Phân tích đa thức thành nhân tử :
4x4 + 4x3 + 5x2 + 2x + 1
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Bài 1:
\(1,Sửa:x^3-2x^2+x=x\left(x^2-2x+1\right)=x\left(x-1\right)^2\\ 2,=6\left(x^2+2xy+y^2\right)=6\left(x+y\right)^2\\ 3,=2y\left(y^2+4y+4\right)=2y\left(y+2\right)^2\\ 4,=5\left(x^2-2xy+y^2\right)=5\left(x-y\right)^2\)
Bài 2:
\(1,=x\left(x^2-64\right)=x\left(x-8\right)\left(x+8\right)\\ 2,=2y\left(4x^2-9\right)=2y\left(2x-3\right)\left(2x+3\right)\\ 3,=3\left(x^3-1\right)=3\left(x-1\right)\left(x^2+x+1\right)\)
Bài 3:
\(a,=5\left(x^2+2x+1-y^2\right)=5\left[\left(x+1\right)^2-y^2\right]=5\left(x-y+1\right)\left(x+y+1\right)\\ b,=3x\left(x^2-2x+1-4y^2\right)=3x\left[\left(x-1\right)^2-4y^2\right]\\ =3x\left(x-2y-1\right)\left(x+2y-1\right)\\ c,=ab\left(a-b\right)\left(a+b\right)+\left(a+b\right)^2\\ =\left(a+b\right)\left(a^2b-ab^2+a+b\right)\\ d,=2x\left(x^2-y^2-4x+4\right)=2x\left[\left(x-2\right)^2-y^2\right]\\ =2x\left(x-y-2\right)\left(x+y-2\right)\)
1) \(x^2-7x+6=x^3+1-7x-7=\left(x^3+1\right)-7\left(x+1\right)=\left(x+1\right)\left(x^2-x-6\right)\)
2) \(x^3-9x^2+6x+16\)
\(\left(x^3+1\right)-\left[\left(9x^2-6x+1\right)-16\right]\)
\(=\left(x^3+1\right)-\left[\left(3x-1\right)^2-16\right]=\left(x^3+1\right)-\left(3x-1+4\right)\left(3x-1-4\right)\)\(=\left(x^3+1\right)-3\left(3x-5\right)\left(x+1\right)\)\(=\left(x+1\right)\left[x^2-x+1-9x+15\right]=\left(x+1\right)\left(x^2-10x+16\right)\)
\(=\left(x+1\right)\left[x\left(x-2\right)-8\left(x-2\right)\right]\)\(\left(x+1\right)\left(x-2\right)\left(x-8\right)\)
3) \(x^3-6x^2-x+30\)
\(=x^3-5x^2-x^2+5x-6x+30\)
\(=x^2\left(x-5\right)-x\left(x-5\right)-6\left(x-5\right)\)
\(=\left(x-5\right)\left(x^2-x-1\right)\)
4) \(2x^3-x^2+5x+3=\left(2x^3+x^2\right)-\left(2x^2+x\right)+\left(6x+3\right)\)
\(=x^2\left(2x+1\right)-x\left(2x+1\right)+3\left(2x+1\right)\)
\(=\left(2x+1\right)\left(x^2-x+3\right)\)
5) \(27x^3-27x^2+18x-4=\left(27x^3-1\right)-\left(27x^2-18x+3\right)\)
\(=\left(3x-1\right)\left(9x^2+3x+1\right)-3\left(9x^2-6x+1\right)\)
\(=\left(3x-1\right)\left(9x^2+3x+1\right)-3\left(3x-1\right)^2\)
\(=\left(3x-1\right)\left(9x^2+3x+1-9x+3\right)=\left(3x-1\right)\left(9x^2-6x+4\right)\)
gửi phần này trước còn lại làm sau !!! tk mk nka !!!
Bài 1:
a: \(11x^2-6xy-5y^2\)
\(=11x^2-11xy+5xy-5y^2\)
\(=11x\left(x-y\right)+5y\left(x-y\right)\)
\(=\left(x-y\right)\left(11x+5y\right)\)
b: \(4x^3-16x^2+19x-6\)
\(=4x^3-8x^2-8x^2+16x+3x-6\)
\(=\left(x-2\right)\left(4x^2-8x+3\right)\)
\(=\left(x-2\right)\left(2x-1\right)\left(2x-3\right)\)
Bài 1:
a: \(11x^2-6xy-5y^2\)
\(=11x^2-11xy+5xy-5y^2\)
\(=11x\left(x-y\right)+5y\left(x-y\right)\)
\(=\left(x-y\right)\left(11x+5y\right)\)
b: \(4x^3-16x^2+19x-6\)
\(=4x^3-8x^2-8x^2+16x+3x-6\)
\(=\left(x-2\right)\left(4x^2-8x+3\right)\)
\(=\left(x-2\right)\left(2x-3\right)\left(2x-1\right)\)
\(a,=11x^2-11xy+5xy-5y^2=\left(11x+5y\right)\left(x-y\right)\\ b,=4x^3-8x^2-8x^2+16x+3x-6\\ =\left(x-2\right)\left(4x^2-8x+3\right)\\ =\left(x-2\right)\left(4x^2-2x-6x+3\right)\\ =\left(x-2\right)\left(2x-1\right)\left(2x-3\right)\)
Bài 1:
a: \(11x^2-6xy-5y^2\)
\(=11x^2-11xy+5xy-5y^2\)
\(=11x\left(x-y\right)+5y\left(x-y\right)\)
\(=\left(x-y\right)\left(11x+5y\right)\)
b: \(4x^3-16x^2+19x-6\)
\(=4x^3-8x^2-8x^2+16x+3x-6\)
\(=\left(x-2\right)\left(4x^2-8x+3\right)\)
\(=\left(x-2\right)\left(2x-1\right)\left(2x-3\right)\)
c) \(5x^2+3y+15x+xy=5x\left(x+3\right)+y\left(x+3\right)=\left(x+3\right)\left(5x+y\right)\)
d) \(x^2+6x+9-y^2=\left(x+3\right)^2-y^2=\left(x+3-y\right)\left(x+3+y\right)\)
e) \(x^2-y^2+2x+1=\left(x^2+2x+1\right)-y^2=\left(x+1\right)^2-y^2=\left(x+1-y\right)\left(x+1+y\right)\)
f) \(x^2-2xy-9+y^2=\left(x^2-2xy+y^2\right)-9=\left(x-y\right)^2-3^2=\left(x-y-3\right)\left(x-y+3\right)\)
c: \(5x^2+15x+3y+xy\)
\(=5x\left(x+3\right)+y\left(x+3\right)\)
\(=\left(x+3\right)\left(5x+y\right)\)
d: \(x^2+6x+9-y^2\)
\(=\left(x+3\right)^2-y^2\)
\(=\left(x+3-y\right)\left(x+3+y\right)\)
e: \(x^2+2x+1-y^2\)
\(=\left(x+1\right)^2-y^2\)
\(=\left(x+1-y\right)\left(x+1+y\right)\)
f: \(x^2-2xy+y^2-9\)
\(=\left(x-y\right)^2-9\)
\(=\left(x-y-3\right)\left(x-y+3\right)\)
Lời giải:
$\frac{x}{y}$ không phải đơn thức bạn nhé.
a. $x^2-2x+1=(x-1)^2$
b. $x^2+2xy-25+y^2=(x^2+2xy+y^2)-25=(x+y)^2-5^2=(x+y-5)(x+y+5)$
c. $5x^2-10xy=5x(x-2y)$
d. $x^2-y^2+x-y=(x^2-y^2)+(x-y)=(x-y)(x+y)+(x-y)$
$=(x-y)(x+y+1)$
\(=-5x^2+15x+x-3=-5x\left(x-3\right)+\left(x-3\right)=\left(1-5x\right)\left(x-3\right)\)
x 4 - 5 x 2 + 4 = x 4 - 4 x 2 - x 2 + 4 = x 4 - 4 x 2 - x 2 - 4 = x 2 x 2 - 4 - x 2 - 4 = x 2 - 4 x 2 - 1 = x + 2 x - 2 x + 1 x - 1
ta có :4.x^4 +4.x^3 +5.x^2 +2x+1=(2.x^2)^2 +2.2.(x^2).x +x^2 +4.x^2 +2x +1
=(2.x^2 +x)^2 +2.(2x^2 +x) .1 +1
=(2.x^2 +x +1)^2