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

a) \(\sqrt{x-1}+\sqrt{4x-4}-\sqrt{25x-25}+2=0\) (ĐK: \(x\ge1\)) 

\(\Leftrightarrow\sqrt{x-1}+\sqrt{4\left(x-1\right)}-\sqrt{25\left(x-1\right)}+2=0\)

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

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

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

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

\(\Leftrightarrow x-1=1\)

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

b) \(\sqrt{16x+16}-\sqrt{9x+9}+\sqrt{4x+4}+\sqrt{x+1}=16\) (ĐK: \(x\ge-1\))

\(\Leftrightarrow\sqrt{16\left(x+1\right)}-\sqrt{9\left(x+1\right)}+\sqrt{4\left(x+1\right)}+\sqrt{x+1}=16\)

\(\Leftrightarrow4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}+\sqrt{x+1}=16\)

\(\Leftrightarrow4\sqrt{x+1}=16\)

\(\Leftrightarrow\sqrt{x+1}=4\)

\(\Leftrightarrow x+1=16\)

\(\Leftrightarrow x=15\left(tm\right)\)

\(x=0\)

\(\sqrt{x2+x+25}\) + \(\sqrt{x2+x+16}\)=9

=\(\sqrt{ }\)(x+5)² +\(\sqrt{ }\)(x+4)²=9

= /x+5/ +/x+4/ =9

= x+5+x+4 =9

= 2x+9=9

= 2x=9-9

=2x=0

x= 0:2

x=0

vậy x = 0

Điều kiện: mọi \(x\in R\)

Ta có \(\sqrt{x^2+x+25}=\sqrt{x^2+x+9}+2\)

\(\Leftrightarrow x^2+x+25=x^2+x+9+4.\sqrt{x^2+x+9}+4\)

\(\Leftrightarrow\sqrt{x^2+x+9}=3\)\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=-1\end{cases}}\)

X=0,894427185

tớ bấm máy tính mà

a: \(\Leftrightarrow2\cdot5\sqrt{x-3}-\dfrac{1}{2}\cdot2\sqrt{x-3}+\dfrac{1}{7}\cdot7\sqrt{x-3}=20\)

=>\(10\cdot\sqrt{x-3}=20\)

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

=>x-3=4

=>x=7

b: =>|x-3|=2

=>x-3=2 hoặc x-3=-2

=>x=5 hoặcx=1

\(25\sqrt{\dfrac{x-3}{25}}-7\sqrt{\dfrac{4x-12}{9}}-7\sqrt{x^2-9}+18\sqrt{\dfrac{9x^2-81}{81}}=0\left(x\ge3\right)\)

\(=25\sqrt{\dfrac{1}{25}.\left(x-3\right)}-7\sqrt{\dfrac{4}{9}.\left(x-3\right)}-7\sqrt{x^2-9}+18\sqrt{\dfrac{1}{9}.\left(x^2-9\right)}=0\)

\(=5\sqrt{x-3}-\dfrac{14}{3}\sqrt{x-3}-7\sqrt{x^2-9}+6\sqrt{x^2-9}=0\)

\(\Rightarrow\dfrac{1}{3}\sqrt{x-3}-\sqrt{\left(x-3\right)\left(x+3\right)}=0\Rightarrow\sqrt{x-3}-3\sqrt{\left(x-3\right)\left(x+3\right)}=0\)

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

\(\Rightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{26}{9}\left(l\right)\end{matrix}\right.\)

cảm ơn nhaa<33

Ghi thiếu đề bài nên tl lại 

`sqrt{x-2}+sqrt{6-x}=x^2-8x+16+2sqrt2`

Áp dụng BĐT bunhia ta có:

`sqrt{x-2}+sqrt{6-x}<=sqrt{(1+1)(x-2+6-x)}=2sqrt2`

`=>VT<=2sqrt2(1)`

Mặt khác:

`VP=x^2-8x+16+2sqrt2`

`=(x-4)^2+2sqrt2>=2sqrt2`

`=>VP>=2sqrt2(2)`

`(1)(2)=>VT=VP=2sqrt2`

`<=>x=4`

Vậy `S={4}`

`sqrt{x-2}+sqrt{6-x}=x^2-8x+2sqrt2`

Áp dụng BĐT bunhia ta có:

`sqrt{x-2}+sqrt{6-x}<=sqrt{(1+1)(x-2+6-x)}=2sqrt2`

`=>VT<=2sqrt2(1)`

Mặt khác:

`VP=x^2-8x+16+2sqrt2`

`=(x-4)^2+2sqrt2>=2sqrt2`

`=>VP>=2sqrt2(2)`

`(1)(2)=>VT=VP=2sqrt2`

`<=>x=4`

Vậy `S={4}`

Ta có: \(\sqrt{x^2-6x+9}+\sqrt{x^2+10x+25}=8\)

\(\Rightarrow\sqrt{\left(x-3\right)^2}+\sqrt{\left(x+5\right)^2}=8\)

\(\Rightarrow x-3+x+5=8\)

\(\Rightarrow2x=6\Rightarrow x=3\)

\(\sqrt{x^2-6x+9}+\sqrt{x^2+10x+25}=8\)

\(\Leftrightarrow\sqrt{\left(x-3\right)^2}+\sqrt{\left(x+5\right)^2}=8\Leftrightarrow\left|x-3\right|+\left|x+5\right|=8\) (1)

Nếu \(x< -5\) thì (1) trở thành: 

      \(3-x+\left(-x-5\right)=8\Leftrightarrow-2x-2=8\Leftrightarrow x=-5\) (loại)

-Nếu \(-5\le x< 3\) thì (1) trở thành:

       \(3-x+x+5=8\Leftrightarrow8=8\)

-Nếu \(x>3\) thì (1) trở thành: 

        \(x-3+x+5=8\Leftrightarrow2x+2=8\Leftrightarrow x=3\) (thỏa mãn)

Vậy \(-5\le x\le3\)