Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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
=
1) 4x2 + 5x - 6 = 4x2 + 8x - 3x - 6 = 4x( x + 2 ) - 3( x + 2 ) = ( x + 2 )( 4x - 3 )
2) 5x2 - 18x - 8 = 5x2 - 20x + 2x - 8 = 5x( x - 4 ) + 2( x - 4 ) = ( x - 4 )( 5x + 2 )
3) 2x2 + 3x - 27 = 2x2 - 6x + 9x - 27 = 2x( x - 3 ) + 9( x - 3 ) = ( x - 3 )( 2x + 9 ) < đã sửa ._. >
4) 7x2 + 3xy - 10y2 = 7x2 - 7xy + 10xy - 10y2 = 7x( x - y ) + 10y( x - y ) = ( x - y )( 7x + 10y )
5) x2 + 5x - 2 < sai đề ._. >
6) x8 + x7 + 1 = x8 + x7 + x6 - x6 + 1
= ( x8 + x7 + x6 ) - ( x6 - 1 )
= x6( x2 + x + 1 ) - ( x3 - 1 )( x3 + 1 )
= x6( x2 + x + 1 ) - ( x - 1 )( x2 + x + 1 )( x3 + 1 )
= ( x2 + x + 1 )[ x6 - ( x - 1 )( x3 + 1 ) ]
= ( x2 + x + 1 )( x6 - x4 + x3 - x + 1 )
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?
Giải như sau.
(1)+(2)⇔x2−2x+1+√x2−2x+5=y2+√y2+4⇔(x2−2x+5)+√x2−2x+5=y2+4+√y2+4⇔√y2+4=√x2−2x+5⇒x=3y(1)+(2)⇔x2−2x+1+x2−2x+5=y2+y2+4⇔(x2−2x+5)+x2−2x+5=y2+4+y2+4⇔y2+4=x2−2x+5⇒x=3y
⇔√y2+4=√x2−2x+5⇔y2+4=x2−2x+5, chỗ này do hàm số f(x)=t2+tf(x)=t2+t đồng biến ∀t≥0∀t≥0
Công việc còn lại là của bạn !
\(\left(x+6\right)\left(2x+1\right)=0\)
<=> \(\orbr{\begin{cases}x+6=0\\2x+1=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x=-6\\x=-\frac{1}{2}\end{cases}}\)
Vậy....
hk tốt
^^
a, x^2 + 5x +4
= x^2 + 1x + 4x + 4
= (x^2 + 1x) + (4x + 4)
= x ( x + 1 ) + 4 ( x + 1 )
= (x + 1) (x + 4)
b, x^2 - 6x + 5
= x^2 - 1x - 5x + 5
= (x^2 - 1x) - (5x - 5)
= x (x - 1) - 5 (x - 1)
= (x - 1) (x - 5)
c, x^2 + 7x + 12
= x^2 + 3x + 4x + 12
= (x^2 + 3x) + (4x + 12)
= x (x + 3) + 4 (x + 3)
= (x + 3) (x + 4)
d, 2x^2 - 5x + 3
= 2^x2 - 2x - 3x + 3
= 2x (x - 1) - 3 (x - 1)
= (x-1) (2x - 3)
e, 7x - 3x^2 - 4
= 3x + 4x - 3x^2 - 4
= (3x - 3x^2) + (4x - 4)
= 3x (1 - x) + 4 (x - 1)
= 3x (1-x) - 4 (1 - x)
= (1 - x) (3x - 4)
f, x^2 - 10x + 16
= x^2 - 2x - 8x + 16
= (x^2 - 2x) - (8x - 16)
= x (x - 2) - 8 (x - 2)
= (x - 2) (x - 8)
a, (x+1)(x+4)
b,(x-5)(x-1)
c,(x+3)(x+4)
d,(2x-3)(x-1)
e,(-3x+4)(x-1)
f, (x-8)(x-2)
Mấy câu dễ mình làm trước nhé. Mấy câu khó hơn mình trình bày sau :)
1) 2x2 - 5xy - 3y2 = 2x2 + xy - 6xy - 3y2 = x( 2x + y ) - 3y( 2x + y ) = ( 2x + y )( x - 3y )
2) 7x2 + 3xy - 10y2 = 7x2 - 7xy + 10xy - 10y2 = 7x( x - y ) + 10y( x - y ) = ( x - y )( 7x + 10y )
3) x2 + 5x - 2 = ( x2 + 5x + 25/4 ) - 33/4 = ( x + 5/2 )2 - \(\left(\frac{\sqrt{33}}{2}\right)^2\)= \(\left(x+\frac{5}{2}-\frac{\sqrt{33}}{2}\right)\left(x+\frac{5}{2}+\frac{\sqrt{33}}{2}\right)\)
6) x4 + 324 = ( x4 + 36x2 + 324 ) - 36x2 = ( x2 + 18 )2 - ( 6x )2 = ( x2 - 6x + 18 )( x2 + 6x + 18 )
4) x8 + x7 + 1
= x8 + x7 + x6 - x6 + 1
= x6( x2 + x + 1 ) - ( x6 - 1 )
= x6( x2 + x + 1 ) - ( x3 - 1 )( x3 + 1 )
= x6( x2 + x + 1 ) - ( x - 1 )( x2 + x + 1 )( x3 + 1 )
= ( x2 + x + 1 )( x6 - ( x - 1 )( x3 + 1 ) ]
= ( x2 + x + 1 )( x6 - x4 + x3 - x + 1 )
5) x7 + x5 + 1
= x7 + x6 - x6 + x5 + 1
= x5( x2 + x + 1 ) - ( x6 - 1 )
= x5( x2 + x + 1 ) - ( x3 - 1 )( x3 + 1 )
= x5( x2 + x + 1 ) - ( x - 1 )( x2 + x + 1 )( x3 + 1 )
= ( x2 + x + 1 )[ x5 - ( x - 1 )( x3 + 1 ) ]
= ( x2 + x + 1 )( x5 - x4 + x3 - x + 1 )
7) x5 - 5x3 + 4x
= x5 - x3 - 4x3 + 4x
= x3( x2 - 1 ) - 4x( x2 - 1 )
= ( x2 - 1 )( x3 - 4x )
= ( x - 1 )( x + 1 )x( x2 - 4 )
= x( x - 1 )( x + 1 )( x - 2 )( x + 2 )
8) Xin hàng :)