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ak
x8 + -7x4 + -8 = 0 Reorder the terms: -8 + -7x4 + x8 = 0 Solving -8 + -7x4 + x8 = 0 Solving for variable 'x'. Factor a trinomial. (-1 + -1x4)(8 + -1x4) = 0
Ta có : x4 + 8x2 + 7x + 8
= x4 - x + 8x2 + 8x + 8
= x(x3 - 1) + 8(x2 + x + 1)
= x(x - 1)(x2 + x + 1) + 8(x2 + x + 1)
= (x2 - x)(x2 + x + 1) + 8(x2 + x + 1)
= (x2 + x + 1)(x2 - x + 8)
Học tốt nhé !
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)
\(2x^4+3x^3-7x^2-6x+8\)
\(=2x^4+5x^3-2x^2-8x-2x^3-5x^2+2x+8\)
\(=x\left(2x^3+5x^2-2x-8\right)-\left(2x^3+5x^2-2x-8\right)\)
\(=\left(x-1\right)\left(2x^3+5x^2-2x-8\right)\)
\(=\left(x-1\right)\left(2x^3+x^2-4x+4x^2+2x-8\right)\)
\(=\left(x-1\right)\left[x\left(2x^2+x-4\right)+2\left(2x^2+x-4\right)\right]\)
\(=\left(x-1\right)\left(x+2\right)\left(2x^2+x-4\right)\)
Ta có:
\(\left(x+2\right)\left(x+4\right)\left(x+6\right)\left(x+8\right)+16=\left(x+2\right)\left(x+8\right)\left(x+4\right)\left(x+6\right)+16\)
\(=\left(x^2+8x+2x+16\right)\left(x^2+6x+4x+24\right)+16\)
\(=\left(x^2+10x+16\right)\left(x^2+10x+24\right)+16\)
\(=\left(x^2+10x+16\right)\left(x^2+10x+16+8\right)+16\)
\(=\left(x^2+10x+16\right)\left(x^2+10x+16\right)+8\left(x^2+10x+16\right)+16\)
\(=\left(x^2+10x+16\right)^2+2.\left(x^2+10x+16\right).4+4^2\)
\(=\left(x^2+10x+16+4\right)^2=\left(x^2+10+20\right)^2\)
k nha!!
\(\text{( x + 2 ) ( x + 4 ) ( x + 6 ) ( x + 8 ) + 16}\)
\(\text{Phân tích thành nhân tử :}\)
\(\left(x^2+10x+20\right)^2\)
b: \(\left(x^2+4\right)^2-16x^2\)
\(=\left(x^2-4x+4\right)\left(x^2+4x+4\right)\)
\(=\left(x-2\right)^2\cdot\left(x+2\right)^2\)
c: \(x^5-x^4+x^3-x^2\)
\(=x^4\left(x-1\right)+x^2\left(x-1\right)\)
\(=x^2\left(x-1\right)\left(x^2+1\right)\)
Lời giải:
a. Bạn xem lại đề
b. \((x^2+4)^2-16x^2=(x^2+4)^2-(4x)^2=(x^2+4-4x)(x^2+4+4x)\)
\(=(x-2)^2(x+2)^2\)
c.
\(x^5-x^4+x^3-x^2=x^4(x-1)+x^2(x-1)=(x^4+x^2)(x-1)\)
\(=x^2(x^2+1)(x-1)\)
(x+2)(x+4)(x+6)(x+8)+16
=(x+2)(x+8)(x+4)(x+6)+16
=(x2+10x+16)(x2+10x+24)+16
đặt t=x2+10x+16 ta được:
t.(t+8)+16
=t2+8t+16
=(t+4)2
thay t=x2+10x+16 ta được:
(x2+10x+16)2
=[(x+2)(x+8)]2
=(x+2)2(x+8)2
vậy (x+2)(x+4)(x+6)(x+8)+16 =(x+2)2(x+8)2
(x+2)(x+4)(x+6)(x+8)+16
=(x+2)(x+8)(x+4)(x+6)+16
=(x2+10x+16)(x2+10x+24)+16
đặt t=x2+10x+16 ta được:
t.(t+8)+16
=t2+8t+16
=(t+4)2
thay t=x2+10x+16 ta được:
(x2+10x+16)2
=[(x+2)(x+8)]2
=(x+2)2(x+8)2
vậy (x+2)(x+4)(x+6)(x+8)+16 =(x+2)2(x+8)2
x^8+7x^4+16=x8+8x4+16-x4
=(x4+4)2-x4
=(x4-x2+4)(x4+x2+4)