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

√(x² + x + 1) = 1

⇔ x² + x + 1 = 1

⇔ x² + x = 0

⇔ x(x + 1) = 0

⇔ x = 0 hoặc x + 1 = 0

*) x + 1 = 0

⇔ x = -1

Vậy x = 0; x = -1

--------------------

√(x² + 1) = -3

Do x² ≥ 0 với mọi x

⇒ x² + 1 > 0 với mọi x

⇒ x² + 1 = -3 là vô lý

Vậy không tìm được x thỏa mãn yêu cầu

--------------------

√(x² - 10x + 25) = 7 - 2x

⇔ √(x - 5)² = 7 - 2x

⇔ |x - 5| = 7 - 2x  (1)

*) Với x ≥ 5, ta có 

(1) ⇔ x - 5 = 7 - 2x

⇔ x + 2x = 7 + 5

⇔ 3x = 12

⇔ x = 4 (loại)

*) Với x < 5, ta có:

(1) ⇔ 5 - x = 7 - 2x

⇔ -x + 2x = 7 - 5

⇔ x = 2 (nhận)

Vậy x = 2

--------------------

√(2x + 5) = 5

⇔ 2x + 5 = 25

⇔ 2x = 20

⇔ x = 20 : 2

⇔ x = 10

Vậy x = 10

-------------------

√(x² - 4x + 4) - 2x +5 = 0

⇔ √(x - 2)² - 2x + 5 = 0

⇔ |x - 2| - 2x + 5 = 0 (2)

*) Với x ≥ 2, ta có: 

(2) ⇔  x - 2 - 2x + 5 = 0

⇔ -x + 3 = 0

⇔ x = 3 (nhận)

*) Với x < 2, ta có:

(2) ⇔ 2 - x - 2x + 5 = 0

⇔ -3x + 7 = 0

⇔ 3x = 7

⇔ x = 7/3 (loại)

Vậy x = 3

1)

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

2) Do \(x^2+1>0\forall x\) nên \(x\in\varnothing\)

3) 

\(\Leftrightarrow\sqrt{\left(x-5\right)^2}=7-2x\\ \Leftrightarrow\left|x-5\right|=7-2x\)

Nếu \(x\ge5\) thì

\(\Leftrightarrow x-5-7+2x=0\\ \Leftrightarrow3x-12=0\\ \Leftrightarrow3x=12\\ \Rightarrow x=4\)

=> Loại trường hợp này

Nếu \(x< 5\) thì

\(\Leftrightarrow5-x-7+2x=0\\ \Leftrightarrow x-2=0\\ \Rightarrow x=2\)

=> Nhận trường hợp này

Vậy x = 2 

4)

\(\Leftrightarrow2x+5=5^2=25\\ \Leftrightarrow2x=25-5=20\\ \Rightarrow x=\dfrac{20}{2}=10\)

5)

\(\Leftrightarrow\sqrt{\left(x-2\right)^2}-2x+5=0\\ \Leftrightarrow\left|x-2\right|-2x+5=0\)

Nếu \(x\ge2\) thì

\(\Leftrightarrow x-2-2x+5=0\\ \Leftrightarrow3-x=0\\ \Rightarrow x=3\)

=> Nhận trường hợp này

Nếu \(x< 2\) thì

\(\Leftrightarrow2-x-2x+5=0\\ \Leftrightarrow7-3x=0\\ \Leftrightarrow3x=7\\ \Rightarrow x=\dfrac{7}{3}\)

=> Loại trường hợp này

Vậy x = 3

a) \(7x\left(x+1\right)-3\left(x+1\right)=0\Rightarrow\left(x+1\right)\left(7x-3\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x+1=0\\7x+3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\x=-\dfrac{3}{7}\end{matrix}\right.\)

b) 3(x + 8) - x² - 8x = 0

=> 3(x + 8) - (x² + 8x) = 0

=> 3(x + 8) - x(x + 8) = 0

=> (x + 8)(3 - x) = 0 => \(\left[{}\begin{matrix}x+8=0\\3-x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-8\\x=3\end{matrix}\right.\)

c) \(x^2-10x=-25\Rightarrow x^2-10x+25=0\Rightarrow\left(x-5\right)^2=0\Rightarrow x=5\)

d) Giống câu c

⇒[x+1=07x+3=0⇒⎡⎣x=−1x=−37⇒[x+1=07x+3=0⇒[x=−1x=−37

b) 3(x + 8) - x² - 8x = 0

=> 3(x + 8) - (x² + 8x) = 0

=> 3(x + 8) - x(x + 8) = 0

=> (x + 8)(3 - x) = 0 => [x+8=03−x=0⇒[x=−8x=3[x+8=03−x=0⇒[x=−8x=3

c) ${}^{}-10x=-25\Rightarrow {}^{}-10x+$

\(\sqrt{x^2-25}+\sqrt{x^2+10x+25}=0.\)

\(\Rightarrow\sqrt{x^2-5^2}+\sqrt{x^2+2.5.x+5^2}=0\)

\(\Rightarrow\sqrt{\left(x-5\right).\left(x+5\right)}+\sqrt{\left(x+5\right)^2}=0\)

\(\Rightarrow\sqrt{\left(x+5\right).\left(x-5+1\right)}=0\)

\(\Rightarrow\hept{\begin{cases}x+5=0\\x-5+1=0\end{cases}}\Rightarrow\hept{\begin{cases}x=-5\\x-4=0\end{cases}}\Rightarrow\hept{\begin{cases}x=-5\\x=4\end{cases}}\)

Vậy \(x=\hept{\begin{cases}-5\\4\end{cases}}\)

a: TH1: x>=2

A=x+x-2=2x-2

TH2: x<2

A=x+2-x=2

b: TH1: x>=3

A=x-3-x=-3

TH2: x<3

A=3-x-x=-2x+3

c: TH1: x>=1

C=x-x+1=1

TH2: x<1

C=x+x-1=2x-1

d: TH1: m>=3

C=m-3-2m=-3-m

TH2: m<3

C=-m+3-2m=-3m+3

e: TH1: m>=1

E=m-m+1=1

TH2: m<1

E=m+m-1=2m-1

a, \(\sqrt{x^2-4x+4}=3\Leftrightarrow\sqrt{\left(x-2\right)^2}=3\)

\(\Leftrightarrow x-2=3\Leftrightarrow x=5\)

b, \(\sqrt{x^2-10x+25}=x+3\Leftrightarrow\sqrt{\left(x-5\right)^2}=x+3\)

\(\Leftrightarrow x-5=x+3\Leftrightarrow0\ne8\)( vô nghiệm ) 

câu c nữa bạn!!!!!!!!!!

a) Đk: \(\forall x\in R\)

a) \(\sqrt{x^2-4x+4}=3\) <=> \(\sqrt{\left(x-2\right)^2}=3\) <=> \(\left|x-2\right|=3\)

<=> \(\orbr{\begin{cases}x-2=3\\x-2=-3\end{cases}}\) <=> \(\orbr{\begin{cases}x=5\\x=-1\end{cases}}\)

Vậy S = {5; -1}

b) Đk: \(\forall x\in R\)

Ta có: \(\sqrt{x^2-10x+25}=x+3\)

<=> \(\sqrt{\left(x-5\right)^2}=x+3\)

<=> \(\left|x-5\right|=x+3\)

<=> \(\orbr{\begin{cases}x-5=x+3\\5-x=x+3\end{cases}}\)

<=> \(\orbr{\begin{cases}0x=8\left(vl\right)\\2=2x\end{cases}}\) <=> x = 1

Vậy S = {1}

c)Đk: x \(\ge\)0

 \(\sqrt{x+1+2\sqrt{x}}-\sqrt{x+16-8\sqrt{x}}=3\)

<=> \(\sqrt{\left(\sqrt{x}+1\right)^2}-\sqrt{\left(\sqrt{x}-4\right)^2}=3\)

<=> \(\left|\sqrt{x}+1\right|-\left|\sqrt{x}-4\right|=3\)

Do \(x\ge0\) => \(\sqrt{x}+1>0\)

<=> \(\sqrt{x}+1-\left|\sqrt{x}-4\right|=3\)

<=> \(\sqrt{x}-2=\left|\sqrt{x}-4\right|\)

<=> \(\orbr{\begin{cases}\sqrt{x}-2=\sqrt{x}-4\left(đk:x\ge16\right)\\\sqrt{x}-2=4-\sqrt{x}\left(đk:0\le x\le16\right)\end{cases}}\)

<=> \(\orbr{\begin{cases}0x=-2\left(vl\right)\\2\sqrt{x}=6\end{cases}}\) <=> \(x=9\)

Vậy S = {9}

1) ĐKXĐ: \(x\ge\dfrac{5}{2}\)

\(\sqrt{x^2}=2x-5\\ \Rightarrow\left|x\right|=2x-5\\ \Rightarrow\left[{}\begin{matrix}x=2x-5\\x=5-2x\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=5\left(tm\right)\\x=\dfrac{5}{3}\left(ktm\right)\end{matrix}\right.\)

2) ĐKXĐ: \(x\ge3\)

\(\sqrt{25x^2-10x+1}=2x-6\\ \Rightarrow\left|5x-1\right|=2x-6\\ \Rightarrow\left[{}\begin{matrix}5x-1=2x-6\\5x-1=6-2x\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{5}{3}\left(ktm\right)\\x=1\left(tm\right)\end{matrix}\right.\)

3) ĐKXĐ: \(x\ge\dfrac{5}{2}\)

\(\sqrt{25-10x+x^2}=2x-5\\ \Rightarrow\left|x-5\right|=2x-5\\ \Rightarrow\left[{}\begin{matrix}x-5=2x-5\\x-5=5-2x\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=\dfrac{10}{3}\left(tm\right)\end{matrix}\right.\)

4) ĐKXĐ: \(x\ge\dfrac{1}{2}\)

\(\sqrt{1-2x+x^2}=2x-1\\ \Rightarrow\left|x-1\right|=2x-1\\ \Rightarrow\left[{}\begin{matrix}x-1=2x-1\\x-1=1-2x\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=\dfrac{2}{3}\left(tm\right)\end{matrix}\right.\)

Thiếu soát gì mog bạn thông cảm :]

1) \(\Leftrightarrow3\sqrt{5x}-4\sqrt{5x}+8\sqrt{5x}=21\)

\(\Leftrightarrow7\sqrt{5x}=21\)

\(\Leftrightarrow\sqrt{5x}=3\)

\(\Leftrightarrow5x=9\)

\(\Leftrightarrow x=\frac{9}{5}\)

2)\(\Leftrightarrow x^2-10x+25=16\)

\(\Leftrightarrow x^2-10x+9=0\)

\(\Leftrightarrow\left(x-9\right)\left(x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-9=0\\x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=9\\x=1\end{cases}}\)