^2\right)+3.\left(a+b\right)\left(a+b\right)\\ =5\left(a+b\right)\left(a-b\right)+3\left(a+b\right)\left(a+b\right)\\ =\left(a+b\right)\left(5a-5b+3a+3b\right)\\ =\left(a+b\right)\left(8a-2b\right)\\ =2\left(a+b\right)\left(4a-b\right)\)
\(e.\\ x^3-3x^2+3x-1-y^3\\ =\left(x-1\right)^3-y^3\\ =\left(x-1-y\right)\left[\left(x-1\right)^2+\left(x-1\right).y+y^2\right]\\ =\left(x-y-1\right).\left[\left(x^2-2x+1\right)+y\left(x+y-1\right)\right]\)
Bài 1:
a: Ta có: \(\left(6x+3\right)-\left(2x-5\right)\left(2x+1\right)\)
\(=\left(2x+1\right)\left(3-2x+5\right)\)
\(=\left(2x+1\right)\left(8-2x\right)\)
\(=2\left(4-x\right)\left(2x+1\right)\)
b) Ta có: \(\left(3x-2\right)\left(4x-3\right)-\left(2-3x\right)\left(x-1\right)-2\left(3x-2\right)\left(x+1\right)\)
\(=\left(3x-2\right)\left(4x-3\right)+\left(3x-2\right)\left(x-1\right)-\left(3x-2\right)\left(2x+2\right)\)
\(=\left(3x-2\right)\left(4x-3+x-1-2x-2\right)\)
\(=\left(3x-2\right)\left(3x-6\right)\)
\(=3\left(3x-2\right)\left(x-2\right)\)
Bài 2:
a: Ta có: \(\left(a-b\right)\left(a+2b\right)-\left(b-a\right)\left(2a-b\right)-\left(a-b\right)\left(a+3b\right)\)
\(=\left(a-b\right)\left(a+2b\right)+\left(a-b\right)\left(2a-b\right)-\left(a-b\right)\left(a+3b\right)\)
\(=\left(a-b\right)\left(a+2b+2a-b-a-3b\right)\)
\(=\left(a-b\right)\left(2a-4b\right)\)
\(=2\left(a-b\right)\left(a-2b\right)\)
f: Ta có: \(x^2-6xy+9y^2+4x-12y\)
\(=\left(x-3y\right)^2+4\left(x-3y\right)\)
\(=\left(x-3y\right)\left(x-3y+4\right)\)
1: \(a^2-4b^2-2a-4b\)
\(=\left(a-2b\right)\left(a+2b\right)-2\left(a+2b\right)\)
\(=\left(a+2b\right)\left(a-2b-2\right)\)
2: \(x^3+2x^2-2x-1\)
\(=\left(x-1\right)\left(x^2+x+1\right)+2x\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2+3x+1\right)\)
a) x2-xy+5y-25
= x(2-y)+ 5(y-2)
= x(2-y)-5(2-y)
= (x-5)(2-y)
a) \(8x^3+27=\left(2x+3\right)\left(4x^2-6x+9\right)\)
b) \(4x^2-4x+1-y^2=\left(2x-1\right)^2-y^2=\left(2x-1-y\right)\left(2x-1+y\right)\)
c) \(x^4-2x^3+x^2-2x=x^3\left(x-2\right)+x\left(x-2\right)=x\left(x-2\right)\left(x^2-1\right)=x\left(x-2\right)\left(x-1\right)\left(x+1\right)\)
d) \(x^2-4y^2+2x+4y=\left(x-2y\right)\left(x+2y\right)+2\left(x+2y\right)=\left(x+2y\right)\left(x-2y+2\right)\)
\(a,x^3+x^2-4x-4=x^2\left(x+1\right)-4\left(x+1\right)=\left(x^2-4\right)\left(x+1\right)=\left(x-2\right)\left(x+2\right)\left(x+1\right)\\ b,x^2-2x-15=\left(x^2-5x\right)+\left(3x-15\right)=x\left(x-5\right)+3\left(x-5\right)=\left(x+3\right)\left(x-5\right)\\ c,x^2-4+\left(x-2\right)^2=\left(x-2\right)\left(x+2\right)+\left(x-2\right)^2=\left(x-2\right)\left(x+2+x-2\right)=2x\left(x-2\right)\)
\(d,x^3-2x^2+x-xy^2=x\left(x^2-2x+1-y^2\right)=x\left[\left(x-1\right)^2-y^2\right]=x\left(x-y-1\right)\left(x+y-1\right)\)
Bài 2
a) 5x² + 30y
= 5(x² + 6y)
b) x³ - 2x² - 4xy² + x
= x(x² - 2x - 4y² + 1)
= x[(x² - 2x + 1) - 4y²]
= x[(x - 1)² - (2y)²]
= x(x - 1 - 2y)(x - 1 + 2y)
Bài 3:
a: \(2x\left(x-3\right)-x+3=0\)
=>\(2x\left(x-3\right)-\left(x-3\right)=0\)
=>(x-3)(2x-1)=0
=>\(\left[{}\begin{matrix}x-3=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{2}\end{matrix}\right.\)
b: \(\left(3x-1\right)\left(2x+1\right)-\left(x+1\right)^2=5x^2\)
=>\(6x^2+3x-2x-1-x^2-2x-1=5x^2\)
=>\(5x^2-x-2=5x^2\)
=>-x-2=0
=>-x=2
=>x=-2
câu a sai đề ko bạn