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a) = pq(pm+1 - pm - pq2n+3)
b) = (m-3)2 - (x - 2y)2
= ( m-3 +x -2y)(m-3 -x +2y)
xin lỗi mình chép sai đề, mình vừa mới sửa lại phiền bạn làm lại giùm mình
Lời giải:
$N=p^{m+2}q-pq^{m+3}-p^{m+3}q^{n+4}$
$=pq(p^{m+1}-q^{m+2}-p^{m+2}q^{n+3})$
a) 3x(x + 7)2 - 11x2(x + 7) + 9(x + 7) = (x + 7)[3x(x + 7) - 11x2 + 9) = (x + 7)(3x2 + 21x - 11x2 + 9)
= (x + 7)(-8x2 + 21x + 9)(-8x2 + 24x - 3x + 9) = (x + 7)[-8x(x - 3) - 3(x - 3)] = -(x + 7)(8x + 3)(x - 3)
b) 3x(x - 9)2 - (9 - x)3 = 3x(x - 9)2 + (x - 9)3 = (x - 9)2(3x + x - 9) = (x - 9)2(4x - 9)
c) pm + 2.q - pm + 1.q3 - p2.qn + 1 + p.qn + 3 = (pm + 2.q - p2.qn + 1) - (pm + 1.q3 - p.qn + 3)
= p2.q(pm - qn) - p.q3(pm - qn) = pq(pm - qn)(p - q2)
d) x2y2z + xy2z2 + x2yz = xyz(xy + yz + x)
a) \(3x\left(x+7\right)^2-11x^2\left(x+7\right)+9\left(x+7\right)\)
\(=\left(x+7\right)\left[3x\left(x+7\right)-11x^2+9\right]=\left(x+7\right)\left(3x^2+21x-11x^2+9\right)\)
\(=\left(x+7\right)\left(-8x^2+21x+9\right)=\left(x+7\right)\left[\left(-8x^2+24x\right)-\left(3x-9\right)\right]\)
\(=\left(x+7\right)\left[-8x\left(x-3\right)-3\left(x-3\right)\right]=-\left(x+7\right)\left(x-3\right)\left(8x+3\right)\)
b) \(3x\left(x-9\right)^2-\left(9-x\right)^3=3x\left(x-9\right)^2+\left(x-9\right)^3\)
\(=\left(x-9\right)^2\left(3x+x-9\right)=\left(x-9\right)^2\left(4x-9\right)\)
c) \(p^{m+2}.q-p^{m+1}.q^3-p^2.q^{n+1}+p.q^{n+3}\)
\(=p^{m+1}.q\left(p-q^2\right)-p.q^{n+1}\left(p-q^2\right)\)\(=p.q.\left(p-q^2\right).\left(p^m.q^n\right)\)
d) \(x^2y^2z+xy^2z^2+x^2yz=xyz\left(xy+yz+x\right)\)
\(p^{m+2}.q-p^{m+1}.q^3-p^2.q^{n+1}+p.q^{n+3}\)
\(=pq\left(p^{m+1}-p^mq^2-pq^n+q^{n+2}\right)\)
\(=p^m\left(p-q^2\right)-q^n\left(p-q^2\right)\)
\(=\left(p-q^2\right)\left(p^m-q^n\right)\)
..........
Bài làm:
a) \(x^2-2xy+y^2-zx+yz\)
\(=\left(x-y\right)^2-z\left(x-y\right)\)
\(\left(x-y\right)\left(x-y-z\right)\)
a/ \(x^2-2xy+y^2-zx+yz.\)
\(=\left(x-y\right)^2-z\left(x-y\right)\)
\(=\left(x-y\right)\left(x-y-z\right)\)
c/ \(x^2-y^2-2x-2y.\)
\(=x^2-2x+1-y^2-2y-1\)
\(=\left(x^2-2x+1\right)-\left(y^2+2y+1\right)\)
\(=\left(x-1\right)^2-\left(y+1\right)^2\)
\(=\left(x-1+y+1\right)\left(x-1-y-1\right)\)
\(=\left(x+y\right)\left(x-y-2\right)\)
Ta có:
\(p^{m+2}q-p^{m+1}q^3-p^2q^{n+1}+pq^{n+3}\)
\(=\left(p^{m+2}q-p^{m+1}q^3\right)-\left(p^2q^{n+1}-pq^{n+3}\right)\)
\(=p^{m+1}q\left(p-q^2\right)-pq^{n+1}\left(p-p^2\right)\)
\(=\left(p-p^2\right)\left(p^{m+1}q-pq^{n+1}\right)\)
\(=pq\left(p-p^2\right)\left(p^m-p^n\right)\)
Đề sai nhé .Sửu lại
\(x^2-4x^2y^2+4+4x\)
\(=\left(x^2+4x+4\right)-4x^2y^2\)
\(=\left(x+2\right)^2-\left(2xy\right)^2\)
\(=\left(x+2+2xy\right)\left(x+2-2xy\right)\)
đề bài thiếu bn ơi
đù r mà