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

=>\(\sqrt{\left(x+3\right)^2}\)+ \(\sqrt{\left(x+4\right)^2}\)+\(\sqrt{\left(x+5\right)^2}\)=9x

=> x + 3 + x + 4 + x + 5 = 9x

=> - 6x = - 12

=> x=2

Ủa sao phá đc trị tuyệt đối hay v bạn? (căn a^2 = trị tuyệt đối của a ) 

Vì \(\sqrt{x^2+6x+9}>0\\ \)

\(\sqrt{x^2+8x+16}>0\\ \)

\(\sqrt{x^2+10x+25}>0\\ \)

Suy ra 9x>0. Suy ra x>0 .Nha bạn!

a: ĐKXĐ: x>=5

\(\sqrt{4x-20}+\sqrt{x-5}-\dfrac{1}{3}\cdot\sqrt{9x-45}=4\)

=>\(2\sqrt{x-5}+\sqrt{x-5}-\dfrac{1}{3}\cdot3\sqrt{x-5}=4\)

=>\(2\sqrt{x-5}=4\)

=>\(\sqrt{x-5}=2\)

=>x-5=4

=>x=9(nhận)

b: ĐKXĐ: x>=1/2

\(\sqrt{2x-1}-\sqrt{8x-4}+5=0\)

=>\(\sqrt{2x-1}-2\sqrt{2x-1}+5=0\)

=>\(5-\sqrt{2x-1}=0\)

=>\(\sqrt{2x-1}=5\)

=>2x-1=25

=>2x=26

=>x=13(nhận)

c: \(\sqrt{x^2-10x+25}=2\)

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

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

=>\(\left[{}\begin{matrix}x-5=2\\x-5=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=3\end{matrix}\right.\)

d: \(\sqrt{x^2-14x+49}-5=0\)

=>\(\sqrt{x^2-2\cdot x\cdot7+7^2}=5\)

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

=>|x-7|=5

=>\(\left[{}\begin{matrix}x-7=5\\x-7=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=12\\x=2\end{matrix}\right.\)

\(a,\sqrt{4x-20}+\sqrt{x-5}-\dfrac{1}{3}\sqrt{9x-45}=4\left(đkxđ:x\ge5\right)\\ \Leftrightarrow\sqrt{4\left(x-5\right)}+\sqrt{x-5}-\dfrac{1}{3}\sqrt{9\left(x-5\right)}=4\\ \Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\\ \Leftrightarrow2\sqrt{x-5}=4\\ \Leftrightarrow\sqrt{x-5}=2\\ \Leftrightarrow x-5=4\\ \Leftrightarrow x=9\left(tm\right)\)

\(b,\sqrt{2x-1}-\sqrt{8x-4}+5=0\left(đkxđ:x\ge\dfrac{1}{2}\right)\\ \Leftrightarrow\sqrt{2x-1}-\sqrt{4\left(2x-1\right)}=-5\\ \Leftrightarrow\sqrt{2x-1}-2\sqrt{2x-1}=-5\\ \Leftrightarrow-\sqrt{2x-1}=-5\\ \Leftrightarrow\sqrt{2x-1}=5\\ \Leftrightarrow2x-1=25\\ \Leftrightarrow2x=26\\ \Leftrightarrow x=13\left(tm\right)\)

\(c,\sqrt{x^2-10x+25}=2\\ \Leftrightarrow\sqrt{\left(x-5\right)^2}=2\\ \Leftrightarrow\left|x-5\right|=2\\ \Leftrightarrow\left[{}\begin{matrix}x-5=2\\x-5=-2\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=7\\x=3\end{matrix}\right.\)

\(d,\sqrt{x^2-14x+49}-5=0\\ \Leftrightarrow\sqrt{\left(x-7\right)^2}=5\\ \Leftrightarrow\left|x-7\right|=5\\ \Leftrightarrow\left[{}\begin{matrix}x-7=5\\x-7=-5\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=12\\x=2\end{matrix}\right.\)

\(a)ĐKXĐ:x\ge5\\ \sqrt{4x-20}+\sqrt{x-5}-\dfrac{1}{3}\sqrt{9x-45}=4\\ \Leftrightarrow\sqrt{4\left(x-5\right)}+\sqrt{x-5}-\dfrac{1}{3}\sqrt{9\left(x-5\right)}=4\\ \Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\\ \Leftrightarrow2\sqrt{x-5}=4\\ \Leftrightarrow\sqrt{x-5}=\dfrac{4}{2}\\ \Leftrightarrow\sqrt{x-5}=2\\ \Leftrightarrow\left(\sqrt{x-5}\right)^2=2^2\\ \Leftrightarrow x-5=4\\ \Leftrightarrow x=4+5\\ \Leftrightarrow x=9\left(tmđk\right)\)

Vậy \(S=\left\{9\right\}\)

\(b)ĐKXĐ:x\ge2\\ \sqrt{2x-1}-\sqrt{8x-4}+5=0\\ \Leftrightarrow\sqrt{2x-1}-\sqrt{8x-4}=0-5\\ \Leftrightarrow\sqrt{2x-1}-\sqrt{4\left(2x-1\right)}=-5\\ \Leftrightarrow\sqrt{2x-1}-2\sqrt{2x-1}=-5\\ \Leftrightarrow-\sqrt{2x-1}=-5\\ \Leftrightarrow-\left(\sqrt{2x-1}\right)=\left(-5\right)^2\\ \Leftrightarrow-2x+1=-25\\ \Leftrightarrow-2x=\left(-25\right)-1\\ \Leftrightarrow-2x=-26\\ \Leftrightarrow x=\dfrac{-26}{-2}\\ \Leftrightarrow x=13\left(tmđk\right)\)

Vậy \(S=\left\{13\right\}\)

\(c)\sqrt{x^2-10x+25}=2\\ \Leftrightarrow\sqrt{\left(x-5\right)^2}=2\\ \Leftrightarrow\left|x-5\right|=2\\ \Leftrightarrow\left[{}\begin{matrix}x-5=2\\x-5=-2\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=2+5\\x=\left(-2\right)+5\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=7\\x=3\end{matrix}\right.\)

Vậy: \(S=\left\{7;3\right\}\)

\(d)\sqrt{x^2-14x+49}-5=0\\ \Leftrightarrow\sqrt{x^2-14x+49}=0+5\\ \Leftrightarrow\sqrt{x^2-14x+49}=5\\ \Leftrightarrow\sqrt{\left(x-7\right)^2}=5\\ \Leftrightarrow\left|x-7\right|=5\\ \Leftrightarrow\left[{}\begin{matrix}x-7=5\\x-7=-5\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=5+7\\x=\left(-5\right)+7\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=12\\x=2\end{matrix}\right.\)

Vậy \(S=\left\{12;2\right\}.\)

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

\(\Leftrightarrow\left|x+3\right|=4\) \(\Leftrightarrow\left[{}\begin{matrix}x+3=4\\x+3=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-7\end{matrix}\right.\) ( TM )

b) \(\Leftrightarrow\sqrt{\left(2x-1\right)^2}=5x+3\)

\(\Leftrightarrow\left|2x-1\right|=5x+3\)

\(\Leftrightarrow\left\{{}\begin{matrix}5x+3\ge0\\\left[{}\begin{matrix}2x-1=5x+3\\2x-1=-5x-3\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-\frac{3}{5}\\\left[{}\begin{matrix}x=-\frac{4}{3}\left(KTM\right)\\x=-\frac{2}{7}\left(TM\right)\end{matrix}\right.\end{matrix}\right.\)

a \(\sqrt{x^2+6x+9}=4\Leftrightarrow\sqrt{\left(x+3\right)^2=4}\)

\(\Leftrightarrow x+3=4\)

\(\Rightarrow x=1\)

Thiếu soát gì mog bạn thông cảm :]

Gửi em

\(---\begin{gathered} a)\sqrt {1 - 6x + 9{x^2}} = 5 \hfill \\ \Leftrightarrow \sqrt {{{\left( {1 - 3x} \right)}^2}} = 5 \hfill \\ \Leftrightarrow \left| {1 - 3x} \right| = 5 \hfill \\ T{H_1}:1 - 3x \geqslant 0 \Rightarrow x \leqslant \frac{1}{3} \hfill \\ 1 - 3x = 5 \hfill \\ \Leftrightarrow - 3x = 5 - 1 \hfill \\ \Leftrightarrow - 3x = 4 \hfill \\ \Leftrightarrow x = - \frac{4}{3}\left( {TM} \right) \hfill \\ T{H_2}:1 - 3x < 0 \Rightarrow x > \frac{1}{3} \hfill \\ - \left( {1 - 3x} \right) = 5 \hfill \\ \Leftrightarrow - 1 + 3x = 5 \hfill \\ \Leftrightarrow 3x = 5 + 1 \hfill \\ \Leftrightarrow 3x = 6 \hfill \\ \Leftrightarrow x = \frac{6}{3} \hfill \\ \Leftrightarrow x = 2\left( {TM} \right) \hfill \\ b)\sqrt {{x^2} - 4x + 4} = 7 \hfill \\ \Leftrightarrow \sqrt {{{\left( {x - 2} \right)}^2}} = 7 \hfill \\ \Leftrightarrow \left| {x - 2} \right| = 7 \hfill \\ T{H_1}:x - 2 \geqslant 0 \Rightarrow x \geqslant 2 \hfill \\ x - 2 = 7 \hfill \\ \Leftrightarrow x = 7 + 2 \hfill \\ \Leftrightarrow x = 9\left( {TM} \right) \hfill \\ T{H_2}:x - 2 < 0 \Rightarrow x < 2 \hfill \\ - \left( {x - 2} \right) = 7 \hfill \\ \Leftrightarrow - x + 2 = 7 \hfill \\ \Leftrightarrow - x = 7 - 2 \hfill \\ \Leftrightarrow - x = 5 \hfill \\ \Leftrightarrow x = - 5\left( {TM} \right) \hfill \\ c)\sqrt {25 - 10x + {x^2}} = 7 - 2x \hfill \\ \Leftrightarrow \sqrt {{{\left( {5 - x} \right)}^2}} = 7 - 2x \hfill \\ \Leftrightarrow \left| {5 - x} \right| = 7 - 2x \hfill \\ \Leftrightarrow \left| {5 - x} \right| + 2x = 7 \hfill \\ T{H_1}:5 - x \geqslant 0 \Rightarrow x \leqslant 5 \hfill \\ 5 - x + 2x = 7 \hfill \\ \Leftrightarrow 5 + x = 7 \hfill \\ \Leftrightarrow x = 7 - 5 \hfill \\ \Leftrightarrow x = 2\left( {TM} \right) \hfill \\ T{H_2}:5 - x < 0 \Rightarrow x > 5 \hfill \\ - \left( {5 - x} \right) + 2x = 7 \hfill \\ \Leftrightarrow - 5 + x + 2x = 7 \hfill \\ \Leftrightarrow 3x = 7 + 5 \hfill \\ \Leftrightarrow 3x = 12 \hfill \\ \Leftrightarrow x = 4\left( {KTM} \right) \hfill \\ d)\sqrt {{x^2} + 6x + 9} = 3x - 1 \hfill \\ \Leftrightarrow \sqrt {{{\left( {x + 3} \right)}^2}} = 3x - 1 \hfill \\ \Leftrightarrow \left| {x + 3} \right| = 3x - 1 \hfill \\ \Leftrightarrow \left| {x + 3} \right| - 3x = - 1 \hfill \\ T{H_1}:x + 3 \geqslant 0 \Rightarrow x \geqslant - 3 \hfill \\ x + 3 - 3x = - 1 \hfill \\ \Leftrightarrow - 2x = - 1 - 3 \hfill \\ \Leftrightarrow - 2x = - 4 \hfill \\ \Leftrightarrow x = \frac{{ - 4}}{{ - 2}} \hfill \\ \Leftrightarrow x = 2\left( {TM} \right) \hfill \\ T{H_2}:x + 3 < 0 \Rightarrow x < - 3 \hfill \\ - \left( {x + 3} \right) - 3x = - 1 \hfill \\ \Leftrightarrow - x - 3 - 3x = - 1 \hfill \\ \Leftrightarrow - 4x = - 1 + 3 \hfill \\ \Leftrightarrow - 4x = 2 \hfill \\ \Leftrightarrow x = \frac{2}{{ - 4}} \hfill \\ \Leftrightarrow x = - \frac{1}{2}\left( {KTM} \right) \hfill \\ \end{gathered} \)

Những câu này bạn chỉ cần quy vế trước về hđthức xong bỏ căn đi ra giá trị tuyệt đối r tính bình thường thoi ạ

a)\(\sqrt{3x+2}=2-\sqrt{3}\)

\(\Leftrightarrow3x+2=\left(2-\sqrt{3}\right)^2\)

\(\Leftrightarrow3x+2=7-4\sqrt{3}\)

\(\Leftrightarrow3x=7-2-4\sqrt{3}\)

\(\Leftrightarrow3x=5-4\sqrt{3}\)

\(\Leftrightarrow x=\dfrac{5}{3}-\dfrac{4\sqrt{3}}{3}\)

\(\Leftrightarrow x=\dfrac{5-4\sqrt{3}}{3}\)

b) \(\sqrt{x^2-4x+4}=49\)

\(\Leftrightarrow\sqrt{\left(x-2\right)^2}=49\)

\(\Leftrightarrow\left|x-2\right|=49\)\

\(\Leftrightarrow\left[{}\begin{matrix}x-2=49\\-x+2=49\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=51\\x=-47\end{matrix}\right.\)

c) \(\sqrt{x+1}=x-1\)

ĐKXĐ: \(x-1\ge0\Rightarrow x\ge1\)

\(\Leftrightarrow x+1=\left(x-1\right)^2\)

\(\Leftrightarrow x+1=x^2-2x+1\)

\(\Leftrightarrow-x^2+2x+x=-1+1\)

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

\(\Leftrightarrow x\left(3-x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\3-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(lo\text{ại}\right)\\x=3\left(nh\text{ậ}n\right)\end{matrix}\right.\)

d)e) lát mình làm sau

\(\sqrt{9x^2-6x+1}+\sqrt{25-30x+9x^2}\)

\(=\sqrt{\left(3x-1\right)^2}+\sqrt{\left(5-3x\right)^2}\)

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

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