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a) \(x^2+12x+35\)
\(=x^2+5x+7x+35\)
\(=\left(x^2+5x\right)+\left(7x+35\right)\)
\(=x\left(x+5\right)+7\left(x+5\right)\)
\(=\left(x+5\right)\left(x+7\right)\)
b)\(x^2-x-56\)
\(=x^2+7x-8x-56\)
\(=\left(x^2+7x\right)-\left(8x+56\right)\)
\(=x\left(x+7\right)-8\left(x+7\right)\)
\(=\left(x+7\right)\left(x-8\right)\)
c)\(5x^2-x-4\)
\(=5x^2-5x+4x-4\)
\(=\left(5x^2-5x\right)+\left(4x-4\right)\)
\(=5x\left(x-1\right)+4\left(x-1\right)\)
\(=\left(x-1\right)\left(5x+4\right)\)
TL:
a)\(x^2+5x+7x+35\)
=\(x\left(x+5\right)+7\left(x+5\right)\)
=\(\left(x+7\right)\left(x+5\right)\)
b) \(x^2-x-56\)
=\(x^2+7x-8x-56\)
=\(x\left(x+7\right)-8\left(x+7\right)\)
=\(\left(x-8\right)\left(x+7\right)\)
d)\(4x^4+1=\left(2x^2\right)^2+4x^2+1-4x^2\)
=\(\left(2x^2+1\right)^2-4x^2\)
=\(\left(2x^2+1+4x\right)\left(2x^2+1-4x\right)\)
.......................(tự lm)
hc tốt
(x - 4)(x2 + 4x + 16) - x(x2 - 6) = 2
x3 - 64 - x3 + 6x = 2
6x = 2 + 64
6x = 66
x = 66 : 6
x = 11
x3 - 27 + 3x(x - 3)
= (x - 3)(x2 + 3x + 9) + 3x(x - 3)
= (x - 3)(x2 + 3x + 9 + 3x)
= (x - 3)(x2 + 6x + 9)
= (x - 3)(x + 3)2
5x3 - 7x2 + 10x - 14
= 5x(x2 + 2) - 7(x2 + 2)
= (x2 + 2)(5x - 7)
a/ \(x^4+16\)
\(=x^4+4x^2+16-4x^2\)
\(=\left(x^4+4x^2+16\right)-4x^2\)
\(=\left(x^2+4\right)^2-\left(2x\right)^2\)
\(=\left(x^2+4-2x\right)\left(x^2+4+2x\right)\)
b/ \(64x^4+y^4\)
\(=64x^4+y^4+16x^2y^2-16x^2y^2\)
\(=\left(64x^4+y^4+16x^2y^2\right)-16x^2y^2\)
\(=\left(8x^2+y^2\right)^2-\left(4xy\right)^2\)
\(=\left(y^2+8x^2-4xy\right)\left(8x^2+y^2-4xy\right)\)
\(x^4-x^3-x^2+1\)
\(\text{ Phân tích thành nhân tử}\)
\(\left(x-1\right)\left(x^3-x-1\right)\)
\(-x-y^2+x^2-y\)
\(\text{ Phân tích thành nhân tử}\)
\(\left(-\left(y-x+1\right)\right)\left(y+x\right)\)
\(x^2-y^2-x-y\)
\(\text{ Phân tích thành nhân tử}\)
\(\left(-\left(y-x+1\right)\right)\left(y+x\right)\)
\(x^2-y^2+4-4x\)
\(\text{ Phân tích thành nhân tử}\)
\(\left(-\left(y-x+2\right)\right)\left(y-x+2\right)\)
a)\(\frac{1}{64}x^6-125y^3=\left(\frac{1}{4}x^2\right)^3-\left(5y\right)^3\)\(=\left(\frac{1}{4}x^2-5y\right)\left(\frac{1}{16}x^4+\frac{5}{4}x^2y+25y^2\right)\)
b)\(x^6+1=\left(x^2\right)^3+1^3=\left(x^2+1\right)\left(x^4+x^2+1\right)\)
c)\(x^6-y^6=\left(x^3\right)^2-\left(y^3\right)^2=\left(x^3+y^3\right)\left(x^3-y^3\right)\)
\(=\left(x+y\right)\left(x^2-xy+y^2\right)\left(x-y\right)\left(x^2+xy+y^2\right)\)
d)\(x^9+1=\left(x^3\right)^3+1=\left(x^3+1\right)\left(x^6-x^3+1\right)\)
\(=\left(x+1\right)\left(x^2-x+1\right)\left(x^6-x^3+1\right)\)
\(=x^3\left(x+1\right)\left(x^2-x+1\right)\left(x^2-x+1\right)\)
a: =xy(x^2-4xy^2+4y^4)
=xy(x-2y^2)^2
b:=(x^3-y)^2
c: =(a^2-b^2)(a^2+b^2)
=(a^2+b^2)(a-b)(a+b)
d: 64x^6-27y^6
=(4x^2-3y^2)(16x^4+12x^2y^2+9y^4)
e: =(2x)^3+(3y)^3
=(2x+3y)(4x^2-6xy+9y^2)
a, \(=\left(2x^2\right)^2+2.9.2x^2+9^2-36x^2\)
\(=\left(2x^2+9\right)^2-\left(6x\right)^2\)
\(=\left(2x^2-6x+9\right)\left(2x^2+6x+9\right)\)
b, \(=x^4+2x^2+1-2x^2\)
\(=\left(x^2+1\right)^2-\left(x\sqrt{2}\right)^2\)
\(=\left(x^2+x\sqrt{2}+1\right)\left(x^2-x\sqrt{2}+1\right)\)
c, \(=\left(8x^2\right)^2+8x^2.2.y^2+y^4-16x^2y^2\)
\(=\left(8x^2+y^2\right)^2-\left(4xy\right)^2\)
\(=\left(8x^2-4xy+y^2\right)\left(8x^2+4xy+y^2\right)\)
d, \(=x^2+x-6=0\)
\(=x^2-2x+3x-6\)
\(=x\left(x-2\right)+3\left(x-2\right)\)
\(=\left(x-2\right)\left(x+3\right)\)
Bài 3:
a) Ta có: \(4x^4+81\)
\(=4x^4+36x^2+81-36x^2\)
\(=\left(2x^2+9\right)^2-\left(6x\right)^2\)
\(=\left(2x^2-6x+9\right)\left(2x^2+6x+9\right)\)
c) Ta có: \(64x^4+y^4\)
\(=\left(8x^2\right)^2+16x^2y^2+y^4-16x^2y^2\)
\(=\left(8x^2+y^2\right)^2-\left(4xy\right)^2\)
\(=\left(8x^2-4xy+y^2\right)\left(8x^2+4xy+y^2\right)\)