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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: =(2x+1)^2-y^2
=(2x+1+y)(2x+1-y)
3: =x^2(x^2+2x+1)
=x^2(x+1)^2
4: =x^2+6x-x-6
=(x+6)(x-1)
5: =-6x^2+3x+4x-2
=-3x(2x-1)+2(2x-1)
=(2x-1)(-3x+2)
6: =5x(x+y)-(x+y)
=(x+y)(5x-1)
7: =2x^2+5x-2x-5
=(2x+5)(x-1)
8: =(x^2-1)*(x^2-4)
=(x-1)(x+1)(x-2)(x+2)
9: =x^2(x-5)-9(x-5)
=(x-5)(x-3)(x+3)
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)
đề dài nên T giải câu a thôi bn tự làm tiếp mấy câu khác nhé
2x^2 - 2y^2 - 6x - 6y
= 2(x^2-y^2) - 6(x+ y)
= 2(x-y)(x+y) - 6(x+y)
= (2(x-y)-6) (x+y)
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
^^
g) (x+2)(x+3)(x+4)(x+5)-24 = \(\left[\left(x+2\right)\left(x+5\right)\right]\left[\left(x+3\right)\left(x+4\right)\right]-24\)
=\(\left[x^2+7x+10\right]\left[x^2+7x+12\right]\)
đặt \(x^2+7x+10=a\)
ta có \(a\left(a+2\right)-24=a^2+2a-24\)
\(=a^2+2a+1-25\)
\(=\left(a+1\right)^2-5^2\)
\(=\left(a+1-5\right)\left(a+1+5\right)\)
\(=\left(a-4\right)\left(a+6\right)\)
\(\Rightarrow\) \(\left(x^2+7x+10-4\right)\left(x^2+7x+10+6\right)=\left(x^2+7x+6\right)\left(x^2+7x+16\right)\)
a) = (x +5)2 - 22 = (x+5 -2)(x+5 +2) = (x+3)(x+7)
b) = x(x2 -1) -6(x-1)= x(x+1)(x-1) -6(x-1) = (x-1)(x(x+1)-6)
help me!!!
b: \(=x^4+x^2+36-2x^3+12x^2-12x+x^2-6x+9\)
\(=x^4-2x^3+14x^2-18x+45\)
\(=x^4+9x^2-2x^3-18x+5x^2+45\)
\(=\left(x^2+9\right)\left(x^2-2x+5\right)\)
d: \(=2x^4+2x^3+6x^2-x^3-x^2-3x+x^2+x+3\)
\(=\left(x^2+x+3\right)\left(2x^2-x+1\right)\)
e: \(=3x^4-3x^3-3x^2-2x^3+2x^2+2x+2x^2-2x-2\)
\(=\left(x^2-x-1\right)\left(3x^2-2x+1\right)\)