a3+b3+c3-3abc
x2-2xy+y2+3x-3y-10
2x3-12x2+17x-2
x4-6x3+12x2-14x+3
8(x+y+z)3-(x+y)3-(y+z)3-(z+x)3
phân tích ạ
Bài 2 Phân tích đa thức sau thành nhân tử
a. x4 + 2x3 − 4x − 4
b. x2(1 − x2) − 4 − 4x2
c. x2 + y2 − x2y2 + xy − x − y
d* a3 + b3 + c3 − 3abc
a) \(x^4+2x^3-4x-4=\left(x^4+2x^3+x^2\right)-\left(x^2+4x+4\right)\)
\(=\left(x^2+x\right)^2-\left(x+2\right)^2=\left(x^2+x-x-2\right)\left(x^2+x+x+2\right)\)
\(=\left(x^2-2\right)\left(x^2+2x+2\right)\)
a) Ta có: \(x^4+2x^3-4x-4\)
\(=\left(x^4+2x^3+x^2\right)-\left(x^2+4x+4\right)\)
\(=\left(x^2+x\right)^2-\left(x+2\right)^2\)
\(=\left(x^2+x-x-2\right)\left(x^2+x+x+2\right)\)
\(=\left(x^2-2\right)\cdot\left(x^2+2x+2\right)\)
d) Ta có: \(a^3+b^3+c^3-3abc\)
\(=\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc\)
\(=\left(a+b+c\right)\left[\left(a+b\right)^2-c\left(a+b\right)+c^2\right]-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ac-bc+c^2-3ab\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)\)
1. a3 b3 c3 3abc2. a10 a5 13. a8 a 14. a8 a7 15. a16 a8b8 b166. a 1 a 3 a 5 a 7 157. 4x2y2 2x y y2z2 z y x2z2 2x z 8. be a b b c ac b d a c ab c d a b 9. x y 3 y z 3 z x 310. x4 6x3 7x2 6x 1
3 3 3 3 3 3 3 3 3 3 3 3 3
phân tích đa thức:
x4 + 2021x2 + 2020x + 2021
a(b2 - c2) + b(c2 - a2) + c(a2 - b2)
a3(b - c) + b3(c - a) + c3(a - b)
(x + y + z)3 - (x + y - z)3 - (x - y + z)3 - (-x + y + z)3
b) Ta có: \(a\left(b^2-c^2\right)+b\left(c^2-a^2\right)+c\left(a^2-b^2\right)\)
\(=ab^2-ac^2+bc^2-ba^2+ca^2-cb^2\)
\(=\left(ab^2-cb^2\right)+\left(ca^2-c^2a\right)+\left(bc^2-ba^2\right)\)
\(=b^2\left(a-c\right)+ca\left(a-c\right)+b\left(c^2-a^2\right)\)
\(=\left(a-c\right)\left(b^2+ca\right)-b\left(a-c\right)\left(a+c\right)\)
\(=\left(a-c\right)\left(b^2+ca-ba-bc\right)\)
\(=\left(a-c\right)\left[b\left(b-a\right)+c\left(a-b\right)\right]\)
\(=\left(a-c\right)\left[b\left(b-a\right)-c\left(b-a\right)\right]\)
\(=\left(a-c\right)\left(b-a\right)\left(b-c\right)\)
trời ơi cái qq gì í đây
Phân tích đa thức thành nhân tử
1) 2x3–x2+5x+3
2) 27x3−27x2+18x–427x3−27x2+18x–4
3) x2+2xy+y2−x−y–12x2+2xy+y2−x−y–12
4) (x+2)(x+3)(x+4)(x+5)–24(x+2)(x+3)(x+4)(x+5)–24
5) 4x4−32x2+14x4−32x2+1
6) 3(x4+x2+1)−(x2+x+1)23(x4+x2+1)−(x2+x+1)2
7) 64x4+y4
Phân tích đa thức thành nhân tử:
A= x.(y2 - z2) + y.(z2 - x2) + z.(x2 - y2).
B= a.(b3 - c3) + b.(c3 - a3) + c.(a3 - b3).
C= ab.(a + b) - bc.(b + c) + ac. (a - c).
\(A=x\left(y^2-z^2\right)+y\left(z^2-x^2\right)+z\left(x^2-y^2\right)=x\left(y^2-z^2\right)+y\left(-y^2+z^2-x^2+y^2\right)+z\left(x^2-y^2\right)=\left(y^2-z^2\right)\left(x-y\right)+\left(x^2-y^2\right)\left(z-y\right)=\left(y-z\right)\left(y+z\right)\left(x-y\right)-\left(x-y\right)\left(x+y\right)\left(y-z\right)=\left(x-y\right)\left(y-z\right)\left(y+z-x-y\right)=\left(x-y\right)\left(y-z\right)\left(z-x\right)\)
\(B=a\left(b^3-c^3\right)+b\left(c^3-a^3\right)+c\left(a^3-b^3\right)=ab^3-ac^3+bc^3-a^3b+a^3c-b^3c=ab\left(b^2-a^2\right)-c^3\left(a-b\right)+c\left(a^3-b^3\right)=-ab\left(a-b\right)\left(a+b\right)-c^3\left(a-b\right)+c\left(a-b\right)\left(a^2+ab+b^2\right)=\left(a-b\right)\left(-a^2b-ab^2-c^3+a^2c+abc+b^2c\right)\)
\(C=ab\left(a+b\right)-bc\left(b+c\right)+ac\left(a-c\right)=ab\left(a+b\right)-bc\left(a+b-a+c\right)+ac\left(a-c\right)=ab\left(a+b\right)-bc\left(a+b\right)+bc\left(a-c\right)+ac\left(a-c\right)=b\left(a+b\right)\left(a-c\right)+c\left(a-c\right)\left(a+b\right)=\left(a+b\right)\left(c+c\right)\left(a-c\right)\)
Phân tích các đa thức sau thành nhân tử:
a) 3x-6
b) x2-4x+4
c) x2-y2+5x-5y
d) 8(x+y+z)3-(x+y)3-(y+z)3-(y+z)3-(z+x)3
Do câu d mình ko biết làm bởi v mình không làm được
Phân tích đa thức thành nhân tử:
a) x 4 - 6 x 3 + 12 x 2 - 14x + 3.
b) x 4 + 6 x 3 + 7 x 2 -6x + l.
a) ( x 2 – 4x + 1)( x 2 – 2x + 3).
b) ( x 2 + 5x – 1)( x 2 + x – 1).
Bài 1 :Phân tích đa thức sau thành nhân tử
(12x2+6x)(y+z)+(12x2+6x)(y-z)
Bài 2:tìm x:
x(x-6)+10(x-6)=0
1.
\(\left(12x^2+6x\right)\left(y+z\right)+\left(12x^2+6x\right)\left(y-z\right)\\ =\left(12x^2+6x\right)\left(y+z+y-z\right)\\ =2y\left(12x^2+6x\right)\\ =2y.6x\left(2x+1\right)\\ =12xy\left(2x+1\right)\)
2.
\(x\left(x-6\right)+10\left(x-6\right)=0\\ \Leftrightarrow\left(x-6\right)\left(x+10\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=6\\x=-10\end{matrix}\right.\)
Vậy \(x\in\left\{6;-10\right\}\) là nghiệm của pt
Bài 1:
Ta có: \(\left(12x^2+6x\right)\left(y+z\right)+\left(12x^2+6x\right)\left(y-z\right)\)
\(=\left(12x^2+6x\right)\left(y+z+y-z\right)\)
\(=6x\left(2x+1\right)\cdot2y\)
\(=12xy\left(2x+1\right)\)
Bài 2:
Ta có: \(x\left(x-6\right)+10\left(x-6\right)=0\)
\(\Leftrightarrow\left(x-6\right)\left(x+10\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-10\end{matrix}\right.\)
Cứu với ạ
Làm tính chia
1) (x3 – 3x2 + x – 3) : (x – 3) 2) (2x4 – 5x2 + x3 – 3 – 3x) : (x2 – 3)
3) (x – y – z)5 : (x – y – z)3 4) (x2 + 2x + x2 – 4) : (x + 2)
5) (2x3 + 5x2 – 2x + 3) : (2x2 – x + 1) | 6) (2x3 – 5x2 + 6x – 15):(2x – 5) |