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\(a) \sqrt{4x^2− 9} = 2\sqrt{x + 3}\)
\(ĐK:x\ge\dfrac{3}{2}\)
\(pt\Leftrightarrow4x^2-9=4\left(x+3\right)\)
\(\Leftrightarrow4x^2-9=4x+12\)
\(\Leftrightarrow4x^2-4x-21=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1-\sqrt{22}}{2}\left(l\right)\\x=\dfrac{1+\sqrt{22}}{2}\left(tm\right)\end{matrix}\right.\)
\(b)\sqrt{4x-20}+3.\sqrt{\dfrac{x-5}{9}}-\dfrac{1}{3}\sqrt{9x-45}=4\)
\(ĐK:x\ge5\)
\(pt\Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\)
\(\Leftrightarrow2\sqrt{x-5}=4\Leftrightarrow\sqrt{x-5}=2\)
\(\Leftrightarrow x-5=4\Leftrightarrow x=9\left(tm\right)\)
\(c)\dfrac{2}{3}\sqrt{9x-9}-\dfrac{1}{4}\sqrt{16x-16}+27.\sqrt{\dfrac{x-1}{81}}=4\)
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 \(\dfrac{5}{3}+\sqrt{x+3}>0\))
\(\Leftrightarrow x-3=0\Leftrightarrow x=3\left(nhận\right)\)
\(a,\Leftrightarrow x-1=4\Leftrightarrow x=5\\ b,\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{3}{4}\\3x+1=4x-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{3}{4}\\x=4\left(tm\right)\end{matrix}\right.\Leftrightarrow x=4\\ c,ĐK:x\ge-5\\ PT\Leftrightarrow2\sqrt{x+5}-3\sqrt{x+5}+4\sqrt{x+5}=6\\ \Leftrightarrow3\sqrt{x+5}=6\\ \Leftrightarrow\sqrt{x+5}=3\\ \Leftrightarrow x+5=9\\ \Leftrightarrow x=4\left(tm\right)\)
\(d,\Leftrightarrow\sqrt{\left(x-2\right)^2}=\sqrt{\left(\sqrt{5}+1\right)^2}\\ \Leftrightarrow\left|x-2\right|=\sqrt{5}+1\\ \Leftrightarrow\left[{}\begin{matrix}x-2=\sqrt{5}+1\\2-x=\sqrt{5}+1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{5}+3\\x=1-\sqrt{5}\end{matrix}\right.\)
a. \(\Rightarrow2\sqrt{x+5}-3\sqrt{x+5}+4\sqrt{x+5}=6\Rightarrow\sqrt{x+5}\left(2-3+4\right)=6\Rightarrow\sqrt{x+5}=2\Rightarrow x+5=4\Rightarrow x=-1\)
b.\(\Rightarrow5\sqrt{x-1}-\frac{5}{2}\sqrt{x-1}-\sqrt{x-1}=6\Rightarrow\sqrt{x-1}\left(5-\frac{5}{2}-1\right)=6\Rightarrow\sqrt{x-1}=4\Rightarrow x-1=16\Rightarrow x=17\)
a) Ta có: \(\sqrt{4-5x}=12\)
\(\Leftrightarrow4-5x=144\)
\(\Leftrightarrow5x=-140\)
hay x=-28
b) Ta có: \(\sqrt{10+\sqrt{3x}}=2+\sqrt{6}\)
\(\Leftrightarrow\sqrt{3x}+10=10+4\sqrt{6}\)
\(\Leftrightarrow\sqrt{3x}=4\sqrt{6}\)
\(\Leftrightarrow3x=96\)
hay x=32
c) Ta có: \(\sqrt{4x+20}-3\sqrt{x+5}+\dfrac{4}{3}\sqrt{9x+45}=6\)
\(\Leftrightarrow2\sqrt{x+5}-3\sqrt{x+5}+\dfrac{4}{3}\cdot3\sqrt{x+5}=6\)
\(\Leftrightarrow x+5=4\)
hay x=-1
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: \(x\ge-1\))
\(\Leftrightarrow\sqrt{16\left(x+1\right)}-\sqrt{9\left(x+1\right)}+\sqrt{4\left(x+1\right)}+\sqrt{x+1}=16\)
\(\Leftrightarrow4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}+\sqrt{x+1}=16\)
\(\Leftrightarrow4\sqrt{x+1}=16\)
\(\Leftrightarrow\sqrt{x+1}=4\)
\(\Leftrightarrow x+1=16\)
\(\Leftrightarrow x=15\left(tm\right)\)
1) \(ĐK:x\in R\)
2) \(ĐK:x< 0\)
3) \(ĐK:x\in\varnothing\)
4) \(=\sqrt{\left(x+1\right)^2+2}\)
\(ĐK:x\in R\)
5) \(=\sqrt{-\left(a-4\right)^2}\)
\(ĐK:x\in\varnothing\)
a: ĐKXĐ: \(x\in R\)
\(\sqrt{\left(x+3\right)^2}=12\)
=>\(\left|x+3\right|=12\)
=>\(\left[{}\begin{matrix}x+3=12\\x+3=-12\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=9\\x=-15\end{matrix}\right.\)
b: ĐKXĐ: x>=1
\(\sqrt{25x-25}-\sqrt{9x-9}=10\)
=>\(5\sqrt{x-1}-3\sqrt{x-1}=10\)
=>\(2\sqrt{x-1}=10\)
=>x-1=25
=>x=26(nhận)
a) \(\sqrt{ }\)20 + 2\(\sqrt{ }\)45 - 3\(\sqrt{ }\)80 + \(\sqrt{ }\)125
= \(\sqrt{ }\)4.5 +2\(\sqrt{ }\)9.5 - 3\(\sqrt{16.5}\)
= 2\(\sqrt{5}\) + 6\(\sqrt{5}\) - 12\(\sqrt{5}\)
= -4\(\sqrt{5}\)
b) \(\dfrac{2\sqrt{3}+3\sqrt{2}}{\sqrt{3}+\sqrt{2}}\) - \(4\sqrt{\dfrac{3}{2}}\)- \(\dfrac{5}{1-\sqrt{6}}\)
= \(\dfrac{2\left(\sqrt{3}+\sqrt{2}\right)}{\sqrt{3}+\sqrt{2}}\)- \(\sqrt{16.\dfrac{3}{2}}\) - \(\dfrac{5\left(1+\sqrt{6}\right)}{\left(1-\sqrt{6}\right)\left(1+\sqrt{6}\right)}\)
= 2 - \(\sqrt{24}\) - \(\dfrac{5\left(1+\sqrt{6}\right)}{1-6}\)
= 2 - \(\sqrt{4.6}\) + 1+\(\sqrt{ }\)6
= 2 - 2\(\sqrt{ }\)6 + 1+\(\sqrt{ }\)6
= 3 - \(\sqrt{ }\)6
c) (đề bài) với x khác 4...
= \(\dfrac{\sqrt{x}}{\sqrt{x}-2}\)- \(\dfrac{4\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
= \(\dfrac{x}{\sqrt{x}\left(\sqrt{x}-2\right)}\)- ....
= \(x-4\sqrt{x}+4\)/ \(\sqrt{x}\left(\sqrt{x}-2\right)\)
= (căn -2)2/ căn x(căn x -2)
= căn x-2/căn x
\(d,ĐK:x\ge0\\ PT\Leftrightarrow\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=2\\\sqrt{x}=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\left(tm\right)\\x=9\left(tm\right)\end{matrix}\right.\\ e,ĐK:x\ge1\\ PT\Leftrightarrow\sqrt{x-1}+\dfrac{3}{2}\cdot2\sqrt{x-1}-\dfrac{2}{5}\cdot5\sqrt{x-1}=4\\ \Leftrightarrow2\sqrt{x-1}=4\Leftrightarrow\sqrt{x-1}=2\\ \Leftrightarrow x-1=4\Leftrightarrow x=5\left(tm\right)\\ f,ĐK:x\ge5\\ PT\Leftrightarrow\sqrt{x-5}+2\sqrt{x-5}-\dfrac{1}{3}\cdot3\sqrt{x-5}=6\\ \Leftrightarrow2\sqrt{x-5}=6\Leftrightarrow\sqrt{x-5}=3\\ \Leftrightarrow x-5=9\Leftrightarrow x=14\left(tm\right)\)
a)
Đk: \(x+5\ge0\Rightarrow x\ge-5\)
\(\sqrt{4x+20}-2\sqrt{x+5}+\sqrt{9x+45}=12\\ \Leftrightarrow\sqrt{4\left(x+5\right)}-2\sqrt{x+5}+\sqrt{9\left(x+5\right)}=12\\ \Leftrightarrow2\sqrt{x+5}-2\sqrt{x+5}+3\sqrt{x+5}=12\\ \Leftrightarrow3\sqrt{x+5}=12\\ \Leftrightarrow\sqrt{x+5}=4\\ \Leftrightarrow x+5=16\\ \Rightarrow x=11\)
b.
\(\sqrt{x^2-10x+25}=6\)
Đk: \(x^2-10x+25=\left(x-5\right)^2\ge0;\forall x\inℝ\)
\(\sqrt{x^2-10x+25}=6\\ \Leftrightarrow\sqrt{\left(x-5\right)^2}=6\\ \Rightarrow|x-5|=6\)
\(\Rightarrow\left[{}\begin{matrix}x-5=6\\x-5=-6\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=11\\x=-1\end{matrix}\right.\)
Đs....