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Ta có: \(\sqrt{25x-125}-3\cdot\sqrt{\dfrac{x-5}{9}}-\dfrac{1}{3}\sqrt{9x-45}=6\)
\(\Leftrightarrow5\sqrt{x-5}-3\cdot\dfrac{\sqrt{x-5}}{3}-\dfrac{1}{3}\cdot3\sqrt{x-5}=6\)
\(\Leftrightarrow3\sqrt{x-5}=6\)
\(\Leftrightarrow x-5=4\)
hay x=9
\(a,ĐKXĐ:x\ge0\\ \Leftrightarrow\sqrt{\left(x-1\right)^2}=2x\\ \Leftrightarrow\left|x-1\right|=2x\\ \Rightarrow\left[{}\begin{matrix}x-1=2x\\1-x=2x\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\left(kot/mĐKXĐ\right)\\x=\frac{1}{3}\left(t/m\right)\end{matrix}\right.\\ Vậy.....\)
\(b,ĐKXĐ:x\ge5\\ \Leftrightarrow\sqrt{25\left(x-5\right)}-3\cdot\frac{1}{3}\cdot\sqrt{x-5}-\frac{1}{3}\cdot3\cdot\sqrt{x-5}\Leftrightarrow5\sqrt{x-5}-\sqrt{x-5}-\sqrt{x-5}=6\\ \Leftrightarrow\left(5-1-1\right)\sqrt{x-5}=6\\ \Leftrightarrow\sqrt{x-5}=2\\ \Rightarrow x-5=4\\ \Leftrightarrow x=9\left(thỏamãnĐKXĐ\right)\\ Vậy...\)
\(ĐK:x\ge5\\ \Leftrightarrow\sqrt{x-5}+\dfrac{1}{3}\cdot3\sqrt{x-5}=\dfrac{1}{5}\sqrt{25x-119}\\ \Leftrightarrow2\sqrt{x-5}=\dfrac{1}{5}\sqrt{25x-119}\\ \Leftrightarrow4\left(x-5\right)=\dfrac{1}{25}\left(25x-119\right)\\ \Leftrightarrow4x-20=x-\dfrac{119}{25}\\ \Leftrightarrow3x=\dfrac{381}{25}\Leftrightarrow x=\dfrac{127}{25}\)
Ta có: \(\sqrt{x-5}+\dfrac{1}{3}\sqrt{9x-45}=\dfrac{1}{5}\sqrt{25x-125}+6\)
\(\Leftrightarrow x-5=36\)
hay x=41
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/ \(\sqrt{x-1}+\sqrt{4x-4}-\sqrt{25x-25}+2=0\) (ĐKXĐ : \(x\ge1\))
\(\Leftrightarrow\sqrt{x-1}+2\sqrt{x-1}-5\sqrt{x-1}+2=0\)
\(\Leftrightarrow2\sqrt{x-1}=2\Leftrightarrow x-1=1\Leftrightarrow x=2\)
b/ \(\sqrt{9x^2+18}+2\sqrt{x^2+2}-\sqrt{25x^2+50}+3=0\)
\(\Leftrightarrow3\sqrt{x^2+2}+2\sqrt{x^2+2}-5\sqrt{x^2+2}+3=0\)
<=> 3 = 0 (vô lý)
=> pt vô nghiệm.
c/ \(\frac{9x-7}{\sqrt{7x+5}}=\sqrt{7x+5}\) (ĐKXĐ : x>-5/7)
\(\Leftrightarrow9x-7=7x+5\Leftrightarrow2x=12\Leftrightarrow x=6\)
d/ \(\frac{\sqrt{2x-3}}{\sqrt{x-1}}=2\) (ĐKXĐ : \(x\ge\frac{3}{2}\))
\(\Leftrightarrow2x-3=4\left(x-1\Leftrightarrow\right)2x=1\Leftrightarrow x=\frac{1}{2}\) (loại)
Vậy pt vô nghiệm.
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: \(x\ge-1\))
\(\Leftrightarrow\sqrt{16\left(x+1\right)}-\sqrt{9\left(x+1\right)}+\sqrt{4\left(x+1\right)}+\sqrt{x+1}=16\)
\(\Leftrightarrow4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}+\sqrt{x+1}=16\)
\(\Leftrightarrow4\sqrt{x+1}=16\)
\(\Leftrightarrow\sqrt{x+1}=4\)
\(\Leftrightarrow x+1=16\)
\(\Leftrightarrow x=15\left(tm\right)\)
PT <=> \(\sqrt{x-5}+\frac{1}{3}\sqrt{9\left(x-5\right)}=\frac{1}{5}\sqrt{25\left(x-5\right)}+6\)
<=> \(\sqrt{x-5}+\sqrt{x-5}=\sqrt{x-5}+6\)
<=>\(\sqrt{x-5}=6\)
<=> \(x=41\)
KL: \(x\in\left\{41\right\}\)
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