a) Ta có: \(7x^2-28=0\)

\(\Leftrightarrow7\left(x^2-4\right)=0\)

\(\Leftrightarrow7\left(x-2\right)\left(x+2\right)=0\)

mà 7>0

nên (x-2)(x+2)=0

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

Vậy: \(x\in\left\{2;-2\right\}\)

b) Ta có: \(\dfrac{2}{3}x\left(x^2-4\right)=0\)

\(\Leftrightarrow\dfrac{2}{3}x\left(x-2\right)\left(x+2\right)=0\)

mà \(\dfrac{2}{3}>0\)

nên x(x-2)(x+2)=0

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

Vậy: \(x\in\left\{0;-2;2\right\}\)

c) Ta có: \(2x\left(3x-5\right)-\left(5-3x\right)=0\)

\(\Leftrightarrow2x\left(3x-5\right)+\left(3x-5\right)=0\)

\(\Leftrightarrow\left(3x-5\right)\left(2x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-5=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=5\\2x=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=-\dfrac{1}{2}\end{matrix}\right.\)

Vậy: \(x\in\left\{\dfrac{5}{3};-\dfrac{1}{2}\right\}\)

d) Ta có: \(\left(2x-1\right)^2-25=0\)

\(\Leftrightarrow\left(2x-1-5\right)\left(2x-1+5\right)=0\)

\(\Leftrightarrow\left(2x-6\right)\left(2x+4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-6=0\\2x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=6\\2x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)

Vậy: \(x\in\left\{3;-2\right\}\)

a,7x² - 28 = 0

=> 7x² = 28 => x² = 4 => x = 2

b,2/3x(x² - 4) = 0

=>2/3x(x - 2)(x + 2) = 0

=> x ∈ {0 ; 2 ; -2}

c,2x(3x - 5) - (5 - 3x) = 0

= 2x(3x - 5) + (3x - 5)

= (3x - 5)(2x + 1) = 0

=> x ∈ { 5/3 ; -1/2}

d, (2x - 1)² - 25 = 0

=> (2x - 4)(2x - 6) = 0

=> x ∈ {2 ;3}

a,7x² - 28 = 0

=> 7x² = 28 => x² = 4 => x = 2

b,2/3x(x² - 4) = 0

=>2/3x(x - 2)(x + 2) = 0

=> x ∈ {0 ; 2 ; -2}

c,2x(3x - 5) - (5 - 3x) = 0

= 2x(3x - 5) + (3x - 5)

= (3x - 5)(2x + 1) = 0

=> x ∈ { 5/3 ; -1/2}

d, (2x - 1)² - 25 = 0

=> (2x - 4)(2x - 6) = 0

=> x ∈ {2 ;3}

a: \(\Leftrightarrow\left(x+2\right)\left(12-x\right)=0\)

\(\Leftrightarrow x\in\left\{-2;12\right\}\)

b: \(\Leftrightarrow\left(2x+5\right)\left(x-1\right)=0\)

\(\Leftrightarrow x\in\left\{-\dfrac{5}{2};1\right\}\)

a) Ta có: \(36x^3-4x=0\)

\(\Leftrightarrow4x\left(9x^2-1\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{3}\\x=\dfrac{-1}{3}\end{matrix}\right.\)

b) Ta có: \(3x\left(x-2\right)+x-2=0\)

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

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

d) Ta có: \(x^2-6x-16=0\)

\(\Leftrightarrow\left(x-8\right)\left(x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-2\end{matrix}\right.\)

e) Ta có: \(x^4-6x^2-7=0\)

\(\Leftrightarrow\left(x^2-7\right)\left(x^2+1\right)=0\)

\(\Leftrightarrow x\in\left\{\sqrt{7};-\sqrt{7}\right\}\)

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: \(9x^2-30x+25=0\)

\(\Leftrightarrow3x-5=0\)

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) \(\left(3x+5\right)\left(7-2x\right)+6x\left(x+4\right)=0\)

\(\Leftrightarrow21x-6x^2+35-10x+6x^2+24x=0\)

\(\Leftrightarrow35x=-35\Leftrightarrow x=-1\)

b) \(x^3-25x=0\)

\(\Leftrightarrow x\left(x^2-25\right)=0\)

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

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

a: Ta có: \(\left(3x+5\right)\left(7-2x\right)+6x\left(x+4\right)=0\)

\(\Leftrightarrow21x-6x^2+35-10x+6x^2+24x=0\)

\(\Leftrightarrow x=1\)

b: Ta có: \(x^3-25x=0\)

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

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

a. (3x + 5)(7 - 2x) + 6x(x + 4) = 0

<=> 21x - 6x² + 35 - 10x + 6x² + 24x = 0

<=> -6x² + 6x² + 21x - 10x + 24x = -35

<=> 35x = -35

<=> x = \(\dfrac{-35}{35}=-1\)

b. x³ - 25x = 0

<=> x(x² - 5²)

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

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

a: 2x-1=0

nên 2x=1

hay x=1/2

b: 4x²-16=0

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

=>x=2 hoặc x=-2

c: x²-2x=0

=>x(x-2)=0

=>x=0 hoặc x=2

a: 2x-1=0

nên 2x=1

hay x=1/2

b: 4x²-16=0

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

=>x=2 hoặc x=-2

c: x²-2x=0

=>x(x-2)=0

=>x=0 hoặc x=2

 a) \(2x-1=0\)

    \(2x\)        \(=1\)

      \(x\)        \(=1:2\)

      \(x\)        \(=\dfrac{1}{2}\)

Vậy \(x=\dfrac{1}{2}\) là nghiệm của đa thức \(2x-1\)

b) \(4x^2-16=0\)

    \(4x^2\)          \(=16\)

      \(x^2\)          \(=16:4\)

      \(x^2\)          \(=4\)

      \(x\)            \(=\overset{-}{+}\) \(2\)

Vậy \(x=-2\) hoặc \(x=2\) là nghiệm của đa thức \(4x^2-16\)

c) \(x^2-2x=0\)

  \(x.x-2x=0\)

    \(x.\left(x-2\right)=0\)

⇒ \(x=0\) hoặc \(x-2=0\)

⇒ \(x=0\) hoặc \(x\)        \(=0+2=2\)

Vậy \(x=0\) hoặc \(x=2\) là nghiệm của đa thức \(x^2-2x\)

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

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

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

Vậy \(x=1\); \(x=2\) hoặc \(x=-2\) là nghiệm của đa thức  \(\left(x-1\right).\left(x^2-4\right)\)

e) \(x^3+3x=0\)

   \(x.x.x+3x=0\)

     \(x.\left(x^2+3\right)=0\)

⇒ \(x=0\) hoặc \(x^2+3=0\)

⇒ \(x=0\) hoặc \(x^2\)        \(=0+3\)

⇒ \(x=0\) hoặc \(x^2\)         \(=3\)   (Không bằng 0)

Vậy \(x=0\) là nghiệm của đa thức  \(x^3+3x\)

f) \(x^2+3x-4=0\)

⇒ \(x.\left(x+1\right)+4\left(x-1\right)=0\)

⇒   \(\left(x-1\right).\left(x+4\right)=0\)

      ⇔\(\left[{}\begin{matrix}x-1=0=0+1=1\\x+4=0=0-4=-4\end{matrix}\right.\)

Vậy \(x=1\) và \(x=-4\) là nghiệm của đa thức \(x^2+3x-4\)

\(a,\Leftrightarrow x\left(x+9\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-9\end{matrix}\right.\\ b,\Leftrightarrow\left(x+4-4\right)\left(x+4+4\right)=0\\ \Leftrightarrow x\left(x+8\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-8\end{matrix}\right.\\ c,\Leftrightarrow x\left(x-4\right)\left(x+4\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\\ d,\Leftrightarrow\left(x-5\right)^2=0\Leftrightarrow x=5\)

a) \(\Leftrightarrow x\left(x+9\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-9\end{matrix}\right.\)

b) \(\Leftrightarrow x\left(x+8\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-8\end{matrix}\right.\)

c) \(\Leftrightarrow x\left(x-4\right)\left(x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\)

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

a) (x-4)(x+4)-x(x+2)=0

     x²-16-x²-2x = 0

     -16 - 2x = 0

             2x = -16 

               x = -16/2

               x = -8

b) 3x(x-2)-x+2=0

     (3x-1)(x-2)=0

=> x ∈ {1/3 ; 2 }

c) 6x - 12x² = 0

    6x(1-2x) = 0

=> x ∈ {0; 1/2 }

d) mình thấy có vẻ hơi sai đề nên mình ko giải được, bạn thông cảm nha

d/ 4x (3 - ¹/₄ x) + (x -2) ( x+ 2)

câu d bị  sai đề