1, \(\frac{x^2+2x+1}{2x^2-2}=\frac{\left(x+1\right)^2}{2\left(x^2-1\right)}=\frac{\left(x+1\right)^2}{2\left(x+1\right)\left(x-1\right)}=\frac{x+1}{2\left(x-1\right)}\)= \(\frac{x+1}{2x-2}\)

2 \(\frac{x^2-6x+9}{5x^2-45}=\frac{\left(x-3\right)^2}{5\left(x^2-9\right)}=\frac{\left(x-3\right)^2}{5\left(x-3\right)\left(x+3\right)}=\frac{x-3}{5x+15}\)

3 \(\frac{x^2-12x+36}{2x^2-4x}=\frac{\left(x-6\right)^2}{2x\left(x-2\right)}\)

4 \(\frac{x^2-10x+25}{2x^2-50}=\frac{\left(x-5\right)^2}{2\left(x^2-25\right)}=\frac{\left(x-5\right)^2}{2\left(x-5\right)\left(x+5\right)}=\frac{x-5}{2x+10}\)

\(F\left(x\right)=x^2-3x^3-\sqrt{25}+\frac{1}{2}x-\left(-3x^3+\frac{5x}{2}-\sqrt{36}\right)\)

=> \(F\left(x\right)=x^2-3x^3-5+\frac{1}{2}x+3x^3-\frac{5x}{2}+6\)

=> \(F\left(x\right)=x^2+\left(3x^3-3x^3\right)+\left(6-5\right)+\left(\frac{x}{2}-\frac{5x}{2}\right)\)

=> \(F\left(x\right)=x^2+1-2x\)

\(\left(6+5x\right)^{50}=36^{25}\)

\(\left(6+5x\right)^{50}=6^{50}\)

\(\Rightarrow6+5x=6\)

\(\Rightarrow5x=0\)

\(\Rightarrow x=0\)

vay \(x=0\)

(6+5x)⁵⁰=36²⁵

=> [(6+5x)² ]²⁵=36²⁵

=> (6+5x)²=36

=> (6+5x)²=6²

=> 6+5x=6

=> 5x=0

=> x=0

\(\left(6+5x\right)^{50}=36^{25}\)

\(\left(6+5x\right)^{50}=\left(6^2\right)^{25}\)

\(\left(6+5x\right)^{50}=6^{50}\)

=> \(6+5x=6\)

      \(5x=0\)

       \(x=0\)

5x+35=25                      -2 |x| +4=-20                                     4.13.(-125).(-25).8                     -62.25+(-25).36

5x=25-35=-10                 -2|x|=20-4=16                                    =(-25.4).(-125.8).13                   =-25.(62+36)

x=-10/5=-2                      |x|=16/-2=-8                                       =-100.-1000.13                        =-25.98

                                      Vậy x ko có giá trị nào do |x|=-8          =1300000                                =-2450

Bn ghi gì mik ko hiểu 

5x+35=25

5x=25-35

5x=-10

x=-10:5

x=2

1, \(x^2\) - 9 = 0

 (\(x\) - 3)(\(x\) + 3) = 0

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

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

 vậy \(x\) \(\in\) {-3; 3}

7, (3\(x\) + 1)² - 16 = 0

    (3\(x\) + 1 - 4)(3\(x\) + 1 + 4) = 0

    (3\(x\) - 3).(3\(x\) + 5) = 0

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

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

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

Vậy \(x\) \(\in\) {1; - \(\dfrac{5}{3}\)}

10, (\(x\) + 3)² - \(x^2\) = 45

      [(\(x\) + 3) - \(x\)].[(\(x\) + 3) + \(x\)] = 45

                 3.(2\(x\) + 3)  = 45

                     2\(x\) + 3   = 15

                     2\(x\)          = 12

                       \(x\)          = 6

\(1,\\ b,=\left(x-6\right)\left(x+6\right)\\ 3,\\ x^2-2x+1=25\\ \Leftrightarrow\left(x-1\right)^2-25=0\\ \Leftrightarrow\left(x-6\right)\left(x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\)

\(a,5x-5^2=3x+25\)

\(5x-3x=25+5^2\)

\(2x=50\)

\(x=50:2\)

\(x=25\)

\(b,3x+69:23=2x-108:36\)

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

\(3x-2x=-3-3\)

\(x=-6\)

Học tốt

a, 5x - 5² = 3x + 25

=> 5x - 25 = 3x + 25

=> 5x - 3x = 25 + 25

=> 2x = 50

=> x = 25

b, 3x + 69 : 23 = 2x - 108 : 36

=> 3x + 3 = 2x - 3

=> 3x - 2x = - 3 - 3

=> x = - 6

a,
5x-5^2=3x+25
5x-25=3x+25
2x=50 => x = 25
b,
3x+3=2x-3
3x-2x=-6 => x=-6