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1, \(x^2\left(x-3\right)-4x+12=x^2\left(x-3\right)-4\left(x-3\right)\)
\(=\left(x-2\right)\left(x+2\right)\left(x-3\right)\)
2, \(2a\left(x+y\right)-x-y=2a\left(x+y\right)-\left(x+y\right)=\left(2a-1\right)\left(x+y\right)\)
3, \(2x-4+5x^2-10x=2\left(x-2\right)+5x\left(x-2\right)=\left(2+5x\right)\left(x-2\right)\)
4, sửa đề :
\(6x^2-12x-7x+14=6x\left(x-2\right)-7\left(x-2\right)=\left(6x-7\right)\left(x-2\right)\)
5, \(xy-y^2-3x+3y=y\left(x-y\right)-3\left(x-y\right)=\left(y-3\right)\left(x-y\right)\)
a) x2(x-3)-4x+12
=x2(x-3)-4(x-3)
=(x-3)(x2-4)
=(x-3)(x-2)(x+2)
b) 2a(x+y)-x-y
=2a(x+y)-(x+y)
=(x+y)(2a-1)
c) 2x-4+5x2-10x
=2(x-2)+5x(x-2)
=(x-2)(2+5x)
d) 5x2-12x-7x+14
=5x2-19x+14
e) xy-y2-3x+3y
=y(x-y)-3(x-y)
=(x-y)(y-3)
#H
a.( 5x2-4x).(x-3)=5x3-15x2-4x2+12x=5x3-11x2+12x
b.(2-3xy).(3x4+4y2+5xy)=6x4+8y2+10xy-9x5y-12xy3-15x2y2
c.(-3x2+x+1).(x2+x-5)=-3x4-3x3+15+x3+x2-5x+x2+x-5=-3x4-2x3+2x-4x+10
a, \(5x^2-10xy+5y^2=5\left(x^2-2xy+y^2\right)=5.\left(x-y\right)^2\)
b, \(x^2-4x+4-y^2=\left(x^2-4x+4\right)-y^2=\left(x-2\right)^2-y^2\)
\(=\left(x-2-y\right)\left(x-2+y\right)\)
c, \(3x^2-2x-5=3x^2-5x+3x-5=x\left(3x-5\right)+3x-5\)
\(=\left(3x-5\right)\left(x+1\right)\)
Bài 1:
a) Ta có: \(\left(15x^2\cdot y^2\cdot z\right):3xyz\)
\(=\dfrac{15x^2y^2z}{3xyz}\)
\(=5xy\)
b) Ta có: \(3x^2\cdot\left(5x^2-4x+3\right)\)
\(=3x^2\cdot5x^2-3x^2\cdot4x+3x^2\cdot3\)
\(=15x^4-12x^3+9x^2\)
c) Ta có: \(\left(2x^2-3x\right):\left(x-4\right)\)
\(=\dfrac{2x^2-8x+5x-20+20}{x-4}\)
\(=\dfrac{2x\left(x-4\right)+5\left(x-4\right)+20}{x-4}\)
\(=2x+5+\dfrac{20}{x-4}\)
d) Ta có: \(-5xy\cdot\left(3x^2y-5xy+y^2\right)\)
\(=-5xy\cdot3x^2y+5xy\cdot5xy-5xy\cdot y^2\)
\(=-15x^3y^2+25x^2y^2-5xy^3\)
`#3107`
`a.`
`4x^2 - 6x = 2x(2x - 3)`
`b.`
`9x^4y^3 + 3x^2y^4 = 3x^2y^2(3x^2y + y^2)`
`c.`
`x^3 - 2x^2 + 5x`
`= x(x^2 - 2x + 5)`
a) 4x² - 6x
= 2x(2x - 3)
b) 9x⁴y³ + 3x²y⁴
= 3x²y³(3x² + 3y)
c) x³ - 2x² + 5x
= x(x² - 2x + 5)
a: Ta có: \(5\left(4x-1\right)+2\left(1-3x\right)-6\left(x+5\right)=10\)
\(\Leftrightarrow20x-5+2-6x-6x-30=10\)
\(\Leftrightarrow8x=43\)
hay \(x=\dfrac{43}{8}\)
b: ta có: \(2x\left(x+1\right)+3\left(x-1\right)\left(x+1\right)-5x\left(x+1\right)+6x^2=0\)
\(\Leftrightarrow2x^2+2x+3x^2-3-5x^2-5x+6x^2=0\)
\(\Leftrightarrow6x^2-3x-3=0\)
\(\Leftrightarrow2x^2-x-1=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{2}\end{matrix}\right.\)
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) \(x^2-9x+18=x^2-3x-6x+18\)
\(=x\left(x-3\right)-6\left(x-3\right)=\left(x-3\right)\left(x-6\right)\)
c) \(x^2-6x+5=x^2-x-5x+5\)
\(=x\left(x-1\right)-5\left(x-1\right)=\left(x-1\right)\left(x-5\right)\)
d) \(3x^2+5x-30=3\left(x^2+\dfrac{5x}{3}-10\right)=3\left(x^2+2.x.\dfrac{5}{6}+\dfrac{25}{36}-\dfrac{5347}{500}\right)\)
Câu này bạn xem lại đề nha
e) \(3x^2-5x-2=3x^2-6x+x-2\)
\(3x\left(x-2\right)+x-2=\left(x-2\right)\left(3x+1\right)\)
1, \(5x^2+10x+5-5y^2=5\left(x^2+2x+1-y^2\right)\)
\(=5\left[\left(x+1\right)^2-y^2\right]=5\left(x+1-y\right)\left(x+1+y\right)\)
2, \(3x^3+6x^2+3x-12xy^2=3x\left(x^2+2x+1-4y^2\right)\)
\(=3x\left[\left(x+1\right)^2-4y^2\right]=3x\left(x+1-2y\right)\left(x+1+2y\right)\)
5x(3x2-4x-1)=15x3-20x2-5x