Sửa đề: \(K=\sqrt{x^2+10x+25}+\sqrt{x^2+4x+4}\)

\(=\left|x+5\right|+\left|x+2\right|\)

\(=\left|x+5\right|+\left|-x-2\right|>=\left| x+5-x-2\right|=3\)

Dấu = xảy ra khi -5<=x<=-2

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: \(x\ge\dfrac{1}{2}\)

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

a) giải pt ra ta được  : x=-1

b) giải pt ra ta được  : x=2

c)giải pt ra ta được  : x vô ngiệm

d)giải pt ra ta được  : x=vô ngiệm

~~~~~~~~~~~ai đi ngang qua nhớ để lại k ~~~~~~~~~~~~~

~~~~~~~~~~~~ Chúc bạn sớm kiếm được nhiều điểm hỏi đáp ~~~~~~~~~~~~~~~~~~~

√(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

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√(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, \(\sqrt{4x^2+20x+25}\) + \(\sqrt{x^2-8x+16}\) = \(\sqrt{x^2+18x+81}\)

⇔ 4x² + 20x + 25 + \(2\sqrt{\left(4x^2+20x+25\right)\left(x^2-8x+16\right)}\) = x² + 18x + 81

⇔ 4x² + 20x + 25 - x² - 18x - 81 + \(2\sqrt{\left(2x+5\right)^2.\left(x-4\right)^2}\) = 0

⇔ 3x² + 2x - 56 + 2.(2x + 5) . (x - 4) = 0

⇔ 3x² + 2x - 56 + (4x + 10) . (x - 4) = 0

⇔ 3x² + 2x - 56 + 4x² - 16x + 10x - 40 = 0

⇔ 7x² - 4x - 96 = 0

x₁ = 4 ( nhận )

x₂ = \(\frac{-24}{7}\) ( nhận )

Vậy: S = {4; \(\frac{-24}{7}\)}

ĐK: \(x\ge-\dfrac{5}{2}\)

\(\Leftrightarrow3x^2-4x-4=2x+5\)

\(\Leftrightarrow3x^2-6x-9=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=3\end{matrix}\right.\) (thỏa mãn)

b.

ĐKXĐ: \(3\le x\le8\)

\(\Leftrightarrow-x^2+11x-24-\sqrt{-x^2+11x-24}-2=0\)

Đặt \(\sqrt{-x^2+11x-24}=t\ge0\)

\(\Rightarrow t^2-t-2=0\Rightarrow\left[{}\begin{matrix}t=-1\left(loại\right)\\t=2\end{matrix}\right.\)

\(\Leftrightarrow\sqrt{-x^2+11x-24}=2\)

\(\Leftrightarrow-x^2+11x-28=0\Rightarrow\left[{}\begin{matrix}x=7\\x=4\end{matrix}\right.\)

c) \(\sqrt{\left(x-2\right)^2}=10\)

\(x-2=10\)

\(x=12\)

d) \(\sqrt{9x^2-6x+1}=15\)

\(\sqrt{\left(3x\right)^2-2.3x.1+1^2}=15\)

\(\sqrt{\left(3x-1\right)^2}=15\)

\(3x-1=15\)

\(3x=16\)

\(x=\dfrac{16}{3}\)

a) \(đk:x\ge0\)

\(pt\Leftrightarrow3\sqrt{2x}+4\sqrt{2x}-3\sqrt{2x}=12\)

\(\Leftrightarrow4\sqrt{2x}=12\Leftrightarrow\sqrt{2x}=3\Leftrightarrow2x=9\Leftrightarrow x=\dfrac{9}{2}\left(tm\right)\)

b) \(đk:x\ge-2\)

\(pt\Leftrightarrow3\sqrt{x+2}+12\sqrt{x+2}-2\sqrt{x+2}=26\)

\(\Leftrightarrow13\sqrt{x+2}=26\)

\(\Leftrightarrow\sqrt{x+2}=2\Leftrightarrow x+2=4\Leftrightarrow x=2\left(tm\right)\)

c) \(pt\Leftrightarrow\left|x-2\right|=10\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=10\\x-2=-10\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=12\\x=-8\end{matrix}\right.\)

d) \(pt\Leftrightarrow\sqrt{\left(3x-1\right)^2}=15\)

\(\Leftrightarrow\left|3x-1\right|=15\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-1=15\\3x-1=-15\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{16}{3}\\x=-\dfrac{14}{3}\end{matrix}\right.\)

e) \(đk:x\ge\dfrac{8}{3}\)

\(pt\Leftrightarrow3x+4=9x^2-48x+64\)

\(\Leftrightarrow9x^2-51x+60=0\)

\(\Leftrightarrow3\left(x-4\right)\left(5x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=4\left(tm\right)\\x=\dfrac{5}{3}\left(ktm\right)\end{matrix}\right.\)

a. \(\sqrt{18x}+2\sqrt{8x}-3\sqrt{2x}=12\)      ĐK: \(x\ge0\)

<=> \(\sqrt{9.2x}+2\sqrt{4.2x}-3\sqrt{2x}=12\)

<=> \(3\sqrt{2x}+4\sqrt{2x}-3\sqrt{2x}=12\)

<=> \(\sqrt{2x}\left(3+4-3\right)=12\)

<=> \(4\sqrt{2x}=12\)

<=> \(\sqrt{2x}=12:4\)

<=> \(\sqrt{2x}=3\)

<=> 2x = 3²

<=> 2x = 9

<=> \(x=\dfrac{9}{2}\) (TM)

b. \(\sqrt{9x+18}+2\sqrt{36x+72}-\sqrt{4x+8}=26\)          ĐK: \(x\ge-2\)

<=> \(\sqrt{9\left(x+2\right)}+2\sqrt{36\left(x+2\right)}-\sqrt{4\left(x+2\right)}=26\)

<=> \(3\sqrt{x+2}+72\sqrt{x+2}-2\sqrt{x+2}=26\)

<=> \(\sqrt{x+2}\left(3+72-2\right)=26\)

<=> \(73\sqrt{x+2}=26\)

<=> \(\sqrt{x+2}=\dfrac{26}{73}\)

<=> x + 2 = \(\left(\dfrac{26}{73}\right)^2\)

<=> x + 2 = \(\dfrac{676}{5329}\)

<=> \(x=\dfrac{676}{5329}-2\)

<=> \(x=-1,873146932\) (TM)

c. \(\sqrt{\left(x-2\right)^2}=10\)

<=> \(\left|x-2\right|=10\)

<=> \(\left[{}\begin{matrix}x-2=10\left(x\ge2\right)\\x-2=-10\left(x< 2\right)\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=12\left(TM\right)\\x=-8\left(TM\right)\end{matrix}\right.\)

d. \(\sqrt{9x^2-6x+1}=15\)

<=> \(\sqrt{\left(3x-1\right)^2}=15\)

<=> \(\left|3x-1\right|=15\)

<=> \(\left[{}\begin{matrix}3x-1=15\left(x\ge\dfrac{16}{3}\right)\\3x-1=-15\left(x< \dfrac{16}{3}\right)\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=\dfrac{16}{3}\left(TM\right)\\x=\dfrac{-14}{3}\left(TM\right)\end{matrix}\right.\)

e. \(\sqrt{3x+4}=3x-8\)        ĐK: \(x\ge\dfrac{-4}{3}\)

<=> 3x + 4 = (3x - 8)²

<=> 3x + 4 = 9x² - 48x + 64

<=> 9x² - 3x - 48x + 64 - 4 = 0

<=> 9x² - 51x + 60 = 0

<=> 9x² - 36x - 15x + 60 = 0

<=> 9x(x - 4) - 15(x - 4) = 0

<=> (9x - 15)(x - 4) = 0

<=> \(\left[{}\begin{matrix}9x-15=0\\x-4=0\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=\dfrac{15}{9}\left(TM\right)\\x=4\left(TM\right)\end{matrix}\right.\)

1) \(\sqrt{5-2x}=6\left(đk:x\le\dfrac{5}{2}\right)\)

\(\Leftrightarrow5-2x=36\)

\(\Leftrightarrow2x=-31\Leftrightarrow x=-\dfrac{31}{2}\left(tm\right)\)

2) \(\sqrt{2-x}=\sqrt{x+1}\left(đk:2\ge x\ge-1\right)\)

\(\Leftrightarrow2-x=x+1\)

\(\Leftrightarrow2x=1\Leftrightarrow x=\dfrac{1}{2}\left(tm\right)\)

3) \(\Leftrightarrow\sqrt{\left(2x+1\right)^2}=6\)

\(\Leftrightarrow\left|2x+1\right|=6\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+1=6\\2x+1=-6\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{7}{2}\end{matrix}\right.\)

4) \(\sqrt{x^2-10x+25}=x-2\left(đk:x\ge2\right)\)

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

\(\Leftrightarrow\left|x-5\right|=x-2\)

\(\Leftrightarrow\left[{}\begin{matrix}x-5=x-2\left(x\ge5\right)\\x-5=2-x\left(2\le x< 5\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}5=2\left(VLý\right)\\x=\dfrac{7}{2}\left(tm\right)\end{matrix}\right.\)

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