Phân tích đa thức sau thành nhân tử:
a3m+2a2m+am
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\(9a^3-13a+6=\left(9a^3-6a^2\right)+\left(6a^2-4a\right)-\left(9a-6\right)=3a^2\left(3a-2\right)+2a\left(3a-2\right)-3\left(3a-2\right)=\left(3a-2\right)\left(3a^2+2a-3\right)\)
\(x^4-4x^3+8x+3=\left(x^4+x^3\right)-\left(5x^3+5x^2\right)+\left(5x^2+5x\right)+\left(3x+3\right)=x^3\left(x+1\right)-5x^2\left(x+1\right)+5x\left(x+1\right)+3\left(x+1\right)=\left(x+1\right)\left(x^3-5x^2+5x+3\right)=\left(x+1\right)\left[\left(x^3-3x^2\right)-\left(2x^2-6x\right)-\left(x-3\right)\right]=\left(x+1\right)\left(x-3\right)\left(x^2-2x-1\right)\)
Câu 1)
\(a\left(a+2\right)+b\left(b-2\right)-2ab\)
\(=a^2+2a+b^2-2b-2ab\)
\(=\left(a^2-2ab+b^2\right)+\left(2a-2b\right)\)
\(=\left(a-b\right)^2+2\left(a-b\right)\)
\(=7^2-2.7=35\)
Câu 2)
a) \(a^3m+2a^2m+am\)
\(=am\left(a^2+2a+1\right)\)
\(=am\left(a+1\right)^2\)
b) \(x^8+x^4+1\)
\(=x^8+2x^4+1-x^4\)
\(=\left(x^4+1\right)^2-x^4\)
\(=\left(x^4+1-x^2\right)\left(x^4+x^2+1\right)\)
\(=\left(x^4-x^2+1\right)\left(x^2+x+1\right)\left(x^2-x+1\right)\)
Nha ~ mình không biết đúng sai nhưng mà cảm ơn bạn nhiều lắm nha ~ <3
1A:
a: \(x^3+2x=x\left(x^2+2\right)\)
b: \(3x-6y=3\left(x-2y\right)\)
c: \(5\left(x+3y\right)-15x\left(x+3y\right)\)
\(=5\left(x+3y\right)\left(1-3x\right)\)
d: \(3\left(x-y\right)-5x\left(y-x\right)\)
\(=3\left(x-y\right)+5x\left(x-y\right)\)
\(=\left(x-y\right)\left(5x+3\right)\)
1A. a. x(x2+2)
b. 3(x-2y)
c. 5(x+3y)(1-3x)
d. (x-y) (3-5x)
1B. a. 2x(2x-3)
b.xy(x2-2xy+5)
c. 2x(x+1)(x+2)
d. 2x(y-1)+2y(y-1)=2(y-1)(x-y)
Bài 1:
a: \(4a^2-6b=2\left(2a^2-3b\right)\)
b: \(m^3n-2m^2n^2-mn\)
\(=mn\left(m^2-2mn-1\right)\)
Bài 1:
a) \(4a^2-6b=2\left(a^2-3b\right)\)
b) \(=mn\left(m^2-2mn-1\right)\)
Bài 2:
a) \(=4\left(u-2\right)^2+v\left(u-2\right)=\left(u-2\right)\left(4u-8+v\right)\)
b) \(=a\left(a-b\right)^3-b\left(a-b\right)^2-b^2\left(a-b\right)=\left(a-b\right)\left[a\left(a-b\right)^2-b\left(a-b\right)-b^2\right]=\left(a-b\right)\left(a^3-2a^2b+ab^2-ab+b^2-b^2\right)=\left(a-b\right)\left(a^3-2a^2b+ab^2-ab\right)\)
21) \(=ab\left(x-5\right)+a^2\left(x-5\right)=a\left(x-5\right)\left(a+b\right)\)
22) \(=2a^2\left(x-y\right)+4a\left(x-y\right)=2a\left(x-y\right)\left(a+2\right)\)
23) \(a\left(x-3\right)+a^2\left(x-3\right)=a\left(x-3\right)\left(a+1\right)\)
24) \(=5x^2y\left(x-7\right)+5xy\left(x-7\right)=5xy\left(x-y\right)\left(x+1\right)\)
25) \(=2xy\left(a-1\right)+4x^2y\left(a-1\right)=2xy\left(a-1\right)\left(2x+1\right)\)
26) \(=4a\left(x-3\right)+2\left(x-3\right)=2\left(x-3\right)\left(2a+1\right)\)
27) \(=x^m\left(x-1\right)\)
28) \(=x^m\left(x+1\right)\)
29) \(=x^m\left(x^2-1\right)\)
30) \(=x^{m+1}\left(x-1\right)\)
a3m+2a2m+am = am(a2m+2am+1) = am[(am)2+2am+1] = am(am+1)2
Ta có :
a3m+2a2m+am
= am(a2m+2am+1)
= am[(am)2+2am+1]
= am(am+1)2