phan tich da thuc sau thanh nhan tu
x^3+3x^2+4x+2
6x^4-x^3-7x^2+x+1
giup minh nhe cam on nhieu
phan tich da thuc sau thanh nhan tu
x^3+2x^2+x+2
giup minh nhe cam on nhieu
Ta có:
\(x^3+2x^2+x+2\)
\(=x^2\left(x+2\right)+\left(x+2\right)\)
\(=\left(x^2+1\right)\left(x+2\right)\)
x3+2x2+x+2=x2(x+2)+(x+2)
=(x2+1)(x+2)
Phan tich da thuc thanh nhan tu
A^3 +a^2b-a^2c-abc
A^4+a^3-a^2-a
Tim x biet
X^2+5x+6=0
2(x+3)-x^2-3x=0
X^2+4x+3=0
Giup e nhe mn e cam on nhieu ạ lop 8 . e can gấp giup nhé
phan tich da thuc thanh nhan tu
1) X3+ 6x2 + 12xy + 8
2) x4- 4x3- 8x2 + 8x
3) x4+ 2x3 + x2- y2
giup minh nhe cam on
1.\(x^3+6x^2+12xy+8=x^3+3.2x^2+3.2^2x+2^3=\left(x+2\right)^3\)
3.\(x^4+2x^3+x^2-y^2=\left(x^2\right)^2+2x^2.x+x^2-y^2\)\(=\left(x^2+x\right)^2-y^2=\left(x^2+x-y\right)\left(x^2+x+y\right)\)
k mình nha bn !!!!!!! cái 2 bn xem lại đề đi, rồi mình giải cho
cau so 2 ne ban
x4 - 4x3- 8x2+ 8x
OK bn, 2.\(x^4-4x^3-8x^2+8x=x\left(x^3-4x^2-8x+8\right)\)
z thui bn !!! hì ^.^
phan tich da thuc thanh nhan tu : a) 3x^2 - 22xy + 4x + 8y + 7x^2 + 1 ; b) 12x^2 + 5x - 12y^2 + 12y - 10xy - 3 ; c)x^4 + 6x^3 + 11x^2 + 6x + 1
phan tich da thuc sau thanh nhan tu bang cach ha bac-tách để xuất hiện ntc là (x-4)
x^4-4x^3-7x^2+22x+24
\(x^4-4x^3-7x^2+22x+24\)
\(=\left(x^4-4x^3\right)-\left(7x^2-28x\right)-\left(6x-24\right)\)
\(=x^4.\left(x-4\right)-7x.\left(x-4\right)-6.\left(x-4\right)\)
\(=\left(x-4\right).\left(x^4-7x-6\right)\)
Tham khảo nhé~
phan tich da thuc thanh nhan tu
x(x-4)+5x-20
\(=x\left(x-4\right)+5\left(x-4\right)=\left(x+5\right)\left(x-4\right)\)
Phan tich da thuc thanh nhan tu
a) 3x2 - 16x + 5
b) x3 - 7x + 6
c) 3x3 - 4x2 + 13x - 4
d) x2 + 7x + 12
phan tich cac da thuc sau thanh nhan tu a)x^2+4x+3 b) 4x^2+4x-3 c) x^2-x-12 d)4x^4+4x^2y^2-8y^4
a) x^2+4x+3=x^2+x+3x+3=x(x+1)+3(x+1)=(x+1)(x+3)
b) 4x^2+4x-3=4x^2+4x+1-4=(2x+1)^2-4=(2x+1-2)(2x+1+2)=(2x-1)(2x+3)
c) x^2-x-12=x^2-4x+3x-12=x(x-4)+3(x-4)=(x-4)(x+3)
d) 4x^4+4x^2y^2-8y^4=4(x^4+x^2y^2-2y^4)=4(x^4-x^2y^2+2x^2y^2-2y^4)=4(x^2-y^2)(x^2+2y^2)=4(x-y)(x+y)(x^2+2y^2)
a) \(x^2+4x+3\)
\(=x^2+x+3x+3\)
\(=\left(x^2+x\right)+\left(3x+3\right)\)
\(=x\left(x+1\right)+3\left(x+1\right)\)
\(=\left(x+1\right)\left(x+3\right)\)
c) \(x^2-x-12\)
\(=x^2-4x+3x-12\)
\(=\left(x^2-4x\right)+\left(3x-12\right)\)
\(=x\left(x-4\right)+3\left(x-4\right)\)
\(=\left(x-4\right)\left(x+3\right)\)
\(x^2+4x+3\)
\(=x^2+x+3x+3\)
\(=x\left(x+1\right)+3\left(x+1\right)\)
\(=\left(x+1\right)\left(x+3\right)\)
Rut gon phan thuc: (x^(3)-x^(2)-x-2)/(x^(5)-3x^(4)+4x^(3)-5x^(2)+3x-2)
Minh that su da bo tay vs bai tap nay roi! Cac ban hay giup minh nhe! Minh xin cam on!
\(\frac{x^3-x^2-x-2}{x^5-3x^4+4x^3-5x^2+3x-2}\)
\(=\frac{x^3-2x^2+x^2-2x+x-2}{x^5-2x^4-x^4+2x^3+2x^3-4x^2-x^2+2x+x-2}\)
\(=\frac{\left(x^3-2x^2\right)+\left(x^2-2x\right)+\left(x-2\right)}{\left(x^5-2x^4\right)-\left(x^4-2x^3\right)+\left(2x^3-4x^2\right)-\left(x^2-2x\right)+\left(x-2\right)}\)
\(=\frac{x^2\left(x-2\right)+x\left(x-2\right)+\left(x-2\right)}{x^4\left(x-2\right)-x^3\left(x-2\right)+2x^2\left(x-2\right)-x\left(x-2\right)+\left(x-2\right)}\)
\(=\frac{\left(x-2\right)\left(x^2+x+1\right)}{\left(x-2\right)\left(x^4-x^3+2x^2-x+1\right)}=\frac{x^2+x+1}{x^4-x^3+2x^2-x+1}\)