a: \(x\left(x+7\right)-\left(x-2\right)\left(x+3\right)=0\)

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

hay x=-1

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

\(\Leftrightarrow x+2=0\)

hay x=-2

b. (x + 2)² - x² + 4 = 0

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

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

<=> 4(1 + x) + 4 = 0

<=> 4(1 + x) = -4

<=> 1 + x = -1

<=> x = -1 - 1

<=> x = -2

\(a,\) \(x\left(x+7\right)-\left(x-2\right)\left(x+3\right)\)

\(\Leftrightarrow x^2+7x-x^2-3x+2x+6\\ \Leftrightarrow6x=0\\ \Leftrightarrow x=0\)

    \(Vậy...\)

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

\(\Leftrightarrow x^2+4x+4-x^2+4=0\\ \Leftrightarrow4x+8=0\\ \Leftrightarrow x=-2\)

\(a,\Rightarrow x^2+4x+4+x^2-2x+1+x^2-9-3x^2=-8\\ \Rightarrow2x=-4\\ \Rightarrow x=-2\\ b,\Rightarrow2021x\left(x-2020\right)-\left(x-2020\right)=0\\ \Rightarrow\left(2021x-1\right)\left(x-2020\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-2020=0\\2021x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2020\\x=\dfrac{1}{2021}\end{matrix}\right.\)

a) \(\Rightarrow x^2+4x+4+x^2-2x+1+x^2-9-3x^2=-8\)

\(\Rightarrow2x=-4\Rightarrow x=-2\)

b) \(\Rightarrow2021x\left(x-2020\right)-\left(x-2020\right)=0\)

\(\Rightarrow\left(x-2020\right)\left(2021x-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=2020\\x=\dfrac{1}{2021}\end{matrix}\right.\)

\(a,\Leftrightarrow x^2-6x+9-x^2+4=6\\ \Leftrightarrow-6x=-7\Leftrightarrow x=\dfrac{7}{6}\\ b,\Leftrightarrow x\left(x-12\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=12\end{matrix}\right.\)

a) (5 + x)(x - 5) - x(x + 5) = 10

x² - 25 - x² - 5x = 10

-5x = 10 + 25

-5x = 35

x = 35 : (-5)

x = -7

b) x.(2x + 3) - 2(x² + x) = 2

2x² + 3x - 2x² - 2x = 2

x = 2

a: \(\left(x+5\right)\left(x-5\right)-x\left(x+5\right)=10\)

=>\(x^2-25-x^2-5x=10\)

=>-5x-25=10

=>-5x=35

=>x=-7

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

=>\(2x^2+3x-2x^2-2x=2\)

=>x=2

bài 1 :

x + 678 = 2813

         x  =  2813 - 678

         x  = 2135

4529 + x = 7685

            x = 7685 - 4529

            x = 3156

x - 358 = 4768

        x  = 4768 + 358

        x  =  5126

2495 - x = 698

           x = 2495 - 698 

           x = 1797

x × 23 = 3082

        x = 3082 : 23 

        x = 134

36 × x = 27612

      x =  27612 : 36 

      x = 767

x : 42 = 938

       x =  938 x 42 

       x = 39396

4080 : x = 24

           x = 4080 : 24 

           x =170

bài 2 :

a. x + 6734 = 3478 + 5782

    x + 6734 = 9260

             x    =  9260 - 6734

             x    =  2526

b. 2054 + x = 4725 - 279

    2054 + x  = 4446

                x  =  4446 - 2054

                x  = 2392

c. x - 3254 = 237 x 145

    x - 3254 = 34365

             x   =  34365 + 3254

             x   =   37619

d. 124 - x = 44658 : 54

    124 - x = 827

             x = 827 - 124 

             x = 703

Bài 3: 

a. x × 24 = 3027 + 2589

    X x 24 = 5616

            x  = 5616 : 24 

            x  = 234

b. 42 × x = 24024 - 8274

    42 x X = 15750

            X = 15750 : 42 

            X = 375

c. x : 54 = 246 ×185

    X : 54 = 45510

          X  = 45510 x 54 

          X  = 2457540

d. 134260 : x = 13230 : 54

    134260 : x = 245

                   x = 134260 : 245

                   x  = 548

đúng tick chị nhé em!!! cảm ơn em nà :3

a) x * 2 + x * 3 = 60

x * ( 2 + 3 ) = 60

x = 60 : 5

x = 12

b) x * 15 - x * 9 = 78

x * ( 15 - 9 ) = 78

x = 78 : 6

x = 13

a: Ta có: \(\left(x-5\right)\left(x+3\right)=x\left(x-3\right)\)

\(\Leftrightarrow x^2-2x-15-x^2+3x=0\)

\(\Leftrightarrow x=15\)

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

\(\Leftrightarrow x+2=0\)

hay x=-2

c: Ta có: \(\left(x-6\right)\left(x+6\right)=x^2\)

\(\Leftrightarrow x^2-36=x^2\)(vô lý)

a. (x - 5)(x + 3) = x(x - 3)

<=> x² + 3x - 5x - 15 = x² - 3x

<=> x² - x² + 3x - 5x + 3x - 15 = 0

<=> x = 15

b. (x + 2)² = (x - 1)(x + 2)

<=> x² + 4x + 4 = x² + 2x - x - 2

<=> x² - x² + 4x - 2x + x = -2 - 4

<=> 3x = -5

<=> \(x=\dfrac{-5}{3}\)

c. (x - 6)(x + 6) = x²

<=> x² - 36 - x² = 0

<=> x² - x² = 36

<=> 0 = 36 (vô lí)

Vậy nghiệm của PT là \(S=\varnothing\)

d. (2x - 3)² = 4x² - 8 

<=> 4x² - 12x + 9 - 4x² + 8 = 0

<=> 4x² - 4x² - 12x = -8 - 9

<=> -12x = -17

<=> \(x=\dfrac{17}{12}\)

a: Ta có: \(4\left(2-x\right)+x\left(x+6\right)=x^2\)

\(\Leftrightarrow8-4x+x^2+6x-x^2=0\)

\(\Leftrightarrow2x=-8\)

hay x=-4

b: Ta có: \(x\left(x-7\right)-\left(x-2\right)\left(x+5\right)=0\)

\(\Leftrightarrow x^2-7x-x^2-3x+10=0\)

\(\Leftrightarrow-10x=-10\)

hay x=1

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

\(\Leftrightarrow9-4x^2+4x^2-4x+1=2\)

\(\Leftrightarrow-4x=-8\)

hay x=2