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

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

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

b) Ta có: \(x:0.25+x:0.2+x:0.1+x=34\)

\(\Leftrightarrow4x+5x+x+x=34\)

\(\Leftrightarrow11x=34\)

hay \(x=\dfrac{34}{11}\)

\(a,\Leftrightarrow\left(x-2\right)^3-3x\left(x-2\right)=0\\ \Leftrightarrow\left(x-2\right)\left(x-2-3x\right)=0\\ \Leftrightarrow\left(x-2\right)\left(-2x-2\right)=0\\ \Leftrightarrow\left(x-2\right)\left(x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\\ b,\Leftrightarrow\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)=0\\ \Leftrightarrow\left(x+3\right)\left(x^2-2x\right)=0\\ \Leftrightarrow x\left(x-2\right)\left(x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-3\end{matrix}\right.\)

a) \(\Rightarrow x^2+4x+25-x^2=3\Rightarrow4x=-22\Rightarrow x=-\dfrac{11}{2}\)

b) \(\Rightarrow\left(4x+3-2x+3\right)\left(4x+3+2x-3\right)=0\)

\(\Rightarrow2\left(x+3\right).6x=0\)

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

Lời giải:

a.

$\frac{2}{3}+\frac{1}{3}:3\times x=20\text{%}$
$\frac{2}{3}+\frac{1}{9}\times x=\frac{1}{5}$

$\frac{1}{9}\times x=\frac{1}{5}-\frac{2}{3}=\frac{-7}{15}$

$x=\frac{-7}{15}: \frac{1}{9}=\frac{-21}{5}$
b.

$\frac{3-x}{5-x}=\frac{6}{11}$
$\Rightarrow 6(5-x)=11(3-x)$

$\Rightarrow 30-6x=33-11x$

$\Rightarrow 5x=3$

$\Rightarrow x=\frac{3}{5}$

a) \(\left(x-3\right)^2+\left(4-x\right)\left(x+4\right)=10\)

\(\Leftrightarrow\left(x^2-2\cdot x\cdot3+3^2\right)+\left(4-x\right)\left(4+x\right)=10\)

\(\Leftrightarrow x^2-6x+9+\left(4^2-x^2\right)-10=0\)

\(\Leftrightarrow x^2-6x-1+16-x^2=0\)

\(\Leftrightarrow-6x+15=0\)

\(\Leftrightarrow6x=15\)

\(\Leftrightarrow x=\dfrac{5}{2}\)

b) \(x^2-2x=0\)

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

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

c) \(\left(x^2-9\right)^2-\left(x-3\right)^2=0\)

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

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

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

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

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

a. x² - 6x = -9

<=> x² - 6x + 9 = 0

<=> (x - 3)² = 0

<=> x - 3 = 0

<=> x = 3

b. 2(x + 3) - x² + 3x = 0

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

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

<=> \(\left[{}\begin{matrix}2-x=0\\x+3=0\end{matrix}\right.\)

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

Bài 3: 

a: Ta có: 60-3(x-2)=51

\(\Leftrightarrow x-2=3\)

hay x=5

b: Ta có: \(4x-20=25:2^2\)

\(\Leftrightarrow4x=\dfrac{25}{4}+20=\dfrac{105}{4}\)

hay \(x=\dfrac{105}{16}\)

c: Ta có: \(8\cdot6+288:\left(x-3\right)^2=50\)

\(\Leftrightarrow288:\left(x-3\right)^2=50-48=2\)

\(\Leftrightarrow\left(x-3\right)^2=144\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=12\\x-3=-12\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=15\\x=-9\end{matrix}\right.\)

a: \(\left(x-4\right)^2-\left(x-3\right)\left(x+3\right)=5\)

\(\Leftrightarrow x^2-8x+16-x^2+9=5\)

\(\Leftrightarrow-8x=-20\)

hay \(x=\dfrac{5}{2}\)