Phân tích các đa thức sau thành nhân tử :
a) \(\left(x^2-8\right)^2+36\)
b) \(x^8+x^4+1\)
c) \(x^7+x^5+1\)
d) \(x^7+x^5-1\)
Phân tích các đa thức sau thành nhân tử:
a) \({\left( {x - 1} \right)^2} - 4\)
b) \(4{x^2} + 12x + 9\)
c) \({x^3} - 8{y^6}\)
d) \({x^5} - {x^3} - {x^2} + 1\)
e) \( - 4{x^3} + 4{x^2} + x - 1\)
f) \(8{x^3} + 12{x^2} + 6x + 1\)
\(a,\left(x-1\right)^2-2^2=\left(x-1-2\right)\left(x-1+2\right)=\left(x-3\right)\left(x+1\right)\\ b,=\left(2x\right)^2+2.2x.3+3^2\\ =\left(2x+3\right)^2\\ c,=x^3-\left(2y\right)^3\\ =\left(x-2y\right)\left(x^2+2xy+4y^2\right)\\ d,=x^3\left(x^2-1\right)-\left(x^2-1\right)\\ =\left(x^3-1\right)\left(x^2-1\right)\\ =\left(x-1\right)\left(x^2+x+1\right)\left(x-1\right)\left(x+1\right)\\ =\left(x-1\right)\left(x+1\right)\left(x^2+x+1\right)\)
\(e,=-4x^2\left(x-1\right)+\left(x-1\right)\\ =\left(1-4x^2\right)\left(x-1\right)\\ =\left(1-2x\right)\left(1+2x\right)\left(x-1\right)\)
\(f,=\left(2x\right)^3+3.\left(2x\right)^2.1+3.2x.1^2+1^3\\ =\left(2x+1\right)^3\)
Phân tích các đa thức sau thành nhân tử:
\(B=x^8+2x^5-2x^4+x^2-2x-100+10x\left(x^4+x\right)+\left(5x-1\right)^2\)
\(B=x^8+2x^5-2x^4+x^2-2x-100+10x\left(x^4+x\right)+\left(5x-1\right)^2\)
\(=x^8+2x^5-2x^4+x^2-2x-100+10x^5+25x^2-10x+1\)
\(=x^8+12x^5-2x^4+36x^2-12x-99\)
\(=x^8+6x^5+9x^4+6x^5+36x^2+54x-11x^4-66x-99\)
\(=x^4\left(x^4+6x+9\right)+6x\left(x^4+6x+9\right)-11\left(x^4+6x+9\right)\)
\(=\left(x^4+6x+9\right)\left(x^4+6x-11\right)\)
Phân tích đa thức thành nhân tử :
a) C = ( x^2 - 2x + 3 )( x^2 - 2x + 5 ) - 8
b) D = x^8 + x^7 + 1
ủa phần a mình phân tích rồi mà bạn hu hu
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)
Phân tích các đa thức sau thành nhân tử:
a) \(^{4x^4+4x^3-x^2-x}\)
b) \(1-2a+2bc+a^2-b^2-c^2\)
c) \(\left(x-7\right)\left(x-5\right)\left(x-4\right)\left(x-2\right)-72\)
a) \(4x^3\left(x^2+x\right)-\left(x^2+x\right)=\left(x^2+x\right)\left(4x^3-1\right)\)
b)\(\left(1-2a+a^2\right)-\left(b^2-2bc+c^2\right)=\left(1-a\right)^2-\left(b-c\right)^2=\)\(\left(1-a+b-c\right)\left(1-a-b+c\right)\)
lm tiếp câu c
c) \(C=\left(x-7\right)\left(x-5\right)\left(x-4\right)\left(x-2\right)-72\)
\(=\left[\left(x-7\right)\left(x-2\right)\right]\left[\left(x-5\right)\left(x-4\right)\right]-72\)
\(=\left(x^2-9x+14\right)\left(x^2-9x+20\right)-72\)
Đặt \(x^2-9x+17=a\) ta có:
\(C=\left(a-3\right)\left(a+3\right)-72\)
\(=a^2-9-72\)
\(=a^2-81=\left(a-9\right)\left(a+9\right)\)
Thay trở lại ta được: \(C=\left(x^2-9x++8\right)\left(x^2-9x+26\right)\)
PHÂN TÍCH ĐA THỨC SAU THÀNH NHÂN TỬ BẰNG PHƯƠNG PHÁP ĐẶT BIẾN PHỤ
a) \(\left(x^2+4x+8\right)^2+3x\left(x^2+4x+8\right)+2x^2.\)
b) \(\left(x^2+x+1\right)\left(x^2+x+2\right)-12\)
c) \(\left(x^2+8x+7\right)\left(x^2+8x+15\right)+15\)
d) \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)
b)(x2+x+1)(x2+x+2)-12
Đặt t=x2+x+1
t(t+1)-12=t2+t-12
=(t-3)(t+4)=(x2+x+1-3)(x2+x+1+4)
=(x2+x-2)(x2+x+5)
=(x-1)(x+2)(x2+x+5)
c)(x2+8x+7)(x2+8x+15)+15
Đặt t=x2+8x+7
t(t+8)+15=t2+8t+15
=(t+3)(t+5)
=(x2+8x+7+3)(x2+8x+7+15)
=(x2+8x+10)(x2+8x+22)
d)(x+2)(x+3)(x+4)(x+5)-24
=(x2+7x+10)(x2+7x+12)-24
Đặt t=x2+7x+10
t(t+2)-24=(t-4)(t+6)
=(x2+7x+10-4)(x2+7x+10+6)
=(x2+7x+6)(x2+7x+16)
=(x+1)(x+6)(x2+7x+16)
a/ Đặt x2 + 4x + 8 = a
Thì đa thức ban đầu thành
a2 + 3ax + 2x2 = (a2 + 2ax + x2) + (ax + x2)
= (a + x)2 + x(a + x) = (a + x)(a + 2x)
b/ Đặt x2 + x + 1 = a thì đa thức ban đầu thành
a(a + 1) - 12 = a2 + a - 12 = (a2 - 3a) + (4a - 12)
= (a - 3)(a + 4)
phân tích đa thức thành nhân tử
a) x^7 + x^5 +1
b) x^8 + x^7 +1
c) x^8 + x^7 + 1
các bạn làm ơn giúp mik với!!!
Phân tích đa thức sau thành nhân tử
a) x7+ x2 + 1
b) x5 + x4 + 1
c) x8 + x + 1
d) x7 + x5+1
a) x7+ x2 + 1
=x7-x+x2+x+1
=x.(x6-1)+(x2+x+1)
=x.(x3-1)(x3+1)+(x2+x+1)
=x.(x-1)(x2+x+1)(x3+1)+(x2+x+1)
=(x2+x+1)[x.(x-1)(x3+1)+1]
=(x2+x+1)(x5+x2-x4-x+1)
b) x5 + x4 + 1
=x5+x4+x3+x2+x+1-x3-x2-x
=x3.(x2+x+1)+(x2+x+1)-x.(x2+x+1)
=(x2+x+1)(x3+1-x)
\(x^8+x+1\)
\(=x^8+x^7+x^6-x^7-x^6-x^5+x^5+x^4+x^3-x^4-x^3-x^2+x^2+x+1\)
\(=x^6\left(x^2+x+1\right)-x^5\left(x^2+x+1\right)+x^3\left(x^2+x+1\right)-x^2\left(x^2+x+1\right)+\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^6-x^5+x^3-x^2+1\right)\)
\(x^5+x^4+1\)
\(=x^5+x^4+x^3-x^3-x^2-x+x^2+x+1\)
\(=x^3\left(x^2+x+1\right)-x\left(x^2+x+1\right)+\left(x^2+x+1\right)\)
\(=\left(x^3-x+1\right)\left(x^2+x+1\right)\)
PHÂN TÍCH CÁC ĐA THỨC SAU THÀNH NHÂN TỬ
c) \(\left(x^2+3x+1\right)\left(x^2+3x+2\right)-6\)
d) \(\left(x^2+8x+7\right)\left(x+3\right)\left(x+5\right)+15\)
c) Đặt \(A=\left(x^2+3x+1\right)\left(x^2+3x+2\right)-6\)
Đặt \(x^2+3x+1,5=a\)
\(\Rightarrow A=\left(a-0,5\right)\left(a+0,5\right)-6\)
\(\Rightarrow A=a^2-0,25-6\)
\(\Rightarrow A=a^2-\frac{25}{4}\)
\(\Rightarrow A=\left(a-\frac{5}{2}\right)\left(a+\frac{5}{2}\right)\)
Thay \(a=x^2+3x+0,5\)vào A ta có :
\(A=\left(x^2+3x+0,5-\frac{5}{2}\right)\left(x^2+3x+0,5+\frac{5}{2}\right)\)
\(A=\left(x^2+3x-2\right)\left(x^2+3x+3\right)\)
c, Đặt \(x^2+3x+2=a\)
Ta có : \(\left(a-1\right)a-6=a^2-a-6=\left(a^2-3a\right)+\left(2a-6\right)\)
\(=a\left(a-3\right)+2\left(a-3\right)\)
\(=\left(a+2\right)\left(a-3\right)\)
\(=\left(x^2+3x+4\right)\left(x^2+3x-1\right)\)
Câu d làm tương tự .
Gợi ý : (x+3)(x+5) = x2 + 8x + 15
đặt bằng a rồi giải tiếp
d) Đặt \(B=\left(x^2+8x+7\right)\left(x+3\right)\left(x+5\right)+15\)
\(B=\left(x^2+8x+7\right)\left(x^2+5x+3x+15\right)+15\)
\(B=\left(x^2+8x+7\right)\left(x^2+8x+15\right)+15\)
Đặt \(a=x^2+8x+11\)
\(\Rightarrow B=\left(a-4\right)\left(a+4\right)+15\)
\(\Rightarrow B=a^2-16+15\)
\(\Rightarrow B=a^2-1\)
\(\Rightarrow B=\left(a-1\right)\left(a+1\right)\)
Thay \(a=x^2+8x+11\)vào B ta có :
\(B=\left(x^2+8x+11-1\right)\left(x^2+8x+11+1\right)\)
\(B=\left(x^2+8x+10\right)\left(x^2+8x+12\right)\)
Giup mk nhé
Phân tích đa thức sau thành nhân tử
a) x^4+16
b)64x^4+y^4
c)x^5-x^4-1
d)x^8+x^7+1
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/ \(64x^4+y^4\)
\(=64x^4+y^4+16x^2y^2-16x^2y^2\)
\(=\left(64x^4+y^4+16x^2y^2\right)-16x^2y^2\)
\(=\left(8x^2+y^2\right)^2-\left(4xy\right)^2\)
\(=\left(y^2+8x^2-4xy\right)\left(8x^2+y^2-4xy\right)\)