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a, \(3x^2-9x-2x+6=3x\left(x-3\right)-2\left(x-3\right)=\left(x-3\right)\left(3x-2\right)\)
b. \(8x^2-2x+12x-3=2x\left(4x-1\right)+3\left(4x-1\right)=\left(4x-1\right)\left(2x+3\right)\)
c. đề kiểu gì vậy? -2x-x để thành -3x à? xem lại đi nha
d. \(\left(x^2+10x+25\right)-\left(y^2+6y+9\right)=\left(x+5\right)^2-\left(y+3\right)^2=\left(x+5-y-3\right)\left(x+5+y+3\right)=\left(x-y+2\right)\left(x+y+8\right)\)
e. \(=x^4+2x^2y^2+y^4-x^2y^2=\left(x^2+y^2\right)^2-x^2y^2=\left(x^2+y^2-xy\right)\left(x^2+y^2+xy\right)\)
nhớ L I K E
a: \(=6x^3-12x^2+x^2-2x+x-2\)
\(=\left(x-2\right)\left(6x^2+x+1\right)\)
b: \(=3x^4+3x^3-x^3-x^2-7x^2-7x+5x+5\)
\(=\left(x+1\right)\left(3x^3-x^2-7x+5\right)\)
\(=\left(x+1\right)\left(3x^3-3x^2+2x^2-2x-5x+5\right)\)
\(=\left(x+1\right)\left(x-1\right)\left(3x^2+2x-5\right)\)
\(=\left(x-1\right)^2\cdot\left(x+1\right)\left(3x+5\right)\)
c: \(=4x^3+x^2+4x^2+x+4x+1\)
\(=\left(4x+1\right)\left(x^2+x+1\right)\)
\(a,3x^2-11x+6=3x^2-9x-2x+6=3x\left(x-3\right)-2\left(x-3\right)=\left(3x-2\right)\left(x-3\right)\)
\(b,8x^2+10x-3=8x^2+12x-2x-3=4x\left(2x+3\right)-\left(2x+3\right)=\left(4x-1\right)\left(2x+3\right)\)
\(c,8x^2-2x-1=9x^2-x^2-2x-1=9x^2-\left(x+1\right)^2=\left(3x-x-1\right)\left(3x+x+1\right)\)
\(=\left(2x-1\right)\left(4x+1\right)\)
a) 8x2-2x-1=8x2+4x-2x-2x-1=(8x2+4x)-(2x+1)=4x.(2x+1)-(2x+1)=(2x+1).(4x-1)
b) x2-y2+10x-6y+16x=x2-y2+10x+10y-16y+16x=(x-y)(x+y)+10(x+y)-16(y+x)=(x+y)(x-y+10+16)
phân tích đa thức ->nhân tử:
a)2x2+4x-70
b)x3-5x2+8x-4
c)x2-10+16
rút gọn:
(8x-8x3-10x2+3x4-5):(3x2-2x+1)
Bài 1:
a)2x2+4x-70
=2(x2+2x-35)
=2(x2+7x-5x-35)
=2[x(x+7)-5(x+7)]
=2(x-5)(x+7)
b)x3-5x2+8x-4
=x3-4x2+4x-x2+4x-4
=x(x2-4x+4)-(x2-4x+4)
=(x2-4x+4)(x-1)
=(x-2)2(x-1)
c)x2-10x+16
=x2-2x-8x+16
=x(x-2)-8(x-2)
=(x-8)(x-2)
Bài 2:
\(\frac{8x-8x^3-10x^2+3x^4-5}{3x^2-2x+1}=\frac{\left(x^2-2x-5\right)\left(3x^2-2x+1\right)}{3x^2-2x+1}=x^2-2x-5\)
Bài 1:tìm x ,biết:
a) (2x - 1)(3x + 2) - 6x(x + 1) = 0
\(\Leftrightarrow6x^2+x-2-6x^2-6x=0\)
\(\Leftrightarrow-5x=2\)
\(\Leftrightarrow x=\frac{-2}{5}\)
b) \(\left(4x-1\right)^2-\left(2x+1\right)\left(8x-3\right)=0\)
\(\Leftrightarrow16x^2-8x+1-16x^2-2x+3=0\)
\(\Leftrightarrow-10x=-4\)
\(\Leftrightarrow x=\frac{2}{5}\)
c) \(4x^2-1=2\left(2x+1\right)\)
\(\Leftrightarrow\left(2x+1\right)\left(2x-1\right)-2\left(2x+1\right)=0\)
\(\Leftrightarrow\left(2x+1\right)\left(2x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\x=\frac{3}{2}\end{cases}}\)
2a) \(4x^2-9y^2-6y-1=4x^2-\left(3y+1\right)^2\)
\(=\left(2x-3y-1\right)\left(2x+3y+1\right)\)
b) \(4x^2-1-2x\left(2x-1\right)=\left(2x-1\right)\left(2x+1\right)-2x\left(2x-1\right)\)
\(=1.\left(2x-1\right)\)
c) \(x^2-8x-4y^2+16=\left(x-4\right)^2-4y^2\)
\(=\left(x-4-2y\right)\left(x-4+2y\right)\)
d) \(9x^2-12x-y^2+4=\left(3x-2\right)^2-y^2\)
\(=\left(3x-2-y\right)\left(3x-2+y\right)\)
e) \(4x^2+10x-5=4x^2+2.2.\frac{5}{2}x+\frac{25}{4}-\frac{25}{4}-5\)
\(=\left(2x+\frac{5}{2}\right)^2-\frac{45}{4}\)
\(=\left(2x+\frac{5+3\sqrt{5}}{2}\right)\left(2x+\frac{5-3\sqrt{5}}{2}\right)\)
\(1,4x^4+4x^2y^2-8y^4\)
\(=4\left(x^4+x^2y^2-y^4-y^4\right)\)
\(=4\left[\left(x^4-y^4\right)+\left(x^2y^2-y^4\right)\right]\)
\(=4\left[\left(x^2+y^2\right)\left(x^2-y^2\right)+y^2\left(x^2-y^2\right)\right]\)
\(=4\left(x^2-y^2\right)\left(x^2+y^2+y^2\right)\)
\(=4\left(x-y\right)\left(x+y\right)\left(x^2+2y^2\right)\)
\(2,12x^2y-18xy^2-30y^3\)
\(=6y\left(2x^2-3xy-5y^2\right)\)
\(=6y\left[\left(2x^2+2xy\right)-\left(5xy+5y^2\right)\right]\)
\(=6y\left[2x\left(x+y\right)-5y\left(x+y\right)\right]\)
\(=6y\left(x+y\right)\left(2x-5y\right)\)
a) 8x2 - 2x - 1
=8x2+2x-4x-1
=2x.(4x+1)-(4x+1)
=(4x+1)(2x-1)
b) x2 - y2 + 10x - 6y + 16
=x2+10x+25-y2-6y-9
=(x+5)2-(y+3)2
=(x+5-y-3)(x+5+y+3)
=(x-y+2)(x+y+8)
a.\(3x^2-11x+6\)
= \(3x^2-9x-2x+6\)
=\(3x\left(x-3\right)-2\left(x-3\right)\)
=\(\left(x-3\right)\left(3x-2\right)\)
b\(8x^2+10x-3\)
=.\(8x^2-2x+12x-3\)
=\(2x\left(4x-1\right)+3\left(4x-1\right)\)
=\(\left(4x-1\right)\left(2x+3\right)\)
d.\(x^2-y^2+10x-6y+16\)
=\(\left(x^2+10x+25\right)-\left(y^2+6y+9\right)\)
=\(\left(x+5\right)^2-\left(y+3\right)^2\)
=\(\left(x+5-y-3\right)\left(x+5+y+3\right)\)
=\(\left(x-y+2\right)\left(x+y+8\right)\)
e.\(x^4+x^2y^2+y^4\)
=\(x^4+2x^2y^2+y^4-x^2+y^2\)
=\(\left(x^2+y^2\right)^2-x^2y^2\)
=\(\left(x^2+y^2-xy\right)\left(x^2+y^2+xy\right)\)
a)
\(=3x^2-9x-2x+6=3x\left(x-3\right)-2\left(x-3\right)=\left(x-3\right)\left(3x-2\right)\)