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\(2.a.=5x\left(x-y\right)-9\left(x-y\right)\)
\(=\left(5x-9\right)\left(x-y\right)\)
\(b.=m\left(m^2+4m+3\right)\)
\(=m\left(m+1\right)\left(m+3\right)\)
a) Ta có: \(\left(4x^2-3x-18\right)^2-\left(4x^2+3x\right)^2\)
\(=\left(4x^2-3x-18-4x^2-3x\right)\left(4x^2-3x-18+4x^2+3x\right)\)
\(=\left(-6x-18\right)\left(8x^2-18\right)\)
\(=-6\left(x+3\right)\cdot2\left(4x^2-9\right)\)
\(=-12\left(x+3\right)\left(2x-3\right)\left(2x+3\right)\)
b) Ta có: \(9\left(x+y-1\right)^2-4\left(2x+3y+1\right)^2\)
\(=\left(3x+3y-3\right)^2-\left(4x+6y+2\right)^2\)
\(=\left(3x+3y-3-4x-6y-2\right)\left(3x+3y-3+4x+6y+2\right)\)
\(=-\left(x+3y+5\right)\left(7x+9y-1\right)\)
c) Ta có: \(-4x^2+12xy-9y^2+25\)
\(=-\left(4x^2-12xy+9y^2-25\right)\)
\(=-\left[\left(2x-3y\right)^2-25\right]\)
\(=-\left(2x-3y-5\right)\left(2x-3y+5\right)\)
d) Ta có: \(x^2-2xy+y^2-4m^2+4mn-n^2\)
\(=\left(x^2-2xy+y^2\right)-\left(4m^2-4mn+n^2\right)\)
\(=\left(x-y\right)^2-\left(2m-n\right)^2\)
\(=\left(x-y-2m+n\right)\left(x-y+2m-n\right)\)
mk ghi đáp án, ko phân tích đc thì IB mk
a) \(x^2+6xy+9y^2=\left(x+3y\right)^2\)
b) \(4a^4-4a^2b^2+b^4=\left(2a^2-b^2\right)^2\)
c) \(x^6+y^2-2x^3y=\left(x^3-y\right)^2\)
d) \(\left(x+y\right)^3-\left(x-y\right)^3=2y\left(3x^2+y^2\right)\)
e) \(25x^4-10x^2y^2+y^4=\left(5x^2-y^2\right)^2\)
f) \(-a^2-2a-1=-\left(a+1\right)^2\)
g) \(27b^3-8a^3=\left(3b-2a\right)\left(9b^2+6ab+4a^2\right)\)
h) \(x^3+9x^2y+27xy^2+27y^3=\left(x+3y\right)^3\)
i) \(16x^2-9\left(x+y\right)^2=\left(x-3y\right)\left(7x+3y\right)\)
Bài làm :
\(\text{a)}9\left(x+y-1\right)^2-4\left(2x+3y+1\right)^2\)
\(=\left(3x+3y-3\right)^2-\left(4x+6y+2\right)^2\)
\(=\left(3x+3y-3-4x-6y-2\right)\left(3x+3y-3+4x+6y+2\right)\)
\(=\left(-x-3y-5\right)\left(7x+9y-1\right)\)
\(\text{b)}3x^4y^2+3x^3y^2+3xy^2+3y^2\)
\(=\left(3x^4y^2+3xy^2\right)+\left(3x^3y^2+3y^2\right)\)
\(=3xy^2\left(x^3+1\right)+3y^2\left(x^3+1\right)\)
\(=\left(3xy^2+3y^2\right)\left(x^3+1\right)\)
\(=3y^2\left(x+1\right)\left(x+1\right)\left(x^2-x+1\right)\)
\(=3y^2\left(x+1\right)^2\left(x^2-x+1\right)\)
\(\text{c)}\left(x+y\right)^3-1-3xy\left(x+y-1\right)\)
\(=\left(x+y-1\right)\left[\left(x+y\right)^2+x+y+1\right]-3xy\left(x+y-1\right)\)
\(=\left(x+y-1\right)\left(x^2+2xy+y^2+x+y+1-3xy\right)\)
\(=\left(x+y-1\right)\left(x^2+x+y^2+y+1-xy\right)\)
\(d ) x^3+3x^2+3x+1-27z^3\)
\(=\left(x+1\right)^3-\left(3z\right)^3\)
\(=\left(x+1-3z\right)\left(x^2+2x+1+3xz+3z+9z^2\right)\)
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
\(x\left(x-y\right)+y\left(y-x\right)=x^2-xy+y^2-xy=x^2-2xy+y^2=\left(x-y\right)^2\)
\(2x^3y-2xy^3-4xy^2-2xy\)
\(=2xy.\left(x^2-y^2-2y-1\right)\)
\(=2xy.[x^2-\left(y^2+2y+1\right)]\)
\(=2xy.[x^2-\left(y+1\right)^2]\)
\(=2xy.\left(x+y+1\right).\left(x-y-1\right)\)
Vậy chọn đáp án A
a) x3-2x2-x+2
=x(x2-1)+2(-x2+1)
=x(x2-1)-2(x2-1)
=(x2-1)(x-2)
b)
x2+6x-y2+9
=x2+6x+9-y2
=(x+3)2-y2
=(x+3-y)(x+3+y)