x² - 4x + 4 = 25

x² - 4x       = 25 - 4

x² - 4x       =  21

x²              = 21 + 4x

x.x             = 21 + 4x

\(\Rightarrow\) 3x        = 21

        x        = 21 : 3

        x        = 7

pt<=>(x+2)²=25=>x+2=5 hoac x+2=-5=>x=3 hoac x=-7

x^2-4x+4=25

x^2-4x-21=0

x^2-7x+3x-21=0

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

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

x-7=0 hoặc x+3=0

x=7 ,x=-3

vì \(\left(4x^2-4x+1\right)^{2022}\ge0\left(\forall x\right)\),\(\left(y^2-\dfrac{4}{5}y+\dfrac{4}{25}\right)^{2022}\ge0\left(\forall y\right)\),\(\left|x+y+z\right|\ge0\)

mà \(\left(4x^2-4x+1\right)^{2022}+\left(y^2+\dfrac{4}{5}y+\dfrac{4}{25}\right)^{2022}+\left|x+y-z\right|=0\)

=>\(\left\{{}\begin{matrix}4x^2-4x+1=0\\y^2+\dfrac{4}{5}y+\dfrac{4}{25}=0\\x+y-z=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}2x-1=0\\y+\dfrac{2}{5}=0\\x+y-z=0\end{matrix}\right.\)

<=>\(\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=\dfrac{-2}{5}\\\dfrac{1}{2}-\dfrac{2}{5}-z=0\end{matrix}\right.\)

<=>\(\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=\dfrac{-2}{5}\\z=\dfrac{1}{10}\end{matrix}\right.\)

KL: vậy \(\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=\dfrac{-2}{5}\\z=\dfrac{1}{10}\end{matrix}\right.\)

a  x = \(\dfrac{-1}{12}\)

b  x = \(\dfrac{-4}{3}\)

c  x = \(\dfrac{-1}{6}\)

d  x = \(\dfrac{-1}{4}\)

\(\left(4x+1\right)^2=\dfrac{4}{9}\)

\(\left(4x+1\right)=\perp\left(\dfrac{2}{3}\right)^2\)

\(\text{Vậy }4x+1=\dfrac{2}{3}\)

       \(4x\)        \(=\dfrac{2}{3}+\left(-1\right)=\dfrac{-1}{3}\)

        \(x\)         \(=\left(\dfrac{-1}{3}\right).\dfrac{1}{4}=\dfrac{-1}{12}\)

\(\text{hoặc }4x+1=\dfrac{-2}{3}\)

        \(4x\)        \(=\left(\dfrac{-2}{3}\right)+\left(-1\right)=\dfrac{-5}{3}\)

         \(x\)         \(=\left(\dfrac{-5}{3}\right).\dfrac{1}{4}=\dfrac{-5}{12}\)

\(\Rightarrow x\in\left\{\dfrac{-1}{12};\dfrac{-5}{12}\right\}\)

\(\left(3x-1\right)^2=25\)

\(\left(3x-1\right)^2=\perp\left(5\right)^2\)

\(\text{Vậy }3x-1=5\)

       \(3x\)        \(=5+1=6\)

        \(x\)         \(=6:3=2\)

\(\text{hoặc }3x-1=-5\)

        \(3x\)       \(=\left(-5\right)+1=-4\)

         \(x\)        \(=\left(-4\right):3=\dfrac{-4}{3}\)

\(\Rightarrow x\in\left\{2;\dfrac{-4}{3}\right\}\)

\(\left(x-\dfrac{1}{3}\right)^2=\dfrac{1}{4}\)

\(\left(x-\dfrac{1}{3}\right)^2=\perp\left(\dfrac{1}{2}\right)^2\)

\(\text{Vậy }x-\dfrac{1}{3}=\dfrac{1}{2}\)

       \(x\)         \(=\dfrac{1}{2}+\dfrac{1}{3}=\dfrac{5}{6}\)

\(\text{hoặc }x-\dfrac{1}{3}=\dfrac{-1}{2}\)

        \(x\)         \(=\left(\dfrac{-1}{2}\right)+\dfrac{1}{3}=\dfrac{-1}{6}\)

\(\Rightarrow x\in\left\{\dfrac{5}{6};\dfrac{-1}{6}\right\}\)

\(\left(4x-3\right)^2=16\)

\(\left(4x-3\right)=\perp\left(4\right)^2\)

\(\text{Vậy }4x-3=4\) 

        \(4x\)       \(=4+3=7\)

          \(x\)       \(=7:4=\dfrac{7}{4}\)

\(\text{hoặc }4x-3=-4\)

        \(4x\)        \(=\left(-4\right)+3=-1\)

          \(x\)        \(=\left(-1\right):4=\dfrac{-1}{4}\)

\(\Rightarrow x\in\left\{\dfrac{7}{4};\dfrac{-1}{4}\right\}\)

a: =>25-4x=1

=>4x=24

hay x=6

b: =>2x-4=0

hay x=2

c: =>x-35=115

hay x=150

d: =>x-2=12

hay x=14

e: =>x-36=216

hay x=252

8x²+30x+7=0

 8x²+16x+14x+7=0

8x(x+2) +7(x+2)=0

(8x+7)(x+2)=0

=>\(\orbr{\begin{cases}8x+7=0\\x+2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-\frac{7}{8}\\x=-2\end{cases}}}\)

a)

4x²-8x+4=2(1-x)(x+1)

4x²-8x+4-2+2x²=0

6x²-8x+2=0

2(3x²-4x+1)=0

3x²-3x-x+1=0

3x(x-1) -(x-1)=0

(3x-1)(x-1)=0

\(\Rightarrow\orbr{\begin{cases}3x-1=0\\x-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{1}{3}\\x=1\end{cases}}}\)

d,

x²+3x-18=0

=> x²-3x+6x-18=8

=> x(x-3)+6(x-3)=0

=> (x-3)(x+6)=0

=> \(\orbr{\begin{cases}x=3\\x=-6\end{cases}}\)