Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a: \(x^3z+x^2yz-x^2z^2-xyz^2\)
\(=x^2z\left(x+y\right)-xz^2\left(x+y\right)\)
\(=xz\left(x+y\right)\left(x-z\right)\)
\(a.a\left(m-n\right)+m-n\)
\(=a\left(m-n\right)+\left(m-n\right)\)
\(=\left(a+1\right)\left(m-n\right)\)
\(b.ma+mb-a-b\)
\(=m\left(a+b\right)-\left(a+b\right)\)
\(=\left(m-1\right)\left(a+b\right)\)
\(c.4x+by+4y+bx\)
\(=\left(4x+4y\right)+\left(bx+by\right)\)
\(=4\left(x+y\right)+b\left(x+y\right)\)
\(=\left(b+4\right)\left(x+y\right)\)
\(d.1-ax-x+a\)
\(=\left(a-ax\right)+\left(1-x\right)\)
\(=a\left(1-x\right)+\left(1-x\right)\)
\(=\left(a+1\right)\left(1-x\right)\)
1.a(m-n)+m-n=am-an+m-n=(am+m)-(an+n)=m(a+1)-n(a+1)=(a+1)(m-n)
2.ma+mb-a-b=(ma-a)+(mb-b)=a(m-1)+b(m-1)=(m-1)(a+b)
3.4x+by+4y+bx=(4x+bx)+(4y+by)=x(4+b)+y(4+b)=(4+b)(x+y)
4.1-ax-x+a=(1+a)-(ax+x)=(1+a)-x(a+1)=(1+a)(1-x)
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 ) \(x^3z+x^2yz-x^2z^2-xyz^2=\left(x^3z-x^2z^2\right)+\left(x^2yz-xyz^2\right)\)
\(=\left(x-z\right)\left(x^2z+xyz\right)\)
\(=xz\left(x-z\right)\left(x+y\right)\)
b ) \(p^{m+2}.q-p^{m+1}q^3-p^2q^{n+1}+pq^{n+3}\)
\(=p^{m+1}q\left(p-q^2\right)-pq^{n+1}\left(p-q^2\right)\)
\(=\left(p-q^2\right)\left(p^{m+1}q-pq^{n+1}\right)\)
\(=pq\left(p-q^2\right)\left(p^m-q^n\right)\)
Bài 1:
e: Ta có: \(x\left(y-x\right)^2-x^2+2xy-y^2\)
\(=x\left(x-y\right)^2-\left(x-y\right)^2\)
\(=\left(x-y\right)^2\cdot\left(x-1\right)\)
Bài 2:
a: Ta có: \(M=m^2\left(m+n\right)-n^2m-n^3\)
\(=m^2\left(m+n\right)-n^2\left(m+n\right)\)
\(=\left(m+n\right)^2\cdot\left(m-n\right)\)
\(=\left(-2017+2017\right)^2\cdot\left(-2017-2017\right)\)
=0
\(a,\left(2a+3\right)x-\left(2a+3\right)y+\left(2a+3\right)\)
\(=\left(2a+3\right)\left(x-y+1\right)\)
\(b,\left(4x-y\right)\left(a-1\right)-\left(y-4x\right)\left(b-1\right)+\left(4x-y\right)\left(1-c\right)\)
\(=\left(4x-y\right)\left(a-1\right)+\left(4x-y\right)\left(b-1\right)+\left(4x-y\right)\left(1-c\right)\)
\(=\left(4x-y\right)\left(a-1+b-1+1-c\right)\)
\(=\left(4x-y\right)\left(a+b-c-1\right)\)
\(c,x^k+1-x^k-1\)
\(=0?!?!\)
\(d,x^m+3-x^m+1\)
\(=4\)
\(e,3\left(x-y\right)^3-2\left(x-y\right)^2\)
\(=\left(x-y\right)^2\left(3\left(x-y\right)-2\right)\)
\(=\left(x-y\right)^2\left(3x-3y-2\right)\)
\(f,81a^2+18a+1\)
\(=\left(9a\right)^2+2.9a+1\)
\(=\left(9a+1\right)^2\)
\(g,25a^2.b^2-16c^2\)
\(=\left(5ab\right)^2-\left(4c\right)^2\)
\(=\left(5ab+4c\right)\left(5ab-4c\right)\)
\(h,\left(a-b\right)^2-2\left(a-b\right)c+c^2\)
\(=\left(a-b-c\right)^2\)
\(i,\left(ax+by\right)^2-\left(ax-by\right)^2\)
\(=\left(ax+by-ax+by\right)\left(ax+by+ax-by\right)\)
\(=2by.2ax\)
\(=4axby\)
Bài 4:
Ta có: \(\left(x^3-x^2\right)-4x^2+8x-4=0\)
\(\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
4 x 2 + 4 x – y 2 + 1 = ( ( 2 x ) 2 + 2 . 2 x + 1 ) – y 2
= ( 2 x + 1 ) 2 – y 2
= (2x + 1 – y)(2x + 1 + y)
= (2x – y + 1)(2x + y + 1)
Vậy đa thức trong chỗ trống là 2x – y + 1
Đáp án cần chọn là: B