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a)x^2-(a+b)x+ab
= x^2 - ax - bx + ab
= (x^2 - ax) - (bx - ab)
= x(x-a) - b(x-a)
= (x-b)(x-a)
b)7x^3-3xyz-21x^2+9z
=
c)4x+4y-x^2(x+y)
= 4(x + y) - x^2(x+y)
= (4-x^2) (x+y)
= (2-x)(2+x)(x+y)
d) y^2+y-x^2+x
= (y^2 - x^2) + (x+y)
= (y-x)(y+x)+ (x+y)
= (y-x+1) (x+y)
e)4x^2-2x-y^2-y
= [(2x)^2 - y^2] - (2x +y)
= (2x-y)(2x+y) - (2x+y)
= (2x -y -1)(2x+y)
f)9x^2-25y^2-6x+10y
=
2. a. \(A=2x^2-8x-10=2\left(x^2-4x+4\right)-18\)
\(=2\left(x-2\right)^2-18\)
Vì \(\left(x-2\right)^2\ge0\forall x\)\(\Rightarrow2\left(x-2\right)^2-18\ge-18\)
Dấu "=" xảy ra \(\Leftrightarrow2\left(x-2\right)^2=0\Leftrightarrow x-2=0\Leftrightarrow x=2\)
Vậy minA = - 18 <=> x = 2
b. \(B=9x-3x^2=-3\left(x^2-3x+\frac{9}{4}\right)+\frac{27}{4}\)
\(=-3\left(x-\frac{3}{2}\right)^2+\frac{27}{4}\)
Vì \(\left(x-\frac{3}{2}\right)^2\ge0\forall x\)\(\Rightarrow-3\left(x-\frac{3}{2}\right)^2+\frac{27}{4}\le\frac{27}{4}\)
Dấu "=" xảy ra \(\Leftrightarrow-3\left(x-\frac{3}{2}\right)^2=0\Leftrightarrow x-\frac{3}{2}=0\Leftrightarrow x=\frac{3}{2}\)
Vậy maxB = 27/4 <=> x = 3/2
a)x3-6x2+9x=x(x2-6x+9)=x(x-3)2
b)x2-2x-4y2-4y=(x2-2x+1)-(4y2+4y+1)=(x-1)2-(2y+1)2=(x-1-2y-1)(x-1+2y+1)=(x-2y-2)(x+2y)
c)x2-x+xy-y=x(x-1)+y(x-1)=(x-1)(x+y)
d)3x2-6xy-75+3y2=3[(x2-2xy+y2)-25]=3[(x-y)2-52]=3(x-y-5)(x-y+5)
e)2x2-5x-7=(2x2+2x)-(7x+7)=2x(x+1)-7(x+1)=(x+1)(2x-7)
f)x4+36=x4+12x2+36-12x2=(x2+6)2-12x2=(x2-\(\sqrt{12}x\)+6)(x2+\(\sqrt{12}x\)+6)
h)x4+4y4=x4+4x2y2+4y2-4x2y2=(x2+2y2)-4x2y2=(x2+2y2-2xy)(x2+2y2+2xy)
a) \(45+x^3-5x^2-9x\)
\(=\left(x^3-5x^2\right)-\left(9x-45\right)\)
\(=x^2\left(x-5\right)-9\left(x-5\right)\)
\(=\left(x-5\right)\left(x^2-9\right)=\left(x-5\right)\left(x-3\right)\left(x+3\right)\)
\(a,45+x^3-5x^2-9x\)
\(=\left(x^3-5x^2\right)+\left(45-9x\right)\)
\(=x^2\left(x-5\right)+9\left(5-x\right)\)
\(=x^2\left(x-5\right)-9\left(x-5\right)\)
\(=\left(x-5\right)\left(x^2-9\right)\)
\(=\left(x-5\right)\left(x^2-3^2\right)\)
\(=\left(x-5\right)\left(x+3\right)\left(x-3\right)\)
\(c,2x^2+3x-5\)
\(=2x^2-2x+5x-5\)
\(=2x\left(x-1\right)+5\left(x-1\right)\)
\(=\left(x-1\right)\left(2x+5\right)\)
\(e,\left(x+2\right)\left(x+4\right)\left(x+6\right)\left(x+8\right)+16\)
\(=\left[\left(x+2\right)\left(x+8\right)\right]\left[\left(x+4\right)\left(x+6\right)\right]+16\)
\(=\left(x^2+8x+2x+16\right)\left(x^2+6x+4x+24\right)+16\)
\(=\left(x^2+10x+16\right)\left(x^2+10x+24\right)+16\left(1\right)\)
\(\text{Đặt }x^2+10x+\frac{16+24}{2}=t\)
\(\text{hay }x^2+10x+20=t\left(2\right)\)
(1)\(\Leftrightarrow\left(t-4\right)\left(t+4\right)+16\)
\(=t^2-4^2+16\)
\(=t^2-16+16\)
\(=t^2\left(3\right)\)
Thay (3) vào (2),ta được:\(\left(x^2+10x+20\right)^2\)
c. \(x^4-2x^3-2x^2-2x-3=x^3\left(x-3\right)+x^2\left(x-3\right)+x\left(x-3\right)+x-3\)
\(=\left(x-3\right)\left(x^3+x^2+x+1\right)=\left(x-3\right)\left(x+1\right)\left(x^2+1\right)\)
a) \(2x^2-2y^2\)
\(=2\left(x^2-y^2\right)\)
\(=2\left(x-y\right)\left(x+y\right)\)
b) \(x^2-4x+4\)
\(=x^2-2\cdot x\cdot2+2^2\)
\(=\left(x-2\right)^2\)
c) \(x^2+2x+1-y^2\)
\(=\left(x+1\right)^2-y^2\)
\(=\left(x-y+1\right)\left(x+y+1\right)\)
d) \(x^2-4x\)
\(=x\left(x-4\right)\)
e) \(x^2+10x+25\)
\(=x^2+2\cdot x\cdot5+5^2\)
\(=\left(x+5\right)^2\)
g) \(x^2-2xy+y^2-9\)
\(=\left(x-y\right)^2-3^2\)
\(=\left(x-y-3\right)\left(x-y+3\right)\)
h) \(2x^2-2\)
\(=2\left(x^2-1\right)\)
\(=2\left(x-1\right)\left(x+1\right)\)
i) \(5x^2-5xy+9x-9y\)
\(=5x\left(x-y\right)+9\left(x-y\right)\)
\(=\left(x-y\right)\left(5x+9\right)\)
k) \(y^2-4y+4-x^2\)
\(=\left(y-2\right)^2-x^2\)
\(=\left(y-x-2\right)\left(y+x-2\right)\)
l) \(x^2-16\)
\(=x^2-4^2\)
\(=\left(x-4\right)\left(x+4\right)\)
m) \(3x^2-3xy+2x-2y\)
\(=3x\left(x-y\right)+2\left(x-y\right)\)
\(=\left(x-y\right)\left(3x+2\right)\)
o) \(3x^4-6x^3+3x^2\)
\(=3x^2\left(x^2-2x+1\right)\)
\(=3x^2\left(x-1\right)^2\)
a) 2x2 - 2y2
= (2x - 2y)(2x + 2y)
= 4(x - y)(x + y)
b) x2 - 4x + 4
= (x - 2)2
c) x2 + 2x + 1 - y2
= (x + 1)2 - y2
= (x + 1 - y)(x + 1 + y)
d) x2 - 4x
= x(x - 4)
e) x2 +10x + 25
= (x + 5)2
g) x2 - 2xy + y2 - 9
= (x - y)2 - 32
= (x - y - 3)(x - y + 3)
h) 2x2 - 2
= 2(x2 - 1)
= 2(x - 1)(x + 1)
i) 5x2 - 5xy + 9x - 9y
= 5x(x - y) + 9(x- y)
= (5x + 9)(x - y)
k) y2 - 4y + 4 - x2
= (y - 2)2 - x2
= (y - 2 - x)(y - 2 + x)
l) x2 - 16
= x2 - 42
= (x - 4)(x + 4)
m) 3x2 - 3xy + 2x -2y
= 3x(x - y) +2(x-y)
= (3x + 2)(x - y)
o) 3x4 - 6x3 + 3x2
= 3x4 - 3x3 - 3x3 + 3x2
= 3x3(x - 1) - 3x2(x - 1)
= (3x3 - 3x2)(x - 1)
= 3x2(x - 1)(x - 1)
= 3x2.(x - 1)2
a) 5xy ( x - y ) - 2x + 2y
= 5xy ( x - y ) - 2 ( x - y )
= ( x - y ) ( 5xy - 2 )
b) 6x-2y-x(y-3x)
= 2 ( y - 3x ) - x ( y - 3x )
= ( y - 3x ( ( 2 - x )
c) x2 + 4x - xy-4y
= x ( x + 4 ) - y ( x + 4 )
( x + 4 ) ( x - y )
d) 3xy + 2z - 6y - xz
= ( 3xy - 6y ) + ( 2z - xz )
= 3y ( x - 2 ) + z ( x - 2 )
= ( x - 2 ) ( 3y + z )
a,5xy(x-y)-2x+2y=5xy(x-y)-2(x-y)=(x-y)(5xy-2)
b,6x-2y-x(y-3x)=-2(y-3x)-x(y-3x)=(y-3x)(-2-x)
c,x^2+4x-xy-4y=x(x+4)-y(x+4)=(x+4)(x-y)
d,3xy+2z-6y-xz=(3xy-6y)+(2z-xz)=3y(x-2)+z(2-x)=3y(x-2)-z(x-2)=(x-2)(3y-z)
11)
a,4-9x^2=0
(2-3x)(2+3x)=0
2-3x=0=>x=2/3 hoặc 2+3x=0=>x=-2/3
b,x^2 +x+1/4=0
(x+1/2)^2 =0
x+1/2=0
x=-1/2
c,2x(x-3)+(x-3)=0
(x-3)(2x+1)=0
x-3=0=>x=3 hoặc 2x+1=0=>x=-1/2
d,3x(x-4)-x+4=0
3x(x-4)-(x-4)=0
(x-4)(3x-1)=0
x-4=0=>x=4 hoặc 3x-1=0=>x=1/3
e,x^3-1/9x=0
x(x^2-1/9)=0
x(x+1/3)(x-1/3)=0
x=0 hoặc x+1/3=0=>x=-1/3 hoặc x-1/3=0=>x=1/3
f,(3x-y)^2-(x-y)^2 =0
(3x-y-x+y)(3x-y+x-y)=0
2x(4x-2y)=0
4x(2x-y)=0
x=0hoặc 2x-y=0=>x=y/2
Mình mới lớp 7 nên chỉ giải được 1 bài thôi!
\(3x^2-7x+2=3x^2-\left(6x+1x\right)+2=3x^2-6x-1x+2\)
\(3x\left(x-2\right)-1\left(x-2\right)=\left(x-2\right)\left(3x-1\right)=3\left(x-2\right)\left(x-\frac{1}{3}\right)\)