a) \(\Rightarrow\left(2x-3\right)^2=49\)

\(\Rightarrow\left[{}\begin{matrix}2x-3=7\\2x-3=-7\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)

b) \(\Rightarrow\left(x-5\right)\left(2x+7\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=5\\x=-\dfrac{7}{2}\end{matrix}\right.\)

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

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

a, ⇒ (2x - 3)² = 49

    ⇒  (2x - 3)² = \(\left(\pm7\right)^2\)

    ⇒ \(\left[{}\begin{matrix}2x-3=7\\2x-3=-7\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=10\\2x=-4\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)

b, ⇒ 2x.(x - 5) + 7.(x - 5) = 0

    ⇒ (x - 5).(2x + 7)  = 0

    ⇒ \(\left[{}\begin{matrix}x-5=0\\2x+7=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\2x=-7\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=5\\x=-\dfrac{7}{2}\end{matrix}\right.\)

c, ⇒ x² - 5x + 2x - 10 = 0

    ⇒ (x² - 5x) + (2x - 10) = 0

    ⇒ x.(x - 5) +2.(x - 5)    = 0

    ⇒ (x - 5).(x + 2)=0

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

a) \(7x\left(2x-3\right)-\left(4x^2-9\right)=0\Rightarrow7x\left(2x-3\right)-\left(2x-3\right)\left(2x+3\right)=0\Rightarrow\left(2x-3\right)\left(7x-2x+3\right)=0\Rightarrow\left[{}\begin{matrix}2x-3=0\\5x+3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{5}\end{matrix}\right.\)

b) \(\left(2x-7\right).\left(x-2\right)\left(x^2-4\right)=0\Rightarrow\left(2x-7\right)\left(x-2\right)^2\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}2x-7=0\\\left(x-2\right)^2=0\\x+2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=2\\x=-2\end{matrix}\right.\)

c)\(\left(9x^2-25\right)-\left(6x-10\right)=0\Rightarrow\left(3x-5\right)\left(3x+5\right)-2\left(3x-5\right)=0\Rightarrow\left(3x-5\right)\left(3x+5-2\right)=0\Rightarrow\left[{}\begin{matrix}3x-5=0\\3x+3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=1\end{matrix}\right.\)

a: Ta có: \(7x\left(2x-3\right)-\left(4x^2-9\right)=0\)

\(\Leftrightarrow7x\left(2x-3\right)-\left(2x-3\right)\left(2x+3\right)=0\)

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

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

b: Ta có: \(\left(2x-7\right)\left(x-2\right)\left(x^2-4\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=2\\x=-2\end{matrix}\right.\)

c: Ta có: \(\left(9x^2-25\right)-\left(6x-10\right)=0\)

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

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

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

a: \(\left(2x-3\right)^2-49=0\)

\(\Leftrightarrow\left(2x+4\right)\left(2x-10\right)=0\)

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

a. (2x - 3)² - 49 = 0

<=> (2x - 3)² - 7² = 0

<=> (2x - 3 + 7)(2x - 3 - 7) = 0

<=> (2x + 4)(2x - 10) = 0

<=> \(\left[{}\begin{matrix}2x+4=0\\2x-10=0\end{matrix}\right.\)

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

b. 2x(x - 5) - 7(5 - x) = 0

<=> 2x(x - 5) + 7(x - 5) = 0

<=> (2x + 7)(x - 5) = 0

<=> \(\left[{}\begin{matrix}2x+7=0\\x-5=0\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=-\dfrac{7}{2}\\x=5\end{matrix}\right.\)

c. x² - 3x - 10 = 0

<=> x² - 5x + 2x - 10 = 0

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

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

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

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

a, (2x - 3)² - 49 = 0

(2x - 3)² - 7² = 0

(2x - 3 + 7)( 2x - 3 - 7) = 0

(2x + 4)( 2x - 10) = 0

=> 2x + 4 = 0                => 2x - 10 = 0

     2x       = - 4                   2x         = 10

       x       = - 2                     x         = 5

a. (3x - 1)² - (x + 3)² = 0

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

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

\(\Leftrightarrow4x+2=0\)  hoặc  \(2x-4=0\)

1. \(4x+2=0\Leftrightarrow4x=-2\Leftrightarrow x=-\dfrac{1}{2}\)

2. \(2x-4=0\Leftrightarrow2x=4\Leftrightarrow x=2\)

S=\(\left\{-\dfrac{1}{2};2\right\}\)

b. \(x^3=\dfrac{x}{49}\)

\(\Leftrightarrow49x^3=x\)

\(\Leftrightarrow49x^3-x=0\)

\(\Leftrightarrow x\left(49x^2-1\right)=0\)

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

\(\Leftrightarrow x=0\) hoặc  \(7x+1=0\) hoặc \(7x-1=0\)

1. x=0

2. \(7x+1=0\Leftrightarrow7x=-1\Leftrightarrow x=-\dfrac{1}{7}\)

3. \(7x-1=0\Leftrightarrow7x=1\Leftrightarrow x=\dfrac{1}{7}\)

*Cách khác:

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

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

\(\Leftrightarrow\left[{}\begin{matrix}3x-1=-x-3\\3x-1=x+3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x=-2\\2x=4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=2\end{matrix}\right.\)

Vậy: \(S=\left\{-\dfrac{1}{2};2\right\}\)

b: -7<x<7

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.

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

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

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

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

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

\(a,\Leftrightarrow\left(4x-8\right)\left(x+1\right)=0\\ \Leftrightarrow4\left(x-2\right)\left(x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\\ b,\Leftrightarrow\left(x+1\right)\left(x^2+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\\x^2=-1\left(vô.lí\right)\end{matrix}\right.\Leftrightarrow x=-1\\ c,\Leftrightarrow x^2-2x-4x+8=0\\ \Leftrightarrow\left(x-2\right)\left(x-4\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=4\end{matrix}\right.\\ d,\Leftrightarrow x^3-3x^2+3x-9x+2x-6=0\\ \Leftrightarrow\left(x-3\right)\left(x^2+3x+2\right)=0\\ \Leftrightarrow\left(x-3\right)\left(x^2+x+2x+2\right)=0\\ \Leftrightarrow\left(x-3\right)\left(x+1\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-1\\x=-2\end{matrix}\right.\)

a) \(\Rightarrow4\left(x+1\right)\left(x-2\right)=0\)

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

b) \(\Rightarrow x^2\left(x+1\right)+\left(x+1\right)=0\)

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

\(\Rightarrow x=-1\left(do.x^2+1\ge1>0\right)\)

c) \(\Rightarrow x\left(x-4\right)-2\left(x-4\right)=0\)

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

\(\Rightarrow\left[{}\begin{matrix}x=4\\x=2\end{matrix}\right.\)

d) \(\Rightarrow x^2\left(x-3\right)+3x\left(x-3\right)+2\left(x-3\right)\)

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

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

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

Bài 1:

a) \(\Rightarrow3x^2+3x-2x^2-4x+x+1=0\)

\(\Rightarrow x^2=-1\left(VLý\right)\Rightarrow S=\varnothing\)

b) \(\Rightarrow\left(x-2020\right)\left(2x-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=2020\\x=\dfrac{1}{2}\end{matrix}\right.\)

c) \(\Rightarrow\left(x-10\right)\left(x+2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=10\\x=-2\end{matrix}\right.\)

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

e) \(\Rightarrow\left(x+6\right)\left(x-7\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=-6\\x=7\end{matrix}\right.\)

f) \(\Rightarrow\left(5x-4\right)\left(5x+4\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{4}{5}\\x=-\dfrac{4}{5}\end{matrix}\right.\)

Bài 2:

a) \(\Rightarrow3x\left(x^2-4\right)=0\Rightarrow3x\left(x-2\right)\left(x+2\right)=0\)

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

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

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

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: \(x^2-5x-24=0\)

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

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