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

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

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

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

c) x = 2                          d) x = 1.

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

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

vậy giỏi zữ vậy

a: \(=x^2-36-x^2-14x-49+14x=-85\)

b: \(=\dfrac{5x+35+4x-28-5x-7}{\left(x-7\right)\left(x+7\right)}=\dfrac{4x}{x^2-49}\)

\(a,\left(x+6\right)\left(x-6\right)-\left(x+7\right)^2+14x=x^2-36-x^2-14x-49+14x=-85\\ b,\dfrac{5}{x-7}+\dfrac{4}{x+7}+\dfrac{5x+7}{49-x^2}=\dfrac{5\left(x+7\right)+4\left(x-7\right)-\left(5x+7\right)}{\left(x-7\right)\left(x+7\right)}=\dfrac{5x+35+4x-28-5x-7}{\left(x-7\right)\left(x+7\right)}=\dfrac{4x}{\left(x-7\right)\left(x+7\right)}\)

a)=\(3x^3-15x^2+21x\)

b)\(=-2x^4y-10x^2y+2xy\)

c)\(=-x^3+6x^2+5x-4x^2+24x+20=-x^3+2x^2+29x+20\)

d)\(=2x^4-3x^3+4x^2-2x^2+3x-4=2x^4-3x^32x^2+3x-4\)

e)\(=x^2-4y^2\)

f)\(=-2x^2y^3+y-3\)

g)\(=3xy^4-\dfrac{1}{2}y^2+2x^2y\)

h)\(=9x^2-6x+1-7x^2-14=2x^2-6x-13\)

i)\(=x^2-x-3\)

j)\(=\left(x+2y\right)\left(x^2-2y+4y^2\right):\left(x+2y\right)=x^2-2y+4y^2\)

Tại sao ý b có dấu - trước ngoặc đâu mà đổi dấu mong bn giải đáp

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= -\(\frac{5}{2}\)

a, 2\(x\) + 4 - 5\(x\) = -11

    -(5\(x\) - 2\(x\)) + 4 = -11

    -3\(x\)         + 4 = -11

     3\(x\)              = 11 + 4

     3\(x\)              = 15

       \(x\)              = 15 : 3

        \(x\)              = 5

b, \(x\) - (-5) = 8

    \(x\) + 5 = 8

    \(x\)       = 8 - 5

    \(x\)       = 3