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

3) \(-12+2x-9+x=0\\ -21+3x=0\\ 3x=21\\ x=7\)

4)
\(11+\left(15-x\right)=1\)
\(15-x=1-11\)
\(15-x=-10\)
\(x=15-\left(-10\right)\)
\(x=25\)

5) 
\(4-\left(27-3\right)=x-\left(13-4\right)\)
\(4-24=x-9\)
\(x-9=-20\)
\(x=-20+9\)
\(x=-11\)
 

a/ pt đãcho tương đương với

6x\(^2\)+ 21x -2x-7-6x+5x-6x+5= 16

<=>18x=18

=> x=1

b/ pt đã cho tương đương với

10x\(^2\)+9x-10x\(^2\)-15x+2x+3= 8

<=> -4x=5

<=.> x=-\(\frac{5}{4}\)

c/ pt đã cho tương đương với

21x-15x\(^2\)-35+25x+15x\(^2\)-10x+6x-4-2=0

<=>42x=41

<=> x= \(\frac{41}{42}\)

d/ pt đã cho tương đương với

( x\(^2\)+x )(x+6)-x\(^3\)=5x

<=> x\(^3\)+6x\(^2\)+x\(^2\)+6x-x\(^3\)=5x

<=> 8x\(^2\)+6x-5x=0

<=>8x\(^2\)+16x-10x-5x=0

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

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

<=>x+2=0 hoặc 2x+5=0

=> x=-2 hoặc x= -\(\frac{5}{2}\)

\(x\left(x+1\right)\left(x+6\right)-x^3=5x\\ \Rightarrow x\left(x^2+7x+6\right)-x^3=5x\\ \Rightarrow x^3+7x^2+6-x^3-5x=0\\ \Rightarrow7x^2-5x+6=0\\ \)

đến đây nghiệm xấu lắm bạn xem lại đề đúng không

a) x = 2 7                         b) x = 2.

c) x = 2                          d) 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: \(\left(2x^4-x^3+5x^{}-6x^2-1\right):\left(1-2x\right)\)

\(=\frac{2x^4-x^3-6x^2+5x-1}{-2x+1}\)

\(=\frac{-2x^4+x^3+6x^2-5x+1}{2x-1}=\frac{-x^3\left(2x-1\right)+6x^2-3x-2x+1}{2x-1}\)

\(=\frac{-x^3\left(2x-1\right)+3x\left(2x-1\right)-\left(2x-1\right)}{2x-1}=-x^3+3x-1\)

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

\(=\frac{x^3-x^2+x-2x^2+2x-2}{x^2-x+1}\)

\(=\frac{x\left(x^2-x+1\right)-2\left(x^2-x+1\right)}{x^2-x+1}=x-2\)

c: \(\left(2x^3+5x^2-2x+3\right):\left(2x^2-x+1\right)\)

\(=\frac{2x^3-x^2+x+6x^2-3x+3}{2x^2-x+1}\)

\(=\frac{x\left(2x^2-x+1\right)+3\left(2x^2-x+1\right)}{2x^2-x+1}=x+3\)

d: \(\frac{5xy^2}{x^2y}+\frac{5y-x^2}{xy^2}\)

\(=\frac{y\cdot5xy^2+x\left(5y-x^2\right)}{x^2y^2}=\frac{5x^2y^2+5xy-x^3}{x^2y^2}=\frac{x\left(5xy^2+5y-x^2\right)}{x^2y^2}\)

\(=\frac{5xy^2+5y-x^2}{xy^2}\)

e: \(\frac{27-x^3}{5x^5}:\frac{2x-6}{3x^3}\)

\(=\frac{-\left(x-3\right)\left(x^2+3x+9\right)}{5x^5}\cdot\frac{3x^3}{2\left(x-3\right)}=\frac{-3\left(x^2+3x+9\right)}{5x^2\cdot2}=\frac{-3\left(x^2+3x+9\right)}{10x^2}\)

a) \(\Rightarrow x\left(x+3\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=-3\end{matrix}\right.\)

b) \(\Rightarrow x\left(x^2-4\right)=0\Rightarrow x\left(x-2\right)\left(x+2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\)

c) \(\Rightarrow\left(x-1\right)\left(5x-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{5}\end{matrix}\right.\)

d) \(\Rightarrow2\left(x+5\right)-x\left(x+5\right)=0\Rightarrow\left(x+5\right)\left(2-x\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)

e) \(\Rightarrow2x^2-10x-3x-2x^2=26\)

\(\Rightarrow-13x=26\Rightarrow x=-2\)

f) \(\Rightarrow\left(x-2012\right)\left(5x-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=2012\\x=\dfrac{1}{5}\end{matrix}\right.\)

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