Bài 1
a) \(2x^2y\left(3xy^2-5y\right)\)
= \(6x^3y^3-10x^2y^2\)
b) \(\left(2x-3\right)\left(x^2+2x-4\right)\)
= \(2x^3+4x^2-8x-3x^2-6x+12\)
= \(2x^3+x^2-14x+12\)
Bài 2
\(\left(x-1\right)^2-\left(x+4\right)\left(x-4\right)\)
= \(\left(x^2-2x+1\right)-\left(x^2-16\right)\)
= \(x^2-2x+1-x^2+16\)
= \(-2x+17\)
1) làm tính nhân :
a) 2x2y( 3x2 - 5y)
\(\Rightarrow\) 6x4y - 10x2y2
b) ( 2x - 3 )(x2 + 2x - 4)
\(\Rightarrow\) 2x(x2 + 2x - 4) - 3(x2 + 2x - 4)
\(\Rightarrow\) 2x3 + 4x2 - 8x - 3x2 - 6x + 12
\(\Rightarrow\) 2x3 + x2 - 14x + 12
2) rút gọn :
(x - 1)2 - (x + 4)(x - 4) = 0
x2 - 2x + 1 - (x2 - 16) = 0
\(\Rightarrow\) x2 - 2x + 1 - x2 + 16 = 0
\(\Rightarrow\) -2x + 17 = 0
\(\Rightarrow\) -2x = -17
\(\Rightarrow\) x = \(\dfrac{-17}{-2}=\dfrac{17}{2}\)
Bài 2 : phân tích các đa thức sau đây thành nhân tử :
a) x2 - 3xy
\(\Rightarrow\) x (x - 3y)
b) ( x + 5 )2 - 9
\(\Rightarrow\) (x + 5)2 - 32
\(\Rightarrow\) \(\left[\left(x+5-3\right)\right]\)\(\left[\left(x+5+3\right)\right]\)
\(\Rightarrow\) (x+2)(x+8)
c) xy + xz - 2y - 2z
\(\Rightarrow\) (xy + xz) -(2y+ 2z)
\(\Rightarrow\) x(y + z) - 2(y + z)
\(\Rightarrow\)(y + z) (x - 2)
Bài 3 : tìm x :
a) 3(2x - 4) + 15 = -11
\(\Rightarrow\) 6x - 12 + 15 + 11 = 0
\(\Rightarrow\) 6x + 14 = 0
\(\Rightarrow\) 6x = -14
\(\Rightarrow\) x = \(\dfrac{-14}{6}=\dfrac{-7}{3}\)
b) x(x + 2) - 3x - 6 = 0
\(\Rightarrow\) x(x + 2) - 3(x + 2) = 0
\(\Rightarrow\) (x + 2)(x - 3) = 0
\(\Rightarrow\) x + 2 = 0 hoặc x + 3 = 0
\(\Rightarrow\) x = -2
\(\Rightarrow\) x = -3
Bài 2
a) \(x^2-3xy\)
= \(x\left(x-3y\right)\)
b) \(\left(x+5\right)^2-9\)
= \(\left(x+5\right)^2-3^2\)
= \(\left(x+5-3\right)\left(x+5+3\right)\)
= \(\left(x+2\right)\left(x+8\right)\)
c) \(xy+xz-2y-2z\)
= \(\left(xy+xz\right)-\left(2y+2z\right)\)
= \(x\left(y+z\right)-2\left(y+z\right)\)
= \(\left(y+z\right)\left(x-2\right)\)
d) \(4x^3+8x^2y+4xy^2-16x\)
= \(\left(4x^3+8x^2y\right)+\left(4xy^2-16x\right)\)
= \(4x^2\left(x+2y\right)+4x\left(y^2-4\right)\)
= \(4x\left[x\left(x+2y\right)+y^2-4\right]\)
= \(4x\left(x^2+2xy+y^2-2^2\right)\)
= \(4x\left[\left(x+y\right)^2-2^2\right]\)
= \(4x\left(x+y-2\right)\left(x+y+2\right)\)
Bài 3
a) \(3\left(2x-4\right)+15=-11\)
\(6x-12+15+11=0\)
\(6x+14=0\)
\(2\left(3x+7\right)=0\)
\(3x+7=0\)
\(3x=-7\)
\(x=\dfrac{-7}{3}\)
b) \(x\left(x+2\right)-3x-6=0\)
\(x\left(x+2\right)-\left(3x+6\right)=0\)
\(x\left(x+2\right)-3\left(x+2\right)=0\)
\(\left(x+2\right)\left(x-3\right)=0\)
TH1 : \(x+2=0\)
\(x=-2\)
TH2 : \(x-3=0\)
\(x=3\)
