phân tích đa thức thành nhân tử
a, \(100x^2-\left(x^2+25\right)^2\)
b, \(\left(x-y+5\right)^2-2\left(x-y+5\right)+1\)
c, \(\left(x^2+4y^2-5\right)^2-16\left(x^2+y^2+2xy+1\right)\)
Phân tích đa thức thành nhân tử:
a) \(\left(x-y+5\right)^2-2\left(x-y+5\right)+1\)
b) \(\left(x^2+4y^2-5\right)^2-16\left(x^2y^2+2xy+1\right)\)
a)
(x-y+5)2-2.(x-y+5)+1
=(x-y+5-1)2
=(x-y+4)2
b)
(x2+4y2-5)2-16.(x2.y2+2xy+1)
=(x2+4y2-5)2-[4.(xy+1)]2
=(x2+4y2-5-4xy-4)(x2+4y2-5+4xy+4)
=(x2+4y2-4xy-9)(x2+4y2+4xy-1)
=[(x-2y)2-9][(x+2y)2-1]
=(x-2y-3)(x-2y+3)(x+2y-1)(x+2y+1)
=(x2+x-3x-3)((x-2y+3)(x+2y-1)(x+1)2
=[x(x+1)-3(x+1)](x-2y+3)(x+2y-1)(x+1)2
=(x+1)(x-3)(x-2y+3)(x+2y-1)(x+1)2
bài 1: phân tích đa thức thành nhân tử
a,2x+10y
b,x\(^2+4x+4\)
c,\(x^2-y^2+10y-25\)
bài 2 tìm x, biết
a,\(x^2-3x+x-3=0\)
b,\(2x\left(x-3\right)-\dfrac{1}{2}\left(4x^2-3\right)=0\)
c,\(x^2-\left(x-3\right)\left(2x-5\right)=9\)
\(B1\\ a,2x+10y=2\left(x+5y\right)\\ b,x^2+4x+4=x^2+2.2x+2^2=\left(x+2\right)^2\\ c,x^2-y^2+10y-25\\ =\left(x^2-y^2\right)+5\left(2y-5\right)\\ =\left(x-y\right)\left(x+y\right)+5\left(2y-5\right)\\ B2\)
\(a,x^2-3x+x-3=0\\ =>x\left(x-3\right)+\left(x-3\right)=0\\ =>\left(x+1\right)\left(x-3\right)=0\\ =>\left[{}\begin{matrix}x+1=0\\x-3=0\end{matrix}\right.=>\left[{}\begin{matrix}x=-1\\x=3\end{matrix}\right.\\ b,2x\left(x-3\right)-\dfrac{1}{2}\left(4x^2-3\right)=0\\ =>2x^2-6x-2x^2+\dfrac{3}{2}=0\\ =>-6x=-\dfrac{3}{2}\\ =>x=\left(-\dfrac{3}{2}\right):\left(-6\right)\\ =>x=\dfrac{1}{4}\\ c,x^2-\left(x-3\right)\left(2x-5\right)=9\\ =>x^2-2x^2+6x+5x-15=9\\ =>-x^2+11-15-9=0\\ =>-x^2+11x-24=0\\ =>-x^2+8x+3x-24=0\\ =>-x\left(x-8\right)+3\left(x-8\right)=0\\ =>\left(3-x\right)\left(x-8\right)=0\\ =>\left[{}\begin{matrix}3-x=0\\x-8=0\end{matrix}\right.=>\left[{}\begin{matrix}x=3\\x=8\end{matrix}\right.\)
Phân tích các đa thức sau thành nhân tử :
a/ \(10x\left(x-y\right)-6y\left(y-x\right)\)
b/ \(14x^2y-21xy^2+28x^3y^2\)
c/ \(x^2-4+\left(x-2\right)^2\)
d/ \(\left(x+1\right)^2-25\)
e/ \(x^2-4y^2-2x+4y\)
f/ \(x^2-25-2xy+y^2\)
g/ \(x^3-2x^2+x-xy^2\)
h/ \(x^3-4x^2-12x+27\)
i/ \(x^2+5x-6\)
m/ \(6x^2-7x+2\)
n/ \(4x^4+81\)
\(a.10x\left(x-y\right)-6y\left(y-x\right)\\ =10x\left(x-y\right)+6y\left(x-y\right)\\ =\left(10x-6y\right)\left(x-y\right)\\ =2\left(5x-3y\right)\left(x-y\right)\)
\(b.14x^2y-21xy^2+28x^3y^2\\ =7xy\left(x-y+xy\right)\)
\(c.x^2-4+\left(x-2\right)^2\\ =\left(x-2\right)\left(x+2\right)+\left(x-2\right)^2\\ =\left(x-2\right)\left(x+2+x-2\right)\\ =2x\left(x-2\right)\)
\(d.\left(x+1\right)^2-25\\ =\left(x+1-5\right)\left(x+1+5\right)=\left(x-4\right)\left(x+6\right)\)
Phân tích đa thức sau thành nhân tử:
a) \(x^2-2xy+3x-3y+y^2-4\)
b) \(2\left(x^2-6x+1\right)^2+5\left(x^2-6x+1\right)\left(x^2+1\right)+2\left(x^2+1\right)^2\)
a: \(x^2-2xy+y^2+3x-3y-4\)
\(=\left(x-y\right)^2+3\left(x-y\right)-4\)
\(=\left(x-y+4\right)\left(x-y-1\right)\)
Bài 1: Phân tích đa thức thành nhân tử:
1) \(3x^3y^2-6xy\)
2) \(\left(x-2y\right).\left(x+3y\right)-2.\left(x-2y\right)\)
3) \(\left(3x-1\right).\left(x-2y\right)-5x.\left(2y-x\right)\)
4) \(x^2-y^2-6y-9\)
5) \(\left(3x-y\right)^2-4y^2\)
6) \(4x^2-9y^2-4x+1\)
8) \(x^2y-xy^2-2x+2y\)
9) \(x^2-y^2-2x+2y\)
Bài 2: Tìm x:
1) \(\left(2x-1\right)^2-4.\left(2x-1\right)=0\)
2) \(9x^3-x=0\)
3) \(\left(3-2x\right)^2-2.\left(2x-3\right)=0\)
4) \(\left(2x-5\right)\left(x+5\right)-10x+25=0\)
Bài 2:
1: \(\left(2x-1\right)^2-4\left(2x-1\right)=0\)
=>\(\left(2x-1\right)\left(2x-1-4\right)=0\)
=>(2x-1)(2x-5)=0
=>\(\left[{}\begin{matrix}2x-1=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{5}{2}\end{matrix}\right.\)
2: \(9x^3-x=0\)
=>\(x\left(9x^2-1\right)=0\)
=>x(3x-1)(3x+1)=0
=>\(\left[{}\begin{matrix}x=0\\3x-1=0\\3x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{3}\\x=-\dfrac{1}{3}\end{matrix}\right.\)
3: \(\left(3-2x\right)^2-2\left(2x-3\right)=0\)
=>\(\left(2x-3\right)^2-2\left(2x-3\right)=0\)
=>(2x-3)(2x-3-2)=0
=>(2x-3)(2x-5)=0
=>\(\left[{}\begin{matrix}2x-3=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{5}{2}\end{matrix}\right.\)
4: \(\left(2x-5\right)\left(x+5\right)-10x+25=0\)
=>\(2x^2+10x-5x-25-10x+25=0\)
=>\(2x^2-5x=0\)
=>\(x\left(2x-5\right)=0\)
=>\(\left[{}\begin{matrix}x=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{5}{2}\end{matrix}\right.\)
Bài 1:
1: \(3x^3y^2-6xy\)
\(=3xy\cdot x^2y-3xy\cdot2\)
\(=3xy\left(x^2y-2\right)\)
2: \(\left(x-2y\right)\left(x+3y\right)-2\left(x-2y\right)\)
\(=\left(x-2y\right)\cdot\left(x+3y\right)-2\cdot\left(x-2y\right)\)
\(=\left(x-2y\right)\left(x+3y-2\right)\)
3: \(\left(3x-1\right)\left(x-2y\right)-5x\left(2y-x\right)\)
\(=\left(3x-1\right)\left(x-2y\right)+5x\left(x-2y\right)\)
\(=(x-2y)(3x-1+5x)\)
\(=\left(x-2y\right)\left(8x-1\right)\)
4: \(x^2-y^2-6y-9\)
\(=x^2-\left(y^2+6y+9\right)\)
\(=x^2-\left(y+3\right)^2\)
\(=\left(x-y-3\right)\left(x+y+3\right)\)
5: \(\left(3x-y\right)^2-4y^2\)
\(=\left(3x-y\right)^2-\left(2y\right)^2\)
\(=\left(3x-y-2y\right)\left(3x-y+2y\right)\)
\(=\left(3x-3y\right)\left(3x+y\right)\)
\(=3\left(x-y\right)\left(3x+y\right)\)
6: \(4x^2-9y^2-4x+1\)
\(=\left(4x^2-4x+1\right)-9y^2\)
\(=\left(2x-1\right)^2-\left(3y\right)^2\)
\(=\left(2x-1-3y\right)\left(2x-1+3y\right)\)
8: \(x^2y-xy^2-2x+2y\)
\(=xy\left(x-y\right)-2\left(x-y\right)\)
\(=\left(x-y\right)\left(xy-2\right)\)
9: \(x^2-y^2-2x+2y\)
\(=\left(x^2-y^2\right)-\left(2x-2y\right)\)
\(=\left(x-y\right)\left(x+y\right)-2\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y-2\right)\)
bài 11 phân tích đa thức thành nhân tử
a,\(x^2-xy+x\) b,\(x^2-2xy-4+y^2\) c,\(x^3-x^2-16x+16\)
bài 12 tìm x biết :
a,\(2x\left(x-5\right)-x\left(3+2x\right)=26\) b,\(2\left(x+5\right)-x^2-5x=0\)
bài 11
a) \(x^2-xy+x\\ =x\left(x-y+1\right)\)
b)
\(x^2-2xy-4+y^2\\ =\left(x^2-2xy+y^2\right)-4\\ =\left(x-y\right)^2-4\\ =\left(x-y-2\right)\left(x-y+2\right)\)
c)
\(x^3-x^2-16x+16\\ =x^2\left(x-1\right)-16\left(x-1\right)\\ =\left(x-1\right)\left(x-4\right)\left(x+4\right)\)
bài 12
\(2x\left(x-5\right)-x\left(3+2x\right)=26\)
\(2x^2-10x-3x-2x^2=26\)
\(-13x=26\\ x=-2\)
b)
\(2\left(x+5\right)-x^2-5x=0\\ 2\left(x+5\right)-x\left(x+5\right)=0\\ \left(x+5\right)\left(2-x\right)=0\\ \left[{}\begin{matrix}x+5=0\\2-x=0\end{matrix}\right.\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)
Viet cac da thuc sau duoi dang tich :
a, \(100x^2-\left(x^2+25\right)^2\)
b, \(1+\left(x-y+5\right)^2-2\left(x-y+5\right)\)
c, \(\left(x^2+4y^2-5\right)^2-16\left(x^2y^2+2xy+1\right)\)
d, \(\left(x^2+8x-34\right)^2-\left(3x^2-8x-2\right)^2\)
a.100x2-(x2+25)2=(10x)2-(x2+25)2=(10x-x2-25)(10x+x2+25)=(10x-x2-25)(x+5)2
b.1+(x-y+5)2-2(x-y+5)=(x-y+4)2
\(a,100x^2-\left(x^2+25\right)\)
\(=\left(10x-x^2+25\right)\left(10x+x^2+25\right)\)
\(b,1+\left(x-y+5\right)^2-2\left(x-y+5\right)^2\)
\(=\left(1-x+y-5\right)^2\)
phân tích đa thức thành nhân tử :
a) \(x^2+2y^2+2xy-2x-6y+2015\)
b)\(a\left(b^2-c^2\right)-b\left(a^2-c^2\right)+c\left(a^2-b^2\right)\)
c) \(\left(x-y\right)^5+\left(y-z\right)^5+\left(z-x\right)^5\)
1) Đa thức\(\left(x^2+x+1\right)\left(X^2+x+2\right)\)-12 được phân tích thành nhân tử là:
A)\(\left(x^2+x+5\right)\left(x+2\right)\left(x-1\right)\)
B)\(\left(x^2+x-5\right)\left(x+2\right)\left(x-1\right)\)
C)\(\left(x^2-x+5\right)\left(x+2\right)\left(x-1\right)\)
D)\(\left(x^2+x+5\right)\left(x-2\right)\left(x+1\right)\)
2) \(\left(x+a\right)\left(x+2a\right)\left(x+3a\right)\left(x+4a\right)+a^4\) được phân tích thành nhân tử là:
A)\(\left(x^2+5ax-5a^2\right)\left(x^2-5ax+5a^2\right)\)
B)\(\left(x^2-5ax-5a^2\right)\left(x^2+5ax+5a^2\right)\)
C)\(\left(x^2-5ax-5a^2\right)\left(x^2-5ax+5a^2\right)\)
D)\(\left(x^2+5ax+5a^2\right)^{^2}\)
3) Đa thức \(a^3+b^3+c^3-3abc\) được phân tích thành nhân tử là:
A)\(\left(a+b+c\right)\left(a^2+b^2+c^2-ab+bc-ca\right)\)
B)\(\left(a-b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)
C)\(\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)
D)\(\left(a+b+c\right)\left(a^2+b^2+c^2-ab+bc-ca\right)\)
4) Đa thức x(x+1)(x+2)(x+3)+1 được phân tích thành nhân tử là:
A)\(\left(x^2+3x+1\right)\left(x^2+3x-1\right)\)
B)\(\left(x^2+3x+1\right)^{^2}\)
C)\(\left(x^2+3x+1\right)\left(x^2-3x+1\right)\)
D) Cả B và C đều sai
5) Câu trả lời đúng cho M=\(n^2\left(n+1\right)+2n\left(n+1\right)+360\) với \(n\in Z\)
A)M⋮4
B)M⋮5
C)M⋮6
D)M⋮9
6)Cho \(P=\left(2n+5\right)^{^2}-145\) với \(n\in N\)
A) P⋮4 ; B)P⋮3 ; C) P⋮5 ; D)P⋮6
7) Giá trị của biểu thức \(x^2-y^2-2y-1\) tại
x=502 ; y=497 là:
A) 3000
B)5000
C)4500
D) cả A và B đều sai
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