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Bài 4:
Ta có: \(\left(x^3-x^2\right)-4x^2+8x-4=0\)
\(\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
\(a.a\left(m-n\right)+m-n\)
\(=a\left(m-n\right)+\left(m-n\right)\)
\(=\left(a+1\right)\left(m-n\right)\)
\(b.ma+mb-a-b\)
\(=m\left(a+b\right)-\left(a+b\right)\)
\(=\left(m-1\right)\left(a+b\right)\)
\(c.4x+by+4y+bx\)
\(=\left(4x+4y\right)+\left(bx+by\right)\)
\(=4\left(x+y\right)+b\left(x+y\right)\)
\(=\left(b+4\right)\left(x+y\right)\)
\(d.1-ax-x+a\)
\(=\left(a-ax\right)+\left(1-x\right)\)
\(=a\left(1-x\right)+\left(1-x\right)\)
\(=\left(a+1\right)\left(1-x\right)\)
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)
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: \(M=m^2\left(m+n\right)-n^2m-n^3\)
\(=m^2\left(m+n\right)-n^2\left(m+n\right)\)
\(=\left(m+n\right)^2\left(m-n\right)\)
\(=\left(-2017+2017\right)^2\cdot\left(-2017-2017\right)\)
=0
b: \(N=n^3-3n^2-n\left(3-n\right)\)
\(=n^2\left(n-3\right)+n\left(n-3\right)\)
\(=n\left(n-3\right)\left(n+1\right)\)
\(=13\cdot10\cdot14=1820\)
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 ) \(p^{m+2}.q-p^{m+1}q^3-p^2q^{n+1}+pq^{n+3}\)
\(=p^{m+1}q\left(p-q^2\right)-pq^{n+1}\left(p-q^2\right)\)
\(=\left(p-q^2\right)\left(p^{m+1}q-pq^{n+1}\right)\)
\(=pq\left(p-q^2\right)\left(p^m-q^n\right)\)
Lời giải:
$N=p^{m+2}q-pq^{m+3}-p^{m+3}q^{n+4}$
$=pq(p^{m+1}-q^{m+2}-p^{m+2}q^{n+3})$
tìm số nguyên m,n,k nguyên sao cho khi phân tích đa thức (x-10)(x-k)+1 thành nhân tử được (x+m)(x+n)
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