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a) x4+x3+2x2+x+1=(x4+x3+x2)+(x2+x+1)=x2(x2+x+1)+(x2+x+1)=(x2+x+1)(x2+1)
b)a3+b3+c3-3abc=a3+3ab(a+b)+b3+c3 -(3ab(a+b)+3abc)=(a+b)3+c3-3ab(a+b+c)
=(a+b+c)((a+b)2-(a+b)c+c2)-3ab(a+b+c)=(a+b+c)(a2+2ab+b2-ac-ab+c2-3ab)=(a+b+c)(a2+b2+c2-ab-ac-bc)
c)Đặt x-y=a;y-z=b;z-x=c
a+b+c=x-y-z+z-x=o
đưa về như bài b
d)nhóm 2 hạng tử đầu lại và 2hangj tử sau lại để 2 hạng tử sau ở trong ngoặc sau đó áp dụng hằng đẳng thức dề tính sau đó dặt nhân tử chung
e)x2(y-z)+y2(z-x)+z2(x-y)=x2(y-z)-y2((y-z)+(x-y))+z2(x-y)
=x2(y-z)-y2(y-z)-y2(x-y)+z2(x-y)=(y-z)(x2-y2)-(x-y)(y2-z2)=(y-z)(x2-2y2+xy+xz+yz)
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) 3a +3b -a2-ab
= 3.(a+b) -a.(a+b)=(3-a).(a+b)
b) x2 +x +y2-y-2xy
=(x2 - 2xy+y2) +(x-y)
=(x-y).(x-y+1)
c) -x2 +7x -6
= -x2 + x +6x-6
= x.(1-x) -6.(1-x) = (1-x).(x-6)
d) 5x3y -10x2y2 +5xy3
= 5xy.(x2 -2xy +y2) = 5xy.(x-y)2
e) 2x2 +7x -15
= 2x2 -3x +10x -15
=x.(2x-3) + 5.(2x-3)
=(2x-3).(x+5)
g) x2 -2x +2y -xy
=x.(x-2)-y.(x-2)
=(x-y).(x-2)
h) bn go lai de ho mk dc k?
a,15x^2-60=15x^2-15*4=15(x^2-4)
b,=(x^2+2xy+y^2)-25=(x+y)^2-5^2=[(x-y)+5][(x-y)-5]
xog 2 phần
a) \(x^3-16x=0\)
\(\Leftrightarrow x\left(x^2-16\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x^2-16=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm4\end{cases}}\)
Vậy tập nghiệm \(S=\left\{-4;0;4\right\}\)
b) \(x^4-2x^3+10x^2-20x=0\)
\(\Leftrightarrow x^3\left(x-2\right)+10x\left(x-2\right)=0\)
\(\Leftrightarrow\left(x^3+10x\right)\left(x-2\right)=0\)
\(\Leftrightarrow x\left(x^2+10\right)\left(x-2\right)=0\)
Mà \(x^2+10>0\)nên \(\orbr{\begin{cases}x=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)
Vậy tập nghiệm S = { 0;2}
1
a, 2x2+4x+2-2y2 = 2(x2+2x+1-y2)= 2[(x+1)2-y2 ] = 2(x-y+1)(x+y+1)
b, 2x - 2y - x2 + 2xy - y2= 2(x -y) - (x2 - 2xy + y2) = 2(x-y)-(x-y)2=(x-y)(2-x+y)
c, x2-y2-2y-1=x2-(y2+2y+1)=x2-(y+1)2=(x-y-1)(x+y+1)
d, x2-4x-2xy-4y+y2= x2-2xy+y2-4x-4y=(x-y)
2.
a, x2-3x+2=x2-x-2x+2=x(x-1)-2(x-1)=(x-2)(x-1)
b, x2+5x+6=x2+2x+3x+6=x(x+2)+3(x+2)=(x+3)(x+2)
c, x2+6x-6=
a) x^4 - x^3 - x + 1
= x^3 ( x - 1 ) - ( x- 1 )
= ( x^3 - 1 )(x - 1)
= ( x- 1 )^2 (x^2 + x + 1 )
a)x4-x3-x+1
=x3(x-1)-(x-1)
=(x-1)(x3-1)
=(x-1)(x-1)(x2+x+1)
=(x-1)2(x2+x+1)
b)5x2-4x+20xy-8y
(sai đề)
a: 2x^2y-50xy=2xy(x-25)
b: 5x^2-10x=5x(x-2)
c: 5x^3-5x=5x(x^2-1)=5x(x-1)(x+1)
d: \(x^2-xy+x=x\left(x-y+1\right)\)
e: x(x-y)-2(y-x)
=x(x-y)+2(x-y)
=(x-y)(x+2)
f: 4x^2-4xy-8y^2
=4(x^2-xy-2y^2)
=4(x^2-2xy+xy-2y^2)
=4[x(x-2y)+y(x-2y)]
=4(x-2y)(x+y)
f1: x^2ỹ-y^2+y
=(x-y)(x+y)+(x+y)
=(x+y)(x-y+1)
a, x2+2xy+y2+2x+2y-15
<=> (x+y )2+2(x+y)+1-16
Đặt x+y =a
<=> a2+2a+1-42
<=> (a+1)2-42
<=> (a+5)(a-3) =>( x+y+5)(x+y-3)
b, x2-4xy+4y2-2x-4y-35
<=> (x-2y)2-2(x-2y)+1-36
Đặt (x-2y) =b
=> b2-2b+1-62
<=> (b-1)2-62
<=> (b-7)(b+5)=> (x-2y-7)(x-2y+5)
c,
a,A= x^2+2xy+y^2+2x+2y-15
= (x+y)^2+(x+y)-15
Đặt x+y=a, ta có:
A=a^2+2a-15
=a^2+2a+1-16
=(a+1)^2-4^2
=(a+1+4)(a+1-4)
=(a+5)(a-3)
Thay a=x+y, ta có: A=(x+y+5)(x+y-3).