\(\sqrt{3-x}\) - \(\sqrt{12-4x}\) + \(\sqrt{27-9x}\)  = 20 đk \(3-x\) ≥ 0 ⇒ \(x\le3\)

\(\sqrt{3-x}\) - \(\sqrt{4.\left(3-x\right)}\) + \(\sqrt{9.\left(3-x\right)}\) = 20 

\(\sqrt{3-x}\) - 2\(\sqrt{3-x}\) + 3\(\sqrt{3-x}\) = 20

\(\sqrt{3-x}\).( 1 - 2 + 3) = 20

2\(\sqrt{3-x}\) = 20

   \(\sqrt{3-x}\) = 20: 2

    \(\sqrt{3-x}\) = 10

     3 - \(x\) = 100

           \(x\) = 3 - 100 

          \(x\) = -97 (thỏa mãn)

Vậy \(x\) = -97

a: ĐKXĐ: \(x\in R\)

\(\sqrt{x^2-4x+4}=7\)

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

=>|x-2|=7

=>\(\left[{}\begin{matrix}x-2=7\\x-2=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=9\\x=-5\end{matrix}\right.\)

b: ĐKXĐ: x>=-3

\(\sqrt{4x+12}-3\sqrt{x+3}+\dfrac{4}{3}\cdot\sqrt{9x+27}=6\)

=>\(2\sqrt{x+3}-3\sqrt{x+3}+\dfrac{4}{3}\cdot3\sqrt{x+3}=6\)

=>\(3\sqrt{x+3}=6\)

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

=>x+3=4

=>x=1(nhận)

Mn giúp e với ak

a) \(\sqrt{x^2-6x+9}\)

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

\(=\sqrt{\left(x-3\right)^2}\) ≥0,∀x

⇒x∈\(R\)

b) \(\sqrt{x^2-2x+1}\)

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

\(=\sqrt{\left(x-1\right)^2}\) ≥0,∀x

⇒x∈\(R\)

\(a) \sqrt{4x^2− 9} = 2\sqrt{x + 3}\)

\(ĐK:x\ge\dfrac{3}{2}\)

\(pt\Leftrightarrow4x^2-9=4\left(x+3\right)\)

\(\Leftrightarrow4x^2-9=4x+12\)

\(\Leftrightarrow4x^2-4x-21=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1-\sqrt{22}}{2}\left(l\right)\\x=\dfrac{1+\sqrt{22}}{2}\left(tm\right)\end{matrix}\right.\)

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

\(ĐK:x\ge5\)

\(pt\Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\)

\(\Leftrightarrow2\sqrt{x-5}=4\Leftrightarrow\sqrt{x-5}=2\)

\(\Leftrightarrow x-5=4\Leftrightarrow x=9\left(tm\right)\)

\(c)\dfrac{2}{3}\sqrt{9x-9}-\dfrac{1}{4}\sqrt{16x-16}+27.\sqrt{\dfrac{x-1}{81}}=4\)

ĐK:x>=1

\(pt\Leftrightarrow2\sqrt{x-1}-\sqrt{x-1}+3\sqrt{x-1}=4\)

\(\Leftrightarrow4\sqrt{x-1}=4\Leftrightarrow\sqrt{x-1}=1\)

\(\Leftrightarrow x-1=1\Leftrightarrow x=2\left(tm\right)\)

\(d)5\sqrt{\dfrac{9x-27}{25}}-7\sqrt{\dfrac{4x-12}{9}}-7\sqrt{x^2-9}+18\sqrt{\dfrac{9x^2-81}{81}}=0\)

\(ĐK:x\ge3\)

\(pt\Leftrightarrow3\sqrt{x-3}-\dfrac{14}{3}\sqrt{x-3}-7\sqrt{x^2-9}+6\sqrt{x^2-9}=0\)

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

\(\Leftrightarrow(\dfrac{5}{3}+\sqrt{x+3})\sqrt{x-3}=0\)

\(\Leftrightarrow\sqrt{x-3}=0\)    (vì \(\dfrac{5}{3}+\sqrt{x+3}>0\))

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

x thuộc R

=))

\(\sqrt{x-3}-\sqrt{9x-27}+\sqrt{4x-12}=7\)

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

\(0=7\)

bạn Mafia là đúng rồi nhưng hơi tắt mình sẽ giải cụ thể

\(\sqrt{x-3}-\sqrt{9x-27}+\sqrt{4x-12}=7\)

\(\sqrt{x-3}-\sqrt{9\left(x-3\right)}+\sqrt{4\left(x-3\right)=7}\)

\(\sqrt{x-3}-\sqrt{3^2\left(x-3\right)}+\sqrt{2^2\left(x-3\right)}=7\)

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

\(\left(1-3+2\right)\sqrt{x-3}=7\)

\(0\sqrt{x-3}=7\)

\(\sqrt{x-3}=0\)

\(\Rightarrow x-3=0\)

\(\Rightarrow x=3\)

vậy \(x=3\)

\(<=>15\sqrt{x+3} - 3\sqrt{x+3}=4\sqrt{x+3} <=> 15\sqrt{x+3} - 3\sqrt{x+3}-4\sqrt{x+3}=0 <=> 8\sqrt{x+3}=0 <=>\sqrt{x+3}=0 => x+3=0 =>x=-3\)

a: ĐKXĐ: x>=3

Sửa đề: \(\sqrt{4x-12}-\sqrt{9x-27}+\sqrt{\dfrac{25x-75}{4}}-3=0\)

=>\(2\sqrt{x-3}-3\sqrt{x-3}+\dfrac{5}{2}\sqrt{x-3}-3=0\)

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

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

=>x-3=4

=>x=7(nhận)

b: ĐKXĐ: x>=0

\(\dfrac{\sqrt{x}-2}{\sqrt{x}+1}< =-\dfrac{3}{4}\)

=>\(\dfrac{\sqrt{x}-2}{\sqrt{x}+1}+\dfrac{3}{4}< =0\)

=>\(\dfrac{4\sqrt{x}-8+3\sqrt{x}+3}{4\left(\sqrt{x}+1\right)}< =0\)

=>\(7\sqrt{x}-5< =0\)

=>\(\sqrt{x}< =\dfrac{5}{7}\)

=>0<=x<=25/49

c: ĐKXĐ: x>=5

\(\sqrt{9x-45}-14\sqrt{\dfrac{x-5}{49}}+\dfrac{1}{4}\sqrt{4x-20}=3\)

=>\(3\sqrt{x-5}-14\cdot\dfrac{\sqrt{x-5}}{7}+\dfrac{1}{4}\cdot2\cdot\sqrt{x-5}=3\)

=>\(\dfrac{3}{2}\sqrt{x-5}=3\)

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

=>x-5=4

=>x=9(nhận)

a: \(B=3\sqrt{x-3}+\sqrt{x-3}-\dfrac{1}{2}\cdot2\sqrt{x-3}=3\sqrt{x-3}\)

b: B=7 thì \(\sqrt{x-3}=\dfrac{7}{3}\)

=>x-3=49/9

hay x=76/9

Giải:

\(\sqrt{4x-12}+\sqrt{9x-27}-5\sqrt{x-3}+3-x=0\)

\(\Leftrightarrow2\sqrt{x-3}+3\sqrt{x-3}-5\sqrt{x-3}+3-x=0\)

\(\Leftrightarrow3-x=0\)

\(\Leftrightarrow x=3\)

Vậy ...

Mình sửa lại đề chỗ \(4\sqrt{x-3}\) thành \(5\sqrt{x-3}\) để làm ra kết quả tròn, nếu không sửa thì chắc không ra được kết quả