giải giúp mình vs:
a) (x+2).(x^2-2x+4)-x(x^2+2)=15
b) (x-2)^3-(x-3).(x^2+3x+9+6.(x+1)^2=49
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) x12 + 4 = x12 + 4x6 + 4 - 4x6 = (x6 + 2)2 - (2x3)2
= (x6 - 2x3 + 2)(x6 + 2x3 + 2)
b) 4x8 + 1 = 4x8 + 4x4 + 1 - 4x4 = (2x4 + 1)2 - (2x2)2
= (2x4 + 2x2 + 1)(2x4 - 2x2 + 1)
c) x7 + x5 - 1 = x7 - x + x5 + x2 - (x2 - x + 1) = x(x6 - 1) + x2(x3 + 1) - (x2 - x + 1)
= x(x3 - 1)(x3 + 1) + x2(x + 1)(x2 - x + 1) - (x2 - x + 1)
= (x4 - x)(x + 1)(x2 - x + 1) + (x3 + x2)(x2 - x + 1) - (x2 - x + 1)
= (x5 + x4 - x2 - x + x3 + x2 - 1)(x2 -x + 1)
= (x5 + x4 + x3 - x - 1)(x2 - x + 1)
d) x7 + x5 + 1 = x7 - x + x5 - x2 + (x2 + x + 1)
= x(x3 - 1)((x3 + 1) + x2(x3 - 1) + (x2 + x + 1)
= (x4 + x)(x - 1)(x2 + x + 1) + x2(x - 1)((x2 + x + 1) + (x2 + x + 1)
= (x2 + x + 1)(x5 - x4 + x2 - x + x3 - x2 + 1)
= (x2 + x + 1)(x5 - x4 + x3 - x + 1)
a, A=8x6-12x4+6x2-1
=( 2x2-1)3
b,B= \(8\left(\frac{x}{2}+y\right)^3-6\left(x+2y\right)^2x+12\left(x+2y\right)x^2-8x^3\)
=\(8\left(\frac{x^3}{8}+3\frac{x^2}{4}y+3\frac{x}{2}y^2+y^3\right)-6x\left(x^2+4xy+4y^2\right)-8x^3\)
=\(\left[2\left(\frac{x}{2}+y\right)\right]^3-3.2\left(\frac{x}{2}+y\right)^2.2x+3.2\left(\frac{x}{2}+y\right).4x^2-\left(2x\right)^3\)
=\(\left[2\left(\frac{x}{2}+y\right)-2x\right]^3\)
c, C=\(\left(x-y\right)^3-\frac{3\left(x-y\right)^2}{2}y+\frac{3\left(x-y\right)}{4}y^2-\frac{y^3}{8}\)
=\(\left[\left(x-y\right)-\frac{y}{2}\right]^3\)
8.
a,A=(x-y)3+3(x-y)
=(x-y)[(x-y)2+3xy]
=(x-y)(x2-2xy+y2+3xy)
=(x-y)(x2+xy+y2)
b,B=(x+y)3+3(x-y)(x+y)2+3(x-y)2(x+y)+(x-y)3
=(x+y)3+3(x+y)2(x-y)+3(x+y)(x-y)2+(x-y)3
=[(x+y)(x-y)]3
=[(x2-y2)]3
\(\left(x+2\right)^2+4x=\left(x+2\right)\left(4x-3\right)\Leftrightarrow x^2+4x+4+4x=4x^2+8x-3x-6\)
\(\Leftrightarrow3x^2-3x-10=0\Leftrightarrow\orbr{\begin{cases}x=\frac{3+\sqrt{129}}{6}\\x=\frac{3-\sqrt{129}}{6}\end{cases}}\)
\(\left(x+2\right)^2+4x=\left(x+2\right)\left(4x-3\right)\)
\(x^2+4x+4+4x=4x^2+8x-3x-6\)
\(3x^2-3x-10=0\)
\(\Delta=\left(-3\right)^2-\left(4.3.-10\right)=129\)
\(\sqrt{\Delta}=\sqrt{129}\)
\(x_1=\frac{3+\sqrt{129}}{6}\)
\(x_2=\frac{3-\sqrt{129}}{6}\)
(x2 - 3x)(-5x2 + x) - x2(x - 7) + 5x2(x - 2)(x - 4) - (x + 1)(x + 3)
= -5x4 + 16x3 - 3x2 - x3 + 7x2 + 5x2(x2 - 6x + 8) - x2 - 4x - 3
= -5x4 + 15x3 + 3x2 - 4x - 3 + 5x4 - 30x3 + 40x2
= -15x3 + 43x2 - 4x - 3
(-2x2 + 5)(x - 6) - 4x(x2 - 7x + 2) - 2x(x + 3)(x - 2) - (x3 - x2)
= -2x3 + 12x2 + 5x - 30 - 4x3 + 28x2 - 8x - 2x(x2 + x - 6) - x3 + x2
= -8x3 + 41x2 - 3x - 30 - 2x3 - 2x2 + 12x
= -10x3 + 39x2 + 9x - 30
Cái áo rank bàng mùa 11 phải ko xai đừng hcuiwr
a) (x + 2)(x2 - 2x + 4) - -x(x2 + 2) = 15
<=> x3 + 8 - x3 - 2x = 15
<=> 2x = -7
<=> x= -7/2
Vậy S = {-7/2}
b) (x - 2)3 - (x - 3)(x2 + 3x + 9) + 6(x + 1)2 = 49
<=> x3 - 6x2 + 12x - 8 - x3 + 27 + 6x2 + 12x + 6 = 49
<=> 24x = 24
<=> x = 1
Vậy S = {1}