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

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

a) \(7x^2=28\Leftrightarrow x^2=7\Leftrightarrow x=\sqrt{7}\)

c) \(\left(x-1\right)\left(x+\dfrac{5}{2}\right)=0\Leftrightarrow x\in\left\{1;\dfrac{-5}{2}\right\}\)

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) 10+2x=4⁵:4³

10+2x=4²

10+2x=16

2x=16-10

2x=6

  x=6:2

  x=3

vậy x=3

b)130-(100+x)=25

100+x=130-25

100+x=105

       x=105-100

       x=5

vậy x=5

c)2².(x+3²)-5=55

 4.(x+9)=55+5

x+9=60:4

x+9=15

   x=15-9

   x=6

vậy x=9

d)7²-7(13-x)=14

49-7(13-x)=14

7(13-x)=49-14

13-x=35:7

13-x=5

     x=13-5

      x=8

vậy x=8

x + 2x + 3x + 4x + 5x = x + 28 

x + 2x + 3x + 4x + 5x - x = 28 

14x = 28 

x = 28 : 14 

x = 2 

x + 2x + 3x + 4x + 5x = x + 28 

=> x + 2x + 3x + 4x + 5x - x = 28

=> x  x ( 1 + 2 + 3 + 4 + 5 - 1 ) = 28

=> x  x  14 = 28

=> x = 28 : 14

=> x = 2 

Vậy x = 2

Chúc bạn học tốt !!! 

\(\text{x+2x+3x+4x+5x=x+28}\)

\(\Rightarrow x+2x+3x+4x+5x+x=28\)

\(\Rightarrow16x=28\)

\(\Rightarrow x=\frac{28}{16}=\frac{7}{4}\)

\(\Leftrightarrow\sqrt{2\left(x^2-1\right)^2+9}+\sqrt{3\left(x-1\right)^2+25}=-3\left(x-1\right)^2+8\)

Ta có:

\(\left\{{}\begin{matrix}\sqrt{2\left(x^2-1\right)^2+9}+\sqrt{3\left(x-1\right)^2+25}\ge\sqrt{9}+\sqrt{25}=8\\-3\left(x-1\right)^2+8\le8\end{matrix}\right.\)

\(\Rightarrow\sqrt{2\left(x^2-1\right)^2+9}+\sqrt{3\left(x-1\right)^2+25}\ge-3\left(x-1\right)^2+8\)

Đẳng thức xảy ra khi và chỉ khi \(x=1\)

\(a,2x^2-18x+28=0\)

\(\Leftrightarrow2\left(x^2-9x+14\right)=0\)

\(\Leftrightarrow x^2-9x+14=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x-7=0\\x-2=0\end{matrix}\right.\)

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

\(b,\dfrac{x-2}{x^2-9}+\dfrac{3x-1}{x+3}=\dfrac{2x+1}{x-3}+1\left(ĐKXĐ:x\ne\pm3\right)\)

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

\(\Leftrightarrow\dfrac{x-2}{\left(x-3\right)\left(x+3\right)}+\dfrac{3x^2-10x+3}{\left(x-3\right)\left(x+3\right)}-\dfrac{2x^2+7x+3}{\left(x-3\right)\left(x+3\right)}-\dfrac{\left(x-3\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=0\)\(\Rightarrow x-2+3x^2-10x+3-2x^2-7x-3-x^2+9=0\)

\(\Leftrightarrow-16x+7=0\)

\(\Leftrightarrow-16x=-7\)

\(\Leftrightarrow x=\dfrac{7}{16}\left(tm\right)\)

\(VậyS=\left\{\dfrac{7}{16}\right\}\)

a: =>x^2-9x+14=0

=>(x-2)(x-7)=0

=>x=2 hoặc x=7

b: =>x-2+(3x-1)(x-3)=(2x+1)(x+3)+x^2-9

=>x-2+3x^2-9x-x+3=2x^2+7x+3+x^2-9

=>3x^2-9x+1=3x^2+7x-6

=>-16x=-7

=>x=7/16

\(\frac{2x}{42}-\frac{28}{3x}=0\Leftrightarrow\frac{x}{21}-\frac{28}{3x}=0\Leftrightarrow\frac{x^2}{21x}-\frac{196}{21x}=0\Leftrightarrow\left(x-14\right)\left(x+14\right)=0\Leftrightarrow\hept{\begin{cases}x-14=0\\x+14=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=14\\x=-14\end{cases}}}\)