H24
NT
29 tháng 5 2024 lúc 18:28

a: ĐKXĐ: x>=1

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

=>\(\left(\sqrt{3x+1}+\sqrt{x-1}\right)^2=4\)

=>\(3x+1+x-1+2\sqrt{\left(3x+1\right)\left(x-1\right)}=4\)

=>\(2\sqrt{\left(3x+1\right)\left(x-1\right)}=4-4x\)

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

=>\(\left\{{}\begin{matrix}\left(3x+1\right)\left(x-1\right)=\left(2-2x\right)^2=\left(2x-2\right)^2\\x< =1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}4x^2-8x+4=3x^2-3x+x-1\\x< =1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x^2-6x+5=0\\x< =1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\left(x-1\right)\left(x-5\right)=0\\x< =1\end{matrix}\right.\)

=>x=1(nhận)

b: ĐKXĐ: x>=1/5

\(\sqrt{5x-1}-\sqrt{x}=4x-1\)

=>\(\dfrac{5x-1-x}{\sqrt{5x-1}+\sqrt{x}}=4x-1\)

=>\(\sqrt{5x-1}+\sqrt{x}=1\)
=>\(\left(\sqrt{5x-1}+\sqrt{x}\right)^2=1\)

=>\(5x-1+x+2\sqrt{x\left(5x-1\right)}=1\)

=>\(2\sqrt{5x^2-x}=1+1-6x=2-6x\)

=>\(\sqrt{5x^2-x}=1-3x\)

=>\(\left\{{}\begin{matrix}1-3x>=0\\\left(1-3x\right)^2=5x^2-x\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x< =\dfrac{1}{3}\\9x^2-6x+1-5x^2+x=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x< =\dfrac{1}{3}\\4x^2-5x+1=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x< =\dfrac{1}{3}\\\left(x-1\right)\left(4x-1\right)=0\end{matrix}\right.\Leftrightarrow x=\dfrac{1}{4}\left(nhận\right)\)

c: ĐKXĐ: x>=0

\(\sqrt{2x+6}-\sqrt{2x}=6\)

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

=>\(\sqrt{2x+6}+\sqrt{2x}=1\)

=>\(2x+6+2x+2\sqrt{2x\left(2x+6\right)}=1\)

=>\(2\sqrt{4x\left(x+3\right)}=1-4x-6=-4x-5\)

=>\(\sqrt{16x\left(x+3\right)}=-4x-5\)

=>\(\left\{{}\begin{matrix}-4x-5>=0\\16x\left(x+3\right)=\left(-4x-5\right)^2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x< =-\dfrac{5}{4}\\16x^2+48x-16x^2-40x-25=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x< =-\dfrac{5}{4}\\8x-25=0\end{matrix}\right.\)

=>\(x\in\varnothing\)

d: ĐKXĐ: x>=1/10

\(\sqrt{10x-1}+\sqrt{7x-3}=3x+2\)

=>\(\dfrac{10x-1-7x+3}{\sqrt{10x-1}-\sqrt{7x-3}}=3x+2\)

=>\(\sqrt{10x-1}-\sqrt{7x-3}=1\)

=>\(\left(\sqrt{10x-1}-\sqrt{7x-3}\right)^2=1\)

=>\(10x-1+7x-3-2\sqrt{\left(10x-1\right)\left(7x-3\right)}=1\)

=>\(2\sqrt{\left(10x-1\right)\left(7x-3\right)}=17x-5\)

=>\(\sqrt{4\left(10x-1\right)\left(7x-3\right)}=17x-5\)

=>\(\left\{{}\begin{matrix}x>=\dfrac{5}{17}\\\left(17x-5\right)^2=4\left(10x-1\right)\left(7x-3\right)\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>=\dfrac{5}{17}\\289x^2-170x+25=4\left(70x^2-30x-7x+3\right)\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>=\dfrac{5}{17}\\289x^2-170x+25-280x^2+148x-12=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>=\dfrac{5}{17}\\9x^2-22x+13=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>=\dfrac{5}{17}\\\left(9x-13\right)\left(x-1\right)=0\end{matrix}\right.\)

=>\(x\in\left\{\dfrac{13}{9};1\right\}\)

 

e: 

ĐKXĐ: \(x>=-\dfrac{1}{6}\)

\(\sqrt{6x+1}-\sqrt{3x+2}=3x-1\)

=>\(\dfrac{6x+1-3x-2}{\sqrt{6x+1}+\sqrt{3x+2}}=3x-1\)

=>\(\sqrt{6x+1}+\sqrt{3x+2}=1\)

=>\(6x+1+3x+2+2\sqrt{\left(6x+1\right)\left(3x+2\right)}=1\)

=>\(\sqrt{4\left(6x+1\right)\left(3x+2\right)}=1-9x-3=-9x-2\)

=>\(\left\{{}\begin{matrix}x< =-\dfrac{2}{9}\\\left(-9x-2\right)^2=4\left(6x+1\right)\left(3x+2\right)\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x< =-\dfrac{2}{9}\\81x^2+36x+4=4\left(18x^2+15x+2\right)\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x< =-\dfrac{2}{9}\\9x^2-24x-4=0\end{matrix}\right.\)

=>\(x\in\varnothing\)

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