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

\(a,=x^2-4x+4-\dfrac{15}{4}=\left(x-2\right)^2-\dfrac{15}{4}=\left(x-2-\dfrac{\sqrt{15}}{2}\right)\left(x-2+\dfrac{\sqrt{15}}{2}\right)\\ b,=?\\ c,\Rightarrow x^2+7x-8=0\\ \Rightarrow\left(x+8\right)\left(x-1\right)=0\Rightarrow\left[{}\begin{matrix}x=-8\\x=1\end{matrix}\right.\\ d,Sửa:x^3-3x^2=-27+9x\\ \Rightarrow x^3-3x^2+9x-27=0\\ \Rightarrow x^2\left(x-3\right)+9\left(x-3\right)=0\\ \Rightarrow\left(x^2+9\right)\left(x-3\right)=0\Rightarrow\left[{}\begin{matrix}x^2=-9\left(vô.lí\right)\\x=3\end{matrix}\right.\\ \Rightarrow x=3\\ e,\Rightarrow x\left(x-3\right)-7x+21=0\\ \Rightarrow x\left(x-3\right)-7\left(x-3\right)=0\\ \Rightarrow\left(x-7\right)\left(x-3\right)=0\Rightarrow\left[{}\begin{matrix}x=7\\x=3\end{matrix}\right.\\ f,\Rightarrow x^2\left(x-2\right)+\left(x-2\right)=0\\ \Rightarrow\left(x^2+1\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x^2=-1\left(vô.lí\right)\\x=2\end{matrix}\right.\\ \Rightarrow x=2\)

\(g,\Rightarrow x^2-4x+4=0\\ \Rightarrow\left(x-2\right)^2=0\\ \Rightarrow x=2\\ h,Sửa:x^3-x^2+x=1\\ \Rightarrow x^2\left(x-1\right)+\left(x-1\right)=0\\ \Rightarrow\left(x^2+1\right)\left(x-1\right)=0\Rightarrow\left[{}\begin{matrix}x^2=-1\left(vô.lí\right)\\x=1\end{matrix}\right.\\ \Rightarrow x=1\)

a)

 ⇔ \(x^2-16=9\)

⇔ \(x^2=25\)

⇔ \(x=\pm5\)

b)

 ⇔ \(x^2-4x+4-25x^2+20x-4=0\)

⇔ \(16x-24x^2=0\)

⇔ \(8x\left(2-3x\right)=0\)

⇒ \(\left[{}\begin{matrix}x=0\\2-3x=0\end{matrix}\right.\)   ⇔   \(\left[{}\begin{matrix}x=0\\x=\dfrac{2}{3}\end{matrix}\right.\)

Vậy \(x=0\) hoặc \(x=\dfrac{2}{3}\)

c)  

⇔ \(3x^2-10x-20=0\)

⇔ \(x^2-2.x.\dfrac{5}{3}+\dfrac{25}{9}-\dfrac{205}{9}=0\)

⇔ \(\left(x-\dfrac{5}{3}\right)^2=\dfrac{205}{9}\)

⇒ \(\left[{}\begin{matrix}x-\dfrac{5}{3}=\sqrt{\dfrac{205}{9}}\\x-\dfrac{5}{3}=-\sqrt{\dfrac{205}{9}}\end{matrix}\right.\)  ⇔ \(\left[{}\begin{matrix}x=\dfrac{\sqrt{\text{205}}}{\text{3}}+\dfrac{5}{3}\\x=-\dfrac{\sqrt{\text{205}}}{\text{3}}+\dfrac{5}{3}\end{matrix}\right.\)  ⇔ \(\left[{}\begin{matrix}x=\dfrac{15+\text{9}\sqrt{\text{205}}}{\text{9}}\\\text{x}=-\dfrac{15+\text{9}\sqrt{\text{205}}}{\text{9}}\end{matrix}\right.\)

Vậy... 

d) 

⇔ \(\left(x^2+x\right)^2-49=\left(x^2+x\right)^2-7x\)

⇔ 7x = 49

⇔ x=7

Vậy...

1, \(\sqrt{4-4x+x^2}=3\)

\(\Leftrightarrow\sqrt{\left(2+x\right)^2}=3\)

\(\Leftrightarrow\left|2+x\right|=3\)

TH1: \(\left|2-x\right|=2-x\) với \(2-x\ge0\Leftrightarrow x\le2\)

Pt trở thành:

\(2-x=3\) (ĐK: \(x\le2\) )

\(\Leftrightarrow x=2-3\)

\(\Leftrightarrow x=-1\left(tm\right)\)

TH2: \(\left|2-x\right|=-\left(2-x\right)\) với \(2-x< 0\Leftrightarrow x>2\)

Pt trở thành:

\(-\left(2-x\right)=3\) (ĐK: \(x>2\))

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

\(\Leftrightarrow x=3+2\)

\(\Leftrightarrow x=5\left(tm\right)\)

Vậy \(S=\left\{-1;5\right\}\)

2, \(\sqrt{x^2-6x+9}=1\)

\(\Leftrightarrow\sqrt{x^2-2\cdot3\cdot x+3^2}=1\)

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

\(\Leftrightarrow\left|x-3\right|=1\)

TH1: \(\left|x-3\right|=x-3\) với \(x-3\ge0\Leftrightarrow x\ge3\)

Pt trở thành:

\(x-3=1\) (ĐK: \(x\ge3\))

\(\Leftrightarrow x=1+3\)

\(\Leftrightarrow x=4\left(tm\right)\)

TH2: \(\left|x-3\right|=-\left(x-3\right)\) với \(x-3< 0\Leftrightarrow x< 3\)

Pt trở thành:

\(-\left(x-3\right)=1\) (ĐK: \(x< 3\))

\(\Leftrightarrow-x+3=1\)

\(\Leftrightarrow-x=1-3\)

\(\Leftrightarrow-x=-2\)

\(\Leftrightarrow x=2\left(tm\right)\)

Vậy \(S=\left\{2;4\right\}\)

1) √(4 - 4x + x²) = 3

⇔ √(2 - x)² = 3

ĐKXĐ: Với mọi x ∈ R

⇔ |2 - x| = 3 (1)

*) |2 - x| = 2 - x ⇔ 2 - x ≥ 0 ⇔ x ≥ 2

(1) ⇔ 2 - x = 3

⇔ x = 2 - 3

⇔ x = -1 (nhận)

*) |2 - x| = x - 2 ⇔ 2 - x < 0 ⇔ x > 2

(1) ⇔ x - 2 = 3

⇔ x = 5 (nhận)

Vậy x = -1; x = 5

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

(x-2)² = 25 => x- 2 = 5
                  => x = 7

\(x^2-4x+4=25\)

\(\Leftrightarrow\left(x-2\right)^2=25\)

\(\Leftrightarrow\orbr{\begin{cases}x-2=5\\x-2=-5\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=7\\x=-3\end{cases}}\)

B = 2\(x^2\) - 4\(x\) - 8

B = 2(\(x^2\) - 2\(x\) + 4)  - 16

B = 2(\(x-2\))² - 16 

Vì (\(x-2\))² ≥ 0 ∀ \(x\) ⇒ 2(\(x-2\))² ≥ 0 ∀ \(x\)

⇒ 2(\(x-2\))²  - 16 ≥ -16 ∀ \(x\)

Dấu bằng xảy ra khi  (\(x-2\))² = 0 ⇒ \(x-2=0\) ⇒ \(x=2\)

Vậy B_min = -16 khi \(x=2\)

Tìm min của C biết:

C = \(x^2\) - 2\(xy\) + 2y² + 2\(x\) - 10y + 17

C = (\(x^2\) - 2\(xy\) + y²) + 2(\(x\) - y) + y² - 8y + 16 + 1

C = (\(x\) - y)² + 2(\(x\) - y) + 1  + (y² - 8y + 16) 

C = (\(x-y+1\))² + (y - 4)² 

Vì (\(x\) - y + 1)² ≥ 0 ∀ \(x;y\); (y - 4)² ≥ 0 ∀ y

Dấu bằng xảy ra khi: \(\left\{{}\begin{matrix}x-y+1=0\\y-4=0\end{matrix}\right.\) ⇒ \(\left\{{}\begin{matrix}x-y+1=0\\y=4\end{matrix}\right.\)

⇒ \(\left\{{}\begin{matrix}x-4+1=0\\y=4\end{matrix}\right.\) ⇒ \(\left\{{}\begin{matrix}x=-1+4\\y=4\end{matrix}\right.\) ⇒ \(\left\{{}\begin{matrix}x=3\\y=4\end{matrix}\right.\)

Vậy C_min = 0 khi (\(x;y\)) = (3; 4)

D = \(x^2\) - \(xy\) + y² - 2\(x\) - 2y

D=[\(x^2\)-2\(x\)\(\dfrac{y}{2}\)+(\(\dfrac{y}{2}\))²]-(2\(x\)-2\(\dfrac{y}{2}\)) +1 +(\(\dfrac{3}{4}\)y²-2.\(\dfrac{\sqrt{3}}{2}\)y .\(\sqrt{3}\) +3) - 4

D = (\(x-\dfrac{y}{2}\))² - 2(\(x-\dfrac{y}{2}\))+ 1 + (\(\dfrac{\sqrt{3}}{2}\)y - \(\sqrt{3}\))² - 4

D = (\(x-\dfrac{y}{2}\) - 1)² + (\(\dfrac{\sqrt{3}}{2}\)y - \(\sqrt{3}\))² - 4

Vì (\(x-\dfrac{y}{2}\) - 1)² ≥  0 ∀ \(x\);y và (\(\dfrac{\sqrt{3}}{2}\)y - \(\sqrt{3}\))² ≥ 0 ∀ y 

Vậy (\(x-\dfrac{y}{2}\) - 1)² + (\(\dfrac{\sqrt{3}}{2}\)y - \(\sqrt{3}\))² - 4 ≥ - 4 ∀ \(x;y\)

Dấu bằng xảy ra khi: \(\left\{{}\begin{matrix}x-\dfrac{y}{2}-1=0\\\dfrac{\sqrt{3}}{2}y-\sqrt{3}=0\end{matrix}\right.\)

      ⇒ \(\left\{{}\begin{matrix}x-\dfrac{y}{2}-1=0\\\sqrt{3}.\left(\dfrac{1}{2}y-1\right)=0\end{matrix}\right.\)

  ⇒ \(\left\{{}\begin{matrix}x=1+\dfrac{1}{2}y\\\dfrac{1}{2}y=1\end{matrix}\right.\) ⇒ \(\left\{{}\begin{matrix}x=1+1\\y=1:\dfrac{1}{2}\end{matrix}\right.\) ⇒ \(\left\{{}\begin{matrix}x=2\\y=2\end{matrix}\right.\)

Vậy D_min = - 4 khi (\(x;y\)) =(2; 2)

a) \(\Rightarrow9x^2+24x+16-9x^2+1=49\)

\(\Rightarrow24x=32\Rightarrow x=\dfrac{4}{3}\)

b) \(\Rightarrow x^2-13x+22=0\)

\(\Rightarrow\left(x-11\right)\left(x-2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=11\\x=2\end{matrix}\right.\)

c) \(\Rightarrow x^2-3x-10=0\)

\(\Rightarrow\left(x-5\right)\left(x+2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)