\(a,ĐK:x\ge\dfrac{5}{2}\\ PT\Leftrightarrow2x-5=4\Leftrightarrow x=\dfrac{9}{2}\left(tm\right)\\ b,PT\Leftrightarrow\left|x-3\right|=7\Leftrightarrow\left[{}\begin{matrix}x-3=7\\3-x=7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=10\\x=-4\end{matrix}\right.\\ c,ĐK:x\le4\\ PT\Leftrightarrow\left|x-8\right|=4-x\\ \Leftrightarrow\left[{}\begin{matrix}x-8=4-x\left(x\ge8\right)\\8-x=4-x\left(x\le8\right)\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x\in\varnothing\left(trái.vs.ĐK\right)\\0x=4\left(ktm\right)\end{matrix}\right.\Leftrightarrow x\in\varnothing\)

a) \(\sqrt{2x-5}=2\)

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

\(\Leftrightarrow\) \(2x-5=4\)

\(\Leftrightarrow\) 2x = 9

\(\Leftrightarrow\) x = \(\dfrac{9}{2}\)

 Chúc bạn học tốt

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

`\sqrt{8x-4}-2\sqrt{18x-9}+2\sqrt{32x-16}=12`      `ĐK: x >= 1/2`

`<=>2\sqrt{2x-1}-6\sqrt{2x-1}+8\sqrt{2x-1}=12`

`<=>4\sqrt{2x-1}=12`

`<=>\sqrt{2x-1}=3`

`<=>2x-1=9`

`<=>x=5` (t/m)

Vậy `S={5}`.

\(\Leftrightarrow2\sqrt{2x-1}-2\cdot3\sqrt{2x-1}+2\cdot4\sqrt{2x-1}=12\)

=>\(4\sqrt{2x-1}=12\)

=>\(\sqrt{2x-1}=3\)

=>2x-1=9

=>2x=10

=>x=5

\(\sqrt{8x-4}-2\sqrt{18x-9}+2\sqrt{32x-16}=12\) (ĐK: \(x\ge\dfrac{1}{2}\))

\(\Leftrightarrow2\sqrt{2x-1}-2\cdot3\sqrt{2x-1}+2\cdot4\sqrt{2x-1}=12\)

\(\Leftrightarrow2\sqrt{2x-1}-6\sqrt{2x-1}+8\sqrt{2x-1}=12\)

\(\Leftrightarrow\left(2-6+8\right)\sqrt{2x-1}=12\)

\(\Leftrightarrow4\sqrt{2x-1}=12\)

\(\Leftrightarrow\sqrt{2x-1}=12:4\)

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

\(\Leftrightarrow2x-1=9\)

\(\Leftrightarrow2x=9+1\)

\(\Leftrightarrow2x=10\)

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

Vậy \(x=5\)

điều kiện -4<=x<=4x<=4

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

\(A=\left|x+4\right|+\left|x-4\right|\)

KẾT HỢP ĐIỀU KIỆN

\(A=x+4+4-x\)

\(A=8\)

\(B=\sqrt{\left(3x\right)^2-6x+1}+\sqrt{\left(2x\right)^2-12x+3^2}\)

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

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

\(TH1:x>=\frac{3}{2}\)

\(B=3x-1+2x-3\)

\(B=5x-4\)

\(TH2:\frac{1}{3}< =x< \frac{3}{2}\)

\(B=3x-1-2x+3\)

\(B=x+2\)

\(TH3:x< \frac{1}{3}\)

\(B=-3x+1-2x+3\)

\(B=4-5x\)

câu c và câu d tương tự

câu c tách ra: \(C=\sqrt{\left(\sqrt{x}-3\right)^2}-\sqrt{\left(2\sqrt{x}+1\right)^2}\)

còn câu d tách ra :\(D=\sqrt{x-1+2\sqrt{x-1}+1}+\sqrt{x-1-2\sqrt{x-1}+1}\)

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

bạn tự làm nốt câu c, d nha 

a) \(\sqrt{x^2-6x+9}=3\)

⇔ \(\sqrt{\left(x-3\right)^2}=3\)

⇔ \(\left|x-3\right|=3\)

⇔ \(\orbr{\begin{cases}x-3=3\\x-3=-3\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=6\\x=0\end{cases}}\)

b) \(\sqrt{x^2-8x+16}=x+2\)

⇔ \(\sqrt{\left(x-4\right)^2}=x+2\)

⇔ \(\left|x-4\right|=x+2\)

⇔ \(\orbr{\begin{cases}x-4=x+2\left(x\ge4\right)\\4-x=x+2\left(x< 4\right)\end{cases}\Leftrightarrow}x=1\)

c) \(\sqrt{x^2+6x+9}=3x-6\)

⇔ \(\sqrt{\left(x+3\right)^2}=3x-6\)

⇔ \(\left|x-3\right|=3x-6\)

⇔ \(\orbr{\begin{cases}x-3=3x-6\left(x\ge3\right)\\3-x=3x-6\left(x< 3\right)\end{cases}}\Leftrightarrow x=\frac{9}{4}\)

d) \(\sqrt{x^2-4x+4}-2x+5=0\)

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

⇔ \(\left|x-2\right|-2x+5=0\)

⇔ \(\orbr{\begin{cases}x-2-2x+5=0\left(x\ge2\right)\\2-x-2x+5=0\left(x< 2\right)\end{cases}}\Leftrightarrow x=3\)

\(M=\sqrt{x^2-4x+4}+2014\sqrt{x^2-6x+9}+\sqrt{x^2-10x+25}\)

\(M=\left|x-2\right|+2014\left|x-3\right|+\left|x-5\right|\)

\(M=\left|x-2\right|+\left|5-x\right|+2014\left|x-3\right|\)

\(M\ge\left|x-2+5-x\right|+2014\left|x-3\right|=3+2014\left|x-3\right|\ge3\)

\("="\Leftrightarrow x=3\)

a, Với \(-4\le x\le4\)

 \(A=\sqrt{x^2+8x+16}+\sqrt{x^2-8x+16}\)

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

b, \(B=\sqrt{9x^2-6x+1}+\sqrt{4x^2-12x+9}\)

\(=\sqrt{\left(3x\right)^2-2.3x+1}+\sqrt{\left(2x\right)^2-2.2x.3x+3^2}\)

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

\(a,PT\Leftrightarrow\left|3x-1\right|=\left|x-3\right|\Leftrightarrow\left[{}\begin{matrix}3x-1=x-3\\3x-1=3-x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=1\end{matrix}\right.\\ b,PT\Leftrightarrow\left|x-4\right|=4-x\Leftrightarrow\left[{}\begin{matrix}x-4=4-x\left(x\ge4\right)\\x-4=x-4\left(x< 4\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\left(tm\right)\\x=0\left(tm\right)\end{matrix}\right.\)