Bài 4: 

Ta có: \(\left(x^3-x^2\right)-4x^2+8x-4=0\)

\(\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

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: \(M=m^2\left(m+n\right)-n^2m-n^3\)

\(=m^2\left(m+n\right)-n^2\left(m+n\right)\)

\(=\left(m+n\right)^2\left(m-n\right)\)

\(=\left(-2017+2017\right)^2\cdot\left(-2017-2017\right)\)

=0

b: \(N=n^3-3n^2-n\left(3-n\right)\)

\(=n^2\left(n-3\right)+n\left(n-3\right)\)

\(=n\left(n-3\right)\left(n+1\right)\)

\(=13\cdot10\cdot14=1820\)

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: \(x^3z+x^2yz-x^2z^2-xyz^2\)

\(=x^2z\left(x+y\right)-xz^2\left(x+y\right)\)

\(=xz\left(x+y\right)\left(x-z\right)\)

x^m+3+1+x^m+3-(x+1)=x^m+3x+x^m+3-(x+1)=x^m+3(x+1)-(x+1)=(x+1)(x^m+3-1)

 Ta có : x^m+4 + x^m+3,-x-1

<=>x^m. x⁴ + x^m . x³ - (x+1)

<=> x^m+3. (x+1) -( x+1)

<=> (x^m+3-1)(x+1)

\(a.a\left(m-n\right)+m-n\)

\(=a\left(m-n\right)+\left(m-n\right)\)

\(=\left(a+1\right)\left(m-n\right)\)

\(b.ma+mb-a-b\)

\(=m\left(a+b\right)-\left(a+b\right)\)

\(=\left(m-1\right)\left(a+b\right)\)

\(c.4x+by+4y+bx\)

\(=\left(4x+4y\right)+\left(bx+by\right)\)

\(=4\left(x+y\right)+b\left(x+y\right)\)

\(=\left(b+4\right)\left(x+y\right)\)

\(d.1-ax-x+a\)

\(=\left(a-ax\right)+\left(1-x\right)\)

\(=a\left(1-x\right)+\left(1-x\right)\)

\(=\left(a+1\right)\left(1-x\right)\)

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)

= 1.( m³ + n³ + p³ - 1 )

tk nha b

đề bài thiếu bn ơi

đù r mà

\(=p^{m+1}.q.\left(p-q^2\right)-p.q^{n+1}\left(p-q^2\right)\)

\(=p.q.\left(p-q^2\right)\left(p^m-q^n\right)\)