tính phep nhan x phan x +2-x^3 phan x^3+8 . x^2-2x+4 phần x^2 -4
phan tich cac da thuc sau thanh nhan tu
x^2-x-12
x^2+8x+15
x^3-x^2+x+3
x^8+3x^4+4
x^6-x^4-2x^3+2x^2
c)\(x^3-x^2+x+3=x^2+x-2x^2-2x+3x+3\)
\(=x\left(x+1\right)-2x\left(x+1\right)+3\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-2x+3\right)\)
d)\(x^8+3x^4+4=\left(x^8+4x^4+4\right)-x^4=\left(x^4+2\right)^2-\left(x^2\right)^2\)
\(=\left(x^4-x^2+2\right)\left(x^4+x^2+2\right)\)
e)\(x^6-x^4-2x^3+2x^2=x^4\left(x^2-1\right)-2x^2\left(x-1\right)=x^4\left(x-1\right)\left(x+1\right)-2x^2\left(x-1\right)\)
\(=x^2\left(x-1\right)\left(x^3+x^2\right)-2x^2\left(x-1\right)=x^2\left(x-1\right)\left(x^3+x^2-2\right)\)
\(=x^2\left(x-1\right)\left[\left(x^3-1\right)+\left(x^2-1\right)\right]=x^2\left(x-1\right)\left[\left(x-1\right)\left(x^2+x+1\right)+\left(x-1\right)\left(x+1\right)\right]\)
\(=x^2\left(x-1\right)\left(x-1\right)\left(x^2+2x+2\right)=x^2\left(x-1\right)^2\left(x^2+2x+2\right)\)
a)\(x^2-x-12\)
\(=x^2+4x-3x-12\)
\(=x\left(x+4\right)-3\left(x+4\right)\)
\(=\left(x+4\right)\left(x-3\right)\)
b) \(x^2+8x+15\)
\(=x^2+3x+5x+15\)
\(=x\left(x+3\right)+5\left(x+3\right)\)
\(=\left(x+3\right)\left(x+5\right)\)
phan tich da thuc thanh nhan tu 2x(x + 3) + 2(x + 3)
tìm x
a) 5x(x – 2) – x – 2 = 0
b) 4x(x + 1) = 8( x + 1)
c) x(2x + 1) + 1/2 x 3/3 - = 0
d) x(x – 4) + (x – 4)^2 = 0
Bai 1 Tim x
A) x nhan 4 phan 5 = 1 phan 2 nhan 1 phan 3
B ) x : 2 phan 5 = 2phan 3 + 1 phan 2
phan tich da thuc thanh nhan tu
x^4+x^3+2x^2+x+1
x4+x3+2x2+x+1=x4+x3+x2+x2+x+1=(x4+x3+x2)+(x2+x+1)
=x2(x2+x+1)+(x2+x+1)
=(x2+x+1)(x2+1)
=(x^4+2x^2+1)+(x^3+x)
=(x^2+1)^2+x(x^2+1)
(x^+1)*(x^2+1+x0
phan tich da thuc thanh nhan tu; x4-2x2-8
ap dung tim x biet (x4-2x2-8):(x-2)=0
\(x^4-2x^2+8=x^4+2x^2-4x^2+8=\left(x^2-4\right)\left(x^2+2\right)=\left(x-2\right)\left(x+2\right)\left(x^2+2\right)\)\(\left(x^4-2x^2-8\right):\left(x-2\right)=\left(x+2\right)\left(x^2+2\right)=0\)
\(\Rightarrow x=-2\)
Phan tich da thuc sau thanh nhan tu : (x+1)(x+2)(x+3)(x+4)-8
Gợi ý:
Nhóm:\(\left[\left(x+1\right)\left(x+4\right)\right]\left[\left(x+2\right)\left(x+3\right)\right]-8\)
\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-8\)
Đặt \(t=x^2+5x+4\) thì biểu thức trở thành:
\(t\left(t+2\right)-8=t^2+2t-8=\left(t-2\right)\left(t+4\right)\)
Rồi bạn làm tiếp, nếu còn phân tích được thì phải phân tích, mình bận rồi.
(x + 1)(x + 2)(x + 3)(x + 4) - 8
= [(x + 1)(x + 4)][(x + 2)(x + 3)] - 8
= (x2 + 4x + x + 4)(x2 + 3x + 2x + 6) - 8
= (x2 + 5x + 4)(x2 + 5x + 6) - 8
Đặt x2 + 5x + 5 = t
⇒ (x2 + 5x + 5 - 1)(x2 + 5x + 5 + 1) - 8 (1)
Thay t = x2 + 5x + 5 vào (1), ta có:
(t - 1)(t + 1) - 8 = t2 - 1 - 8 = t2 - 9
= (t - 3)(t + 3)
⇔ (x2 + 5x + 5 - 3)(x2 + 5x + 5 + 3)
= (x2 + 5x + 2)(x2 + 5x + 8)
Chúc bạn học tốt !!!!!!!!
(x+1)(x+2)(x+3)(x+4)-8
= [(x+1)(x+4)][(x+2)(x+3)]-8
= (x2+4x+x+4)(x2+3x+2x+6)-8
= (x2+5x+5-1)(x2+5x+5+1)-8
= (x2+5x+5)2-12-8
= (x2+5x+5)2-9
= (x2+5x+5) -32
= (x2+5x+5-3)(x2+5x+5+3) {HĐT số 3}
= (x2+5x+2)(x2+5x+8)
Cach phan tich da thuc thanh nhan tu
a)X^3+2X^2+2X+1
b)X^3-4X^2+12X-27
c)a^6-a^4+2a^3+2a^2
d)x^4+2x^3+2x^2+2x+1
e)x^5+x^4+x^3+x^2+x+1
a)
=x3+x2+x2+x+x+1
=x2(x+1)+x(x+1)+(x+1)
=(x+1)(x2+x+1)
b)
=x3-3x2-x2+3x+9x-27
=x2(x-3)-x(x-3)+9(x-3)
=(x-3)(x2-x+9)
a)
=x3+x2+x2+x+x+1
=x2(x+1)+x(x+1)+(x+1)
=(x+1)(x2+x+1)
b)
=x3-3x2-x2+3x+9x-27
=x2(x-3)-x(x-3)+9(x-3)
=(x-3)(x2-x+9)
d)x^4+2x^3+2x^2+2x+1(no)
e)x^5+x^4+x^3+x^2+x+1 ( no)
Phan tich da thuc thanh nhan tu
1) (2x+1)^2 - (x-1)^2
2) 9(x+3)^2 - 4(x-2)^2
3) 25(2x - y) ^2 - 16(x+2y)^2
4) x^4 + x^3 +x + 1
5) x^3 + 3x^2 + 3x + 1 -8y^3
phan tich da thuc thanh nhan tu:x^4+x^3+2x^2+x+1
\(x^4+x^3+2x^2+x+1=x^4+x^2+x^3+x+x^2+1\)
\(=x^2\left(x^2+1\right)+x\left(x^2+1\right)+1\left(x^2+1\right)\)
\(=\left(x+1\right)\left(x^2+x+1\right)\)
cái cuối là \(\left(x^2+1\right)\left(x^2+x+1\right)\)