Phân tích đa thức thành nhân tử
1) 105ax - 90ay - 63bx + 54by
2) 7x2 + 13x + 6
3) 5x2 - 11x + 6
4) x2 - 3x - 10
5) x2 + x - 42
6) 6x2 - 7x - 3
7) 15x2 + 29x - 14
bài tập thêm bớt hạng tử
1) 4x2+16x-9
2) 6x2+7x+2
3) -5x2-29x-20
4)-7x2+11x+6
\(1,=4x^2-2x+18x-9=2x\left(x-2\right)+9\left(x-2\right)=\left(2x+9\right)\left(x-2\right)\\ 2,=6x^2+3x+4x+2=3x\left(2x+1\right)+2\left(2x+1\right)=\left(3x+2\right)\left(2x+1\right)\\ 3,=-\left(5x^2+4x+25x+20\right)=-\left[x\left(5x+4\right)+5\left(5x+4\right)\right]=-\left(x+5\right)\left(5x+4\right)\\ 4,=-\left(7x^2-14x+3x-6\right)=-\left[7x\left(x-2\right)+3\left(x-2\right)\right]=-\left(7x+3\right)\left(x-2\right)\\ =\left(7x+3\right)\left(2-x\right)\)
1. Phân tích thành nhân tử
A) x4 + 2x3 + x2
B) x3 - x + 3x2y + 3xy2 + y3 - y
C) 5x2 - 10xy +5y2 - 20z2
2. Phân tích thành nhân tử
A) x2 + 5x -6
B) 5x2 + 5xy - x - y
C) 7x - 6x2 - 2
3.Phân tích thành nhân tử
A) x2 + 4 + 3
B) 2x2 + 3x -5
C) 16x - 5x2 - 3
4. Tìm x, bt
A) 5x ( x - 1 ) = x -1
B) 2( x + 5 ) -x2 - 5x = 0
Bài 2:
a: \(x^2+5x-6=\left(x+6\right)\left(x-1\right)\)
b: \(5x^2+5xy-x-y\)
\(=5x\left(x+y\right)-\left(x+y\right)\)
\(=\left(x+y\right)\left(5x-1\right)\)
c:\(-6x^2+7x-2\)
\(=-6x^2+3x+4x-2\)
\(=-3x\left(2x-1\right)+2\left(2x-1\right)\)
\(=\left(2x-1\right)\left(-3x+2\right)\)
1.
a) \(=x^2\left(x^2+2x+1\right)=x^2\left(x+1\right)^2\)
b) \(=\left(x+y\right)^3-\left(x+y\right)=\left(x+y\right)\left[\left(x+y\right)^2-1\right]\)
\(=\left(x+y\right)\left(x+y-1\right)\left(x+y+1\right)\)
c) \(=5\left[\left(x^2-2xy+y^2\right)-4z^2\right]=5\left[\left(x-y\right)^2-4z^2\right]\)
\(=5\left(x-y-2z\right)\left(x-y+2z\right)\)
2.
a) \(=x\left(x+2\right)+3\left(x+2\right)=\left(x+2\right)\left(x+3\right)\)
b) \(=5x\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(5x-1\right)\)
c) \(=-\left[3x\left(2x-1\right)-2\left(2x-1\right)\right]=-\left(2x-1\right)\left(3x-2\right)\)
3.
b) \(=2x\left(x-1\right)+5\left(x-1\right)=\left(x-1\right)\left(2x+5\right)\)
c) \(=-\left[5x\left(x-3\right)-1\left(x-3\right)\right]=-\left(x-3\right)\left(5x-1\right)\)
4.
a) \(\Rightarrow\left(x-1\right)\left(5x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{5}\end{matrix}\right.\)
b) \(\Rightarrow2\left(x+5\right)-x\left(x+5\right)=0\)
\(\Rightarrow\left(x+5\right)\left(2-x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)
Phân tích đa thức thành nhân tử
1)x^2-3x+2
2)x^2-x-6
3)x^2+7x+12
1: \(x^2-3x+2=\left(x-1\right)\left(x-2\right)\)
2: \(x^2-x-6=\left(x-3\right)\left(x+2\right)\)
3: \(x^2+7x+12=\left(x+3\right)\left(x+4\right)\)
Phân tích đa thức thành nhân tử
1)x^2-3x+2
2)x^2-x-6
3)x^2+7x+12
1) \(x^2-3x+2=\left(x^2-x\right)-\left(2x-2\right)=x\left(x-1\right)-2\left(x-1\right)=\left(x-1\right)\left(x-2\right)\)
2) \(x^2-x-6=\left(x^2-3x\right)+\left(2x-6\right)=x\left(x-3\right)+2\left(x-2\right)=\left(x-2\right)\left(x-3\right)\)
3) \(x^2+7x+12=\left(x^2+3x\right)+\left(4x+12\right)=x\left(x+3\right)+4\left(x+3\right)=\left(x+3\right)\left(x+4\right)\)
1: \(x^2-3x+2=\left(x-1\right)\left(x-2\right)\)
2: \(x^2-x-6=\left(x-3\right)\left(x+2\right)\)
3: \(x^2+7x+12=\left(x+3\right)\left(x+4\right)\)
Bài 9: Phân tích đa thức thành nhân tử
1, 5x2 – 10xy + 5y2 – 20z2 2, 16x – 5x2 – 3 3, x2 – 5x + 5y – y2 | 4, 3x2 – 6xy + 3y2 – 12z2 5, x2 + 4x + 3 6, (x2 + 1)2 – 4x2 7, x2 – 4x – 5
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1.\(=5\left(x^2-2xy+y^2-4z^2\right)=5\left[\left(x+y\right)^2-\left(2z\right)^2\right]=5\left(x+y-2z\right)\left(x+y+2z\right)\)
2. \(=\left(-5x^2+15x\right)+\left(x-3\right)=-5x\left(x-3\right)+\left(x-3\right)=\left(1-5x\right)\left(x-3\right)\)
3. \(=\left(x-y\right)\left(x+y\right)-5\left(x-y\right)=\left(x-y\right)\left(x+y-5\right)\)
4.\(=3\left(x^2-2xy+y^2-4z^2\right)=3\left[\left(x-y\right)^2-\left(2z\right)^2\right]=3\left(x-y-2z\right)\left(x-y+2z\right)\)
5. \(=\left(x^2+x\right)+\left(3x+3\right)=x\left(x+1\right)+3\left(x+1\right)=\left(x+1\right)\left(x+3\right)\)
6. \(=\left(x^2-2x+1\right)\left(x^2+2x+1\right)=\left(x-1\right)^2\left(x+1\right)^2\)
7. \(=\left(x^2+x\right)-\left(5x+5\right)=x\left(x+1\right)-5\left(x+1\right)=\left(x-5\right)\left(x+1\right)\)
\(1,=5\left[\left(x-y\right)^2-4z^2\right]=5\left(x-y-2z\right)\left(x-y+2z\right)\\ 2,=-5x^2+15x+x-3=\left(x-3\right)\left(1-5x\right)\\ 3,=\left(x-y\right)\left(x+y\right)-5\left(x-y\right)=\left(x-y\right)\left(x+y-5\right)\\ 4,=3\left[\left(x-y\right)^2-4z^2\right]=3\left(x-y-2z\right)\left(x-y+2z\right)\\ 5,=x^2+x+3x+3=\left(x+3\right)\left(x+1\right)\\ 6,=\left(x^2+2x+1\right)\left(x^2-2x+1\right)=\left(x-1\right)^2\left(x+1\right)^2\\ 7,=x^2+x-5x-5=\left(x+1\right)\left(x-5\right)\)
Phân tích đa thức sau thành nhân tử:
a) 40x4-10x2
b) 16x4-20x2-y2-5y
c) 64a2-9b2-16a+1
d) 5x2+23x-10
e) 7x2-13xy-2y2
f) x4-23x2y2+y4
g) (x2+3x)2+7(x2+3x)+10
h) x(x+1)(x+2)(x+3)-8
a) \(40x^4-10x^2=10x^2\left(4x^2-1\right)=10x^2\left(2x-1\right)\left(2x+1\right)\)
b) \(16x^4-20x^2-y^2-5y=\left(4x^2-\dfrac{5}{2}\right)^2-\left(y-\dfrac{5}{2}\right)^2=\left(4x^2-\dfrac{5}{2}-y+\dfrac{5}{2}\right)\left(4x^2-\dfrac{5}{2}+y-\dfrac{5}{2}\right)=\left(4x^2-y\right)\left(4x^2+y-5\right)\)c)\(64a^2-9b^2-16a+1=\left(8a-1\right)^2-9b^2=\left(8a-1-3b\right)\left(8a-1+3b\right)\)d) \(5x^2+23x-10=5\left(x-\dfrac{2}{5}\right)\left(x+5\right)\)
a: \(40x^4-10x^2\)
\(=10x^2\left(4x^2-1\right)\)
\(=10x^2\cdot\left(2x-1\right)\left(2x+1\right)\)
b: \(16x^4-20x^2-y^2-5y\)
\(=\left(4x^2-y\right)\left(4x^2+y\right)-5\left(4x^2+y\right)\)
\(=\left(4x^2+y\right)\left(4x^2-y-5\right)\)
c: Ta có: \(64a^2-9b^2-16a+1\)
\(=\left(8a-1\right)^2-9b^2\)
\(=\left(8a-1-3b\right)\left(8a-1+3b\right)\)
d: Ta có: \(5x^2+23x-10\)
\(=5x^2+25x-2x-10\)
\(=\left(x+5\right)\left(5x-2\right)\)
Phân tích các đa thức sau thành nhân tử:
1) x3 - 7x + 6
2) x3 - 9x2 + 6x + 16
3) x3 - 6x2 - x + 30
4) 2x3 - x2 + 5x + 3
5) 27x3 - 27x2 + 18x - 4
6) x2 + 2xy + y2 - x - y - 12
7) (x + 2)(x +3)(x + 4)(x + 5) - 24
8) 4x4 - 32x2 + 1
9) 3(x4 + x2 + 1) - (x2 + x + 1)2
10) 64x4 + y4
11) a6 + a4 + a2b2 + b4 - b6
12) x3 + 3xy + y3 - 1
13) 4x4 + 4x3 + 5x2 + 2x + 1
14) x8 + x + 1
15) x8 + 3x4 + 4
16) 3x2 + 22xy + 11x + 37y + 7y2 +10
17) x4 - 8x + 63
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 !!!
6) \(\left(x+y\right)^2-\left(x+y\right)-12\)\(=\left(x+y\right)^2-2\cdot\frac{1}{2}\left(x+y\right)+\frac{1}{4}-\frac{49}{4}\)
\(=\left(x+y-\frac{1}{2}\right)^2-\left(\frac{7}{2}\right)^2\)\(=\left(x+y-\frac{1}{2}-\frac{7}{2}\right)\left(x+y-\frac{1}{2}+\frac{7}{2}\right)\)
\(=\left(x-4\right)\left(x+3\right)\)
7) \(\left(x+2\right)\left(x+5\right)\left(x+3\right)\left(x+4\right)-24\) (NHÂN x + 2 vs x + 5 và x + 3 vs x + 4 )
\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)
ĐẶT \(x^2+7x+11=y\) ta được :
\(\left(y+1\right)\left(y-1\right)-24=y^2-1-24\)
\(=y^2-25=\left(y-5\right)\left(y+5\right)\)
8) \(4x^4-32x^2+1=4x^4+4x^2+1-36x^2\)
\(=\left(2x^2+1\right)^2-\left(6x\right)^2\)\(=\left(2x^2-6x+1\right)\left(2x^2+6x+1\right)\)
9) sai đề rùi bạn ơi ! đề đúng nè
\(3\left(x^4+x^2+1\right)-\left(x^2+x+1\right)^2\)
Ta thấy :
\(x^4+x^2+1=\left(x^4+2x^2+1\right)-x^2\)\(=\left(x^2+1\right)^2-x^2=\left(x^2+x+1\right)\left(x^2-x+1\right)\)
Thay vào biểu thức bài cho ta được :
\(3\left(x^2-x+1\right)\left(x^2+x+1\right)-\left(x^2+x+1\right)^2\)
\(=\left(x^2+x+1\right)\left(3x^2-3x+3-x^2-x-1\right)\)
\(=\left(x^2+x+1\right)\left(2x^2-4x+2\right)\)
\(=2\left(x^2+x+1\right)\left(x-1\right)^2\)
bài ở trên câu 3 : kết luận là \(\left(x-3\right)\left(x^2-x-6\right)\)bạn sửa lại giúp mk nka !!! Th@nk !!! Tk Mk vs
PHÂN TÍCH ĐA THỨC THÀNH NHÂN TỬ
a, 3x(2x - y) + 5y(y - 2x)
b, (x - 5)2 - 9(x + y)2
c, y2 + 2yz + z2 - xy - xz
d, x2 - 9x2y2 + y2 + 2xy
e, x2 - 10x + 24
g, 6x2 + 7x - 5
h, x2 + 4xy - 12y2
k, a4 + 3a2 + 4
a) \(3x\left(2x-y\right)+5y\left(y-2x\right)\)
\(=3x\left(2x-y\right)-5y\left(2x-y\right)\)
\(=\left(3x-5y\right)\left(2x-y\right)\)
b) \(\left(x-5\right)^2-9\left(x+y\right)^2\)
\(=\left(x-5\right)^2-3^2\left(x+y\right)^2\)
\(=\left(x-5\right)^2-\left(3x+3y\right)^2\)
\(=\left(x-5+3x+3y\right)\left(x-5-3x-3y\right)\)
\(=\left(4x+3y-5\right)\left(-2x-3y-5\right)\)
a: \(3x\left(2x-y\right)+5y\left(y-2x\right)=\left(2x-y\right)\left(3x-5y\right)\)
e: \(x^2-10x+24=\left(x-4\right)\left(x-6\right)\)
g) \(6x^2+7x-5\)
=\(6x^2+10x-3x-5\)
=\(\left(6x^2+10x\right)-\left(3x+5\right)\)
=\(2x\left(3x+5\right)-\left(3x+5\right)\)
=\(\left(2x-1\right)\left(3x+5\right)\)
Phân tích các đa thức sau thành nhân tử :
i/ x2+5x−6x2+5x−6
m/ 6x2−7x+26x2−7x+2
n/ 4x4+81
a: =(x+6)(x-1)
n: \(=4x^4+36x^2+81-36x^2\)
\(=\left(2x^2+9-6x\right)\left(2x^2+9+6x\right)\)