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a)\(\left(4x^3-xy^2+y^3\right)\left(x^2y+2xy^2-2y^3\right)\)
\(=x^2y\left(4x^3-xy^2+y^3\right)+2xy^2\left(4x^3-xy^2+y^3\right)\)
\(-2y^3\left(4x^3-xy^2+y^3\right)\)
\(=4x^5y-x^3y^3+x^2y^4+8x^4y^2-2x^2y^4+2xy^5\)
\(-8x^3y^3+2xy^5-2y^6\)
\(=-2y^6+4x^5y+\left(2xy^5+2xy^5\right)+8x^4y^2+\left(x^2y^4-2x^2y^4\right)\)
\(-\left(x^3y^3+8x^3y^3\right)\)
\(=-2y^6+4x^5y+4xy^5+8x^4y^2-x^2y^4-9x^3y^3\)
b)
(!) \(2\left(x+y\right)^2-7\left(x+y\right)+5\)
\(=2\left(x+y\right)^2-2\left(x+y\right)-5\left(x+y\right)+5\)
\(=2\left(x+y\right)\left(x+y-1\right)-5\left(x+y-1\right)\)
\(=\left(2x+2y-5\right)\left(x+y-1\right)\)
(!!) \(\left(x+y+z\right)^2-x^2-y^2-z^2\)
\(=\left(x^2+y^2+z^2+2xy+2yz+2zx\right)-x^2-y^2-z^2\)
\(=2\left(xy+yz+zx\right)\)
a/Dùng hằng đẳng thức A2-B2=(A+B)(A-B) phân tích được ngay
\(\left(x-y+4\right)^2-\left(2x+3y-1\right)^2\)
\(=\left(x-y+4+2x+3y-1\right)\left(x-y+4-2x-3y+1\right)\)
=\(\left(3x-2y+3\right)\left(4-x-4y\right)\)
b/Chắc chỉ phân tích hằng đẳng thức (A-B)2=A2-2ab+B2
\(49\left(y-4\right)^2-9y^2-3y-36=49y^2-392y+784-9y^2-3y-36\)
\(=40y^2-395y+748\)
Mình dùng biệt thức cho ra nghiệm vô tỉ, không biết cho phải tại mình tính sai hay đề thiếu nữa
c/Khai triển biểu thức ban đầu ta được
\(x\left(x-y\right)+y\left(y-x\right)=x^2-xy+y^2-xy=x^2-2xy+y^2=\left(x-y\right)^2\)
\(a,\left(a+b\right)+\left(a+b\right)^2\)
\(=\left(a+b\right)\left(1+a+b\right)\)
\(b,4\left(x-y\right)+3\left(x-y\right)^2\)
\(=\left(x-y\right)\left(4+3\left(x-y\right)\right)\)
\(=\left(x-y\right)\left(4+3x-3y\right)\)
\(c,\left(a-b\right)+\left(b-a\right)^2\)
\(=\left(a-b\right)+\left(a-b\right)^2\)
\(=\left(a-b\right)\left(1+a-b\right)\)
a) \(\left(a+b\right)+\left(a+b\right)^2=\left(a+b\right)\left(1+a+b\right)\)
b) \(4\left(x-y\right)+3\left(x-y\right)^2=\left(x-y\right)\left[4+3.\left(x-y\right)\right]\)
c) \(\left(a-b\right)+\left(b-a\right)^2=\left(a-b\right)+\left(b-a\right)\left(b-a\right)\)
\(=\left(a-b\right)-\left(a-b\right)\left(b-a\right)\)
\(=\left(a-b\right)\left(1-b+a\right)\)
d) \(\left(a-b\right)-\left(b-a\right)^2\)
\(=\left(a-b\right)-\left(b-a\right)\left(b-a\right)\)
\(=\left(a-b\right)+\left(a-b\right)\left(b-a\right)\)
\(=\left(a-b\right)\left(1+b-a\right)\)
e) \(a\left(a-b\right)^2-\left(b-a\right)^3\)
\(=a\left(a-b\right)-\left(a-b\right)\left(b-a\right)^2\)
\(=\left(a-b\right)\left[a-\left(b-a\right)^2\right]\)
f) \(\left(y+z\right)\left(12x^2+6x\right)+\left(y-z\right)\left(12x^2+6x\right)\)
\(=\left(12x^2+6x\right)\left(y+z+y-z\right)\)
\(=\left(12x^2+6x\right)2y\)
a) x4+x3+2x2+x+1=(x4+x3+x2)+(x2+x+1)=x2(x2+x+1)+(x2+x+1)=(x2+x+1)(x2+1)
b)a3+b3+c3-3abc=a3+3ab(a+b)+b3+c3 -(3ab(a+b)+3abc)=(a+b)3+c3-3ab(a+b+c)
=(a+b+c)((a+b)2-(a+b)c+c2)-3ab(a+b+c)=(a+b+c)(a2+2ab+b2-ac-ab+c2-3ab)=(a+b+c)(a2+b2+c2-ab-ac-bc)
c)Đặt x-y=a;y-z=b;z-x=c
a+b+c=x-y-z+z-x=o
đưa về như bài b
d)nhóm 2 hạng tử đầu lại và 2hangj tử sau lại để 2 hạng tử sau ở trong ngoặc sau đó áp dụng hằng đẳng thức dề tính sau đó dặt nhân tử chung
e)x2(y-z)+y2(z-x)+z2(x-y)=x2(y-z)-y2((y-z)+(x-y))+z2(x-y)
=x2(y-z)-y2(y-z)-y2(x-y)+z2(x-y)=(y-z)(x2-y2)-(x-y)(y2-z2)=(y-z)(x2-2y2+xy+xz+yz)
1 ) \(a\left(m+n\right)+b\left(m+n\right)\)
\(=\left(a+b\right)\left(m+n\right)\)
2 ) \(a^2\left(x+y\right)-b^2\left(x+y\right)\)
\(=\left(a^2-b^2\right)\left(x+y\right)\)
\(=\left[\left(a-b\right).\left(a+3\right)\right]\left(x+y\right)\)
3 ) \(6a^2-3a+12ab\)
\(=3a.2a-3a+3a.4b\)
\(=3a.\left(2a-1+4b\right)\)
4 ) \(2x^2y^4-2x^4y^2+6x^3y^3\)
\(=2x^2y^2.y^2-2x^2y^2.x^2+2x^2y^2.3xy\)
\(=2x^2y^2\left(y^2-x^2+3xy\right)\)
5 ) \(\left(x+y\right)^3-x\left(x+y\right)^2\)
\(=\left(x+y\right)^2.\left(x+y-x\right)\)
\(=\left(x+y\right)^2.y\)
1)a(m+n)+b(m+n)
=(a+b)(m+n)
2)a2(x+y)-b2(x+y)
=(a2-b2)(x+y)
3)6a2-3a+12ab
=3a.2a-3a.(1-4b)
=3a.(2a-1+4b)
5)(x+y)3-x(x+y)2
=(x+y)(x+y)2-x(x+y)2
=(x+y)2(x+y-x)
1/
a, x2+36=12x
<=>x2-12x+36=0
<=>(x-6)2=0
<=>x-6=0
<=>x=6
b, 5x(x-3)+3-x=0
<=>5x(x-3)-(x-3)=0
<=>(5x-1)(x-3)=0
<=>\(\orbr{\begin{cases}5x-1=0\\x-3=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{1}{5}\\x=3\end{cases}}}\)
2/ Sửa đề x2z2 = y2z2
Đặt \(A=4x\left(x+y\right)\left(x+y+z\right)\left(x+z\right)+y^2z^2=4x\left(x+y+z\right)\left(x+y\right)\left(x+z\right)+y^2z^2\)
\(=4\left(x^2+xy+xz\right)\left(x^2+xz+xy+yz\right)+y^2z^2\)
Đặt x2+xy+xz=t, ta có
\(A=4t\left(t+yz\right)+y^2z^2=4t^2+4tyz+y^2z^2=\left(2t+yz\right)^2=\left(2x^2+2xy+2xz+y^2z^2\right)^2\ge0\)
a, \(4x^3y^2-8x^2y^3+12x^3y^4\)
\(=4x^2y^2\left(x-2y+3xy^2\right)\)
b, \(x\left(y-z\right)+2\left(z-y\right)\)
\(=x\left(y-z\right)-2\left(y-z\right)\)
\(=\left(y-z\right)\left(x-2\right)\)
c, \(\left(x+y\right)^2-2\left(x+y\right)\)
\(=\left(x+y\right)\left(x+y-2\right)\)
d, \(x\left(2-x\right)^2-\left(2-x\right)^3\)
\(=\left(2-x\right)^2\left(x-2-x\right)=-2\left(x-2\right)^2\)
a, \(4x^3y^2-8x^2y^3+12x^3y^4\)
\(=4x^2y^2\left(x-2y+3xy^2\right)\)
b, \(x\left(y-z\right)+2\left(z-y\right)\)
\(=\left(x-2\right)\left(y-z\right)\)
c, \(\left(x+y\right)^2-2\left(x+y\right)\)
\(=\left(x+y\right)\left(x+y-2\right)\)
d, \(x\left(2-x\right)^2-\left(2-x\right)^3\)
\(=\left(2-x\right)^2\left(x-2+x\right)\)
\(=2\left(x-2\right)^2\left(x-1\right)\)