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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)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a) \sqrt{4x^2− 9} = 2\sqrt{x + 3}\)
\(ĐK:x\ge\dfrac{3}{2}\)
\(pt\Leftrightarrow4x^2-9=4\left(x+3\right)\)
\(\Leftrightarrow4x^2-9=4x+12\)
\(\Leftrightarrow4x^2-4x-21=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1-\sqrt{22}}{2}\left(l\right)\\x=\dfrac{1+\sqrt{22}}{2}\left(tm\right)\end{matrix}\right.\)
\(b)\sqrt{4x-20}+3.\sqrt{\dfrac{x-5}{9}}-\dfrac{1}{3}\sqrt{9x-45}=4\)
\(ĐK:x\ge5\)
\(pt\Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\)
\(\Leftrightarrow2\sqrt{x-5}=4\Leftrightarrow\sqrt{x-5}=2\)
\(\Leftrightarrow x-5=4\Leftrightarrow x=9\left(tm\right)\)
\(c)\dfrac{2}{3}\sqrt{9x-9}-\dfrac{1}{4}\sqrt{16x-16}+27.\sqrt{\dfrac{x-1}{81}}=4\)
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 \(\dfrac{5}{3}+\sqrt{x+3}>0\))
\(\Leftrightarrow x-3=0\Leftrightarrow x=3\left(nhận\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\sqrt{x+3}\) + \(\sqrt{9x+27}\) - \(\sqrt{4x-12}\) = 10 đk \(x+3\) ≥ 0 ⇒ \(x\) ≥ -3
\(\sqrt{x+3}\) + \(\sqrt{9\left(x+3\right)}\) - \(\sqrt{4\left(x+3\right)}\) = 10
\(\sqrt{x+3}\) + 3\(\sqrt{x+3}\) - 2\(\sqrt{x+3}\) = 10
(1 + 3 - 2)\(\sqrt{x+3}\) = 10
2\(\sqrt{x+3}\) = 10
\(\sqrt{x+3}\) = 10: 2
\(\sqrt{x+3}\) = 5
\(x+3\) = 10
\(x\) = 10 - 3
\(x\) = 7 ( thỏa mãn)
Vậy \(x\) = 7
![](https://rs.olm.vn/images/avt/0.png?1311)
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
![](https://rs.olm.vn/images/avt/0.png?1311)
=>\(5\cdot\dfrac{3\sqrt{x-3}}{5}-7\cdot\dfrac{2\sqrt{x-3}}{3}-7\cdot\sqrt{x^2-9}+18\cdot\sqrt{\dfrac{9}{81}\left(x^2-9\right)}=0\)
=>\(3\cdot\sqrt{x-3}-\dfrac{14}{3}\sqrt{x-3}=7\cdot\sqrt{x^2-9}-18\cdot\dfrac{3}{9}\cdot\sqrt{x^2-9}\)
=>\(-\dfrac{5}{3}\sqrt{x-3}=\sqrt{x^2-9}\)
=>\(\sqrt{x-3}\left(\sqrt{x+3}+\dfrac{5}{3}\right)=0\)
=>x-3=0
=>x=3
![](https://rs.olm.vn/images/avt/0.png?1311)
\(<=>15\sqrt{x+3} - 3\sqrt{x+3}=4\sqrt{x+3} <=> 15\sqrt{x+3} - 3\sqrt{x+3}-4\sqrt{x+3}=0 <=> 8\sqrt{x+3}=0 <=>\sqrt{x+3}=0 => x+3=0 =>x=-3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
ĐKXĐ: \(x\ge3\)
\(pt\Leftrightarrow5\sqrt{x-3}+3\sqrt{x-3}-\sqrt{x-3}=7\)
\(\Leftrightarrow7\sqrt{x-3}=7\Leftrightarrow\sqrt{x-3}=1\)
\(\Leftrightarrow x-3=1\Leftrightarrow x=4\left(tm\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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)
x thuộc R
=))
\(\sqrt{x-3}-\sqrt{9x-27}+\sqrt{4x-12}=7\)
\(\sqrt{x-3}-3\sqrt{x-3}+2\sqrt{x-3}=7\)
\(0=7\)