a) => 5x=(x-1):(x-1)

=>5x=1

=>x=1/5

b)=>2x+10-x^2-5x=0

=>10-3x-x^2=0

=>10-5x+2x-x^2=0

=>5(2-x)+x(2-x)=0

=>(2-x)(5+x)=0

=> 2-x=0 hoặc 5+x=0

=> x=2 hoặc x=-5

Chúc bạn học tốt T I C K cho mình nha mình cảm ơn rất nhìu

a)\(5x\left(x-1\right)=\left(x-1\right)\)

=>\(5x\left(x-1\right)-\left(x-1\right)=0\)

=>\(\left(x-1\right)\left(5x-1\right)=0\)

=>\(\orbr{\begin{cases}x-1=0\\5x-1=0\end{cases}}\)

=>\(\orbr{\begin{cases}x=1\\x=\frac{1}{5}\end{cases}}\)

\(S=\left\{\frac{1}{5};1\right\}\)

b) \(2\left(x+5\right)-x^2-5x=0\)

=>\(2\left(x+5\right)-x\left(x+5\right)=0\)

=>\(\left(2-x\right)\left(x+5\right)=0\)

=>\(\orbr{\begin{cases}2-x=0\\x+5=0\end{cases}}\)

=>\(\orbr{\begin{cases}x=2\\x=-5\end{cases}}\)

\(S=\left\{-5;2\right\}\)

x²+5x=0

x(x+5)=0

*x=0

*x+5=0

      x=0-5

      x=-5

Vậy......................

1) \(\left(x^3-8\right):\left(x-2\right)=\left[\left(x-2\right)\left(x^2+2x+4\right)\right]:\left(x-2\right)=x^2+2x+4\)

2) \(\left(x^3-1\right):\left(x^2+x+1\right)=\left[\left(x-1\right)\left(x^2+x+1\right)\right]:\left(x^2+x+1\right)=x-1\)

3) \(\left(x^3+3x^2+3x+1\right):\left(x^2+2x+1\right)=\left(x+1\right)^3:\left(x+1\right)^2=x+1\)

4) \(\left(25x^2-4y^2\right):\left(5x-2y\right)=\left[\left(5x-2y\right)\left(5x+2y\right)\right]:\left(5x-2y\right)=5x+2y\)

Bài 1:

\(a,6x^2-15x^3y\\ b,=-\dfrac{2}{3}x^2y^3+\dfrac{2}{3}x^4y-\dfrac{8}{3}xy\)

Bài 2:

\(a,=20x^3-10x^2+5x-20x^3+10x^2+4x=9x\\ b,=3x^2-6x-5x+5x^2-8x^2+24=24-11x\\ c,=x^5+x^3-2x^3-2x=x^5-x^3-2x\)

câu d của bài 2 là của bài 1 nha mình để nhầm chỗ huhu

a,x(x-2)+x-2=0

⇔ (x-2)(x+1)=0

⇔ x=2;x=-1

b,x³+x²+x+1=0

⇔ x²(x+1)+x+1=0

⇔ (x+1)(x²+1)=0

⇔ x=-1

\(a,\Leftrightarrow\left[{}\begin{matrix}x+5=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=-\dfrac{1}{2}\end{matrix}\right.\\ b,\Leftrightarrow\left(x+2\right)\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=3\end{matrix}\right.\\ c,\Leftrightarrow2x^2-10x-3x-2x^2=26\\ \Leftrightarrow-13x=26\Leftrightarrow x=-2\\ d,\Leftrightarrow x^2-18x+16=0\\ \Leftrightarrow\left(x^2-18x+81\right)-65=0\\ \Leftrightarrow\left(x-9\right)^2-65=0\\ \Leftrightarrow\left(x-9+\sqrt{65}\right)\left(x-9-\sqrt{65}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=9-\sqrt{65}\\9+\sqrt{65}\end{matrix}\right.\)

\(e,\Leftrightarrow x^2-10x-25=0\\ \Leftrightarrow\left(x-5\right)^2-50=0\\ \Leftrightarrow\left(x-5-5\sqrt{2}\right)\left(x-5+5\sqrt{2}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=5+5\sqrt{2}\\x=5-5\sqrt{2}\end{matrix}\right.\\ f,\Leftrightarrow5x\left(x-1\right)-\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(5x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{5}\end{matrix}\right.\\ g,\Leftrightarrow2\left(x+5\right)-x\left(x+5\right)=0\\ \Leftrightarrow\left(2-x\right)\left(x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\\ h,\Leftrightarrow x^2+2x+3x+6=0\\ \Leftrightarrow\left(x+3\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-2\end{matrix}\right.\\ i,\Leftrightarrow4x^2-12x+9-4x^2+4=49\\ \Leftrightarrow-12x=36\Leftrightarrow x=-3\)

\(j,\Leftrightarrow x^2\left(x+1\right)+\left(x+1\right)=0\Leftrightarrow\left(x^2+1\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^2=-1\left(vô.lí\right)\\x=-1\end{matrix}\right.\Leftrightarrow x=-1\\ k,\Leftrightarrow x^2\left(x-1\right)=4\left(x-1\right)^2\\ \Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\\ \Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

a) x = 0                 b) x = - 1 3

c) x = 28 15               d) x = -82.

a: 5x-20xy

\(=5x\cdot1-5x\cdot4y=5x\left(1-4y\right)\)

b: \(x^2-9=\left(x-3\right)\left(x+3\right)\)

c: \(x^2-2xy+y^2-z^2\)

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

=(x-y-z)(x-y+z)

d: \(5x\left(x-1\right)-2\left(x-1\right)=\left(x-1\right)\left(5x-2\right)\)

e; \(x^2+4x+3=x^2+x+3x+3\)

=x(x+1)+3(x+1)

=(x+1)(x+3)

f: \(x^3-x+3x^2y+3xy^2+y^3-y\)

\(=\left(x+y\right)^3-\left(x+y\right)\)

\(=\left(x+y\right)\left\lbrack\left(x+y\right)^2-1\right\rbrack\)

=(x+y)(x+y-1)(x+y+1)

g: \(x^2-x-y^2-y\)

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

=(x-y)(x+y)-(x+y)

=(x+y)(x-y-1)

h: \(16x-5x^2-3\)

\(=-5x^2+15x+x-3\)

=-5x(x-3)+(x-3)

=(x-3)(-5x+1)

i: \(x^3-4x=x\left(x^2-4\right)=x\left(x-2\right)\left(x+2\right)\)

j: \(2x^2-6x=2x\cdot x-2x\cdot3=2x\left(x-3\right)\)

k: \(x^3-3x^2-4x+12\)

\(=x^2\left(x-3\right)-4\left(x-3\right)\)

\(=\left(x-3\right)\left(x^2-4\right)=\left(x-3\right)\cdot\left(x-2\right)\left(x+2\right)\)

l: \(x^2-y^2-5x+5y\)

=(x-y)(x+y)-5(x-y)

=(x-y)(x+y-5)