\(a,25x^2-1=15\)\(< =>x^2=\dfrac{16}{25}< =>x=\pm\dfrac{4}{5}\)

\(b,\left(x-4\right)^2-\left(5x+2\right)^2=0\)\(< =>\left(-4x-6\right)\left(6x-2\right)=0\)

\(< =>\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{1}{3}\end{matrix}\right.\)

\(c,\left(x-1\right)\left(x-9\right)=0< =>\left[{}\begin{matrix}x=1\\x=9\end{matrix}\right.\)

a: Ta có: \(40x^4+5x=0\)

\(\Leftrightarrow5x\left(8x^3+1\right)=0\)

\(\Leftrightarrow x\left(2x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{1}{2}\end{matrix}\right.\)

b: Ta có: \(8x^2-2x-1=0\)

\(\Leftrightarrow8x^2-4x+2x-1=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-\dfrac{1}{4}\end{matrix}\right.\)

a: Ta có: \(\left(3x-2\right)\left(2x-1\right)-\left(6x^2-3x\right)=0\)

\(\Leftrightarrow2x-1=0\)

hay \(x=\dfrac{1}{2}\)

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

\(\Leftrightarrow x^3-x^3-1=x\)

hay x=-1

c: Ta có: \(56x^4+7x=0\)

\(\Leftrightarrow7x\left(8x^3+1\right)=0\)

\(\Leftrightarrow x\left(2x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{1}{2}\end{matrix}\right.\)

d: Ta có: \(x^2-5x-24=0\)

\(\Leftrightarrow\left(x-8\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-3\end{matrix}\right.\)

\(a,\Leftrightarrow\left(x-2\right)\left(5x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{2}{5}\end{matrix}\right.\\ b,\Leftrightarrow2x^2+2x-x^2+4x-4-6=0\\ \Leftrightarrow x^2+6x-10=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-3+\sqrt{19}\\x=-3-\sqrt{19}\end{matrix}\right.\\ c,\Leftrightarrow2x^2-2x+9x-9=0\\ \Leftrightarrow\left(2x+9\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{9}{2}\\x=1\end{matrix}\right.\)

a.

\(\Leftrightarrow\left(3x-1\right)^3=\left(-\dfrac{1}{2}\right)^3\)

\(\Leftrightarrow3x-1=-\dfrac{1}{2}\)

\(\Leftrightarrow3x=\dfrac{1}{2}\)

\(\Leftrightarrow x=\dfrac{1}{6}\)

b.

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

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

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

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

c.

\(\Leftrightarrow3x\left(5x-2\right)-2\left(5x-2\right)=0\)

\(\Leftrightarrow\left(3x-2\right)\left(5x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=\dfrac{2}{5}\end{matrix}\right.\)

`a)(x-6)^2-(x+6)^2=12`

`<=>(x-6-x-6)(x-6+x+6)=12`

`<=>-12.2x=12`

`<=>2x=-1`

`<=>x=-1/2`

Vậy `x=-1/2`

`b)36x^2-12x+1=81`

`<=>(6x-1)^2=81`

`<=>(6x-1-9)(6x-1+9)=0`

`<=>(6x-10)(6x+8)=0`

`<=>(3x-5)(3x+4)=0`

`<=>` \(\left[ \begin{array}{l}x=\dfrac53\\x=-\dfrac43\end{array} \right.\) 

`c)x^2-4x-12=0`

`<=>x^2-6x+2x-12=0`

`<=>x(x-6)+2(x-6)=0`

`<=>(x-6)(x+2)=0`

`<=>` \(\left[ \begin{array}{l}x=-2\\x=6\end{array} \right.\) 

`d)x^2-5x-6=0`

`<=>x^2-6x+x-6=0`

`<=>x(x-6)+x-6=0`

`<=>(x-6)(x+1)=0`

`<=>` \(\left[ \begin{array}{l}x=6\\x=-1\end{array} \right.\) 

a: \(9x^2-30x+25=0\)

\(\Leftrightarrow3x-5=0\)

hay \(x=\dfrac{5}{3}\)

c: \(9x^2-25=0\)

\(\Leftrightarrow\left(3x-5\right)\left(3x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=-\dfrac{5}{3}\end{matrix}\right.\)

a) \(9x^2-30x+25=0\Rightarrow\left(3x-5\right)^2=0\Rightarrow x=\dfrac{5}{3}\)

b) \(25x^2-5x+\dfrac{1}{4}=0\Rightarrow\left(10x-1\right)^2=0\Rightarrow x=\dfrac{1}{10}\)

c) \(9x^2-25=0\Rightarrow\left(3x-5\right)\left(3x+5\right)=0\)

    \(\Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=-\dfrac{5}{3}\end{matrix}\right.\)

d) \(\left(2x-1\right)^2-\left(3x+2\right)^2=0\)

   \(\Rightarrow\left(2x-1+3x+2\right)\left(2x-1-3x-2\right)=0\)

  \(\Rightarrow-\left(5x+1\right)\left(5x+3\right)=0\)

 \(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{5}\\x=-\dfrac{3}{5}\end{matrix}\right.\)

a) 1\(\dfrac{2}{3}\).           b)\(\dfrac{1}{7}\).             c) 1               d )0

a: =>x+5>0

hay x>-5

b: =>2x+1<0

hay x<-1/2

c: =>(x-1)(x-4)>0

=>x>4 hoặc x<1

a) x>-5 ĐKXĐ x\(\ne\)-5

b)x<\(-\dfrac{1}{2}\)

c)x>4 hoặc x<1

d)ĐKXĐ x\(\ne\)1, ko tìm đc x

\(a,\) \(5x\left(4-x\right)+\left(5x^2-12\right)=x+6\)

\(< =>20x-5x^2+5x^2-12-x-6=0\)

\(< =>19x-18=0\)

\(< =>x=\dfrac{18}{19}\)

\(b,\left(2x-7\right)\left(5+4x\right)-8\left(x^2-4x+5\right)=-30\)

\(< =>10x+8x^2-35-28x-8x^2+24x-40+30=0\)

\(< =>6x-45=0< =>x=\dfrac{45}{6}=7,5\)

a) \(5x\left(4-x\right)+\left(5x^2-12\right)=x+\Rightarrow6\\ \Leftrightarrow20x-5x^2+5x^2-12=x+6\\ \Leftrightarrow20x-12=x+6\\\Rightarrow20x-x=6+12\\ \Rightarrow19x=18\\ \Rightarrow x=\dfrac{18}{19}\)

b) \(\left(2x-7\right)\left(5+4x\right)-8\left(x^2-3x+5\right)=-30\\ \Rightarrow10x+8x^2-35-28x-8x^2+24x-40=-30\\ \Rightarrow6x-75=-30\\ \Rightarrow6x=45\\ \Rightarrow x=\dfrac{15}{2}\)

a)x=18/19

b)x=15/2

a.

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

\(\Leftrightarrow2\left(x+5\right)-x\left(x+5\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}2-x=0\\x+5=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\)

b.

\(6x^2-7x+2=0\)

\(\Leftrightarrow6x^2-3x-4x+2=0\)

\(\Leftrightarrow3x\left(2x-1\right)-2\left(2x-1\right)=0\)

\(\Leftrightarrow\left(3x-2\right)\left(2x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=\dfrac{1}{2}\end{matrix}\right.\)