3x+2-\(\sqrt{9x^2+6x+1}\)( x>\(\frac{1}{3}\))
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a/ \(\frac{\sqrt{x^2-4x+4}}{x-2}=\frac{\sqrt{\left(x-2\right)^2}}{x-2}=\frac{\left|x-2\right|}{x-2}\)
có x<2\(\Rightarrow\left|x-2\right|=2-x\)
\(\Rightarrow\frac{2-x}{x-2}\)
b/ \(\frac{\sqrt{9x^2-6x+1}}{9x^2-1}=\frac{\sqrt{\left(3x-1\right)^2}}{\left(3x-1\right)\left(3x+1\right)}=\frac{\left|3x-1\right|}{\left(3x-1\right)\left(3x+1\right)}\)
Có x>\(\frac{1}{3}\Rightarrow\left|3x-1\right|=3x-1\)
\(\Rightarrow\frac{3x-1}{\left(3x-1\right)\left(3x+1\right)}=\frac{1}{3x+1}\)
b)
)\(\sqrt{\frac{4}{\left(2-\sqrt{5}\right)^2}}-\sqrt{\frac{4}{\left(2+\sqrt{5}\right)^2}}\)
= \(\frac{2}{2-\sqrt{5}}-\frac{2}{2+\sqrt{5}}\)
=\(\frac{2\left(2+\sqrt{5}\right)-2\left(2-\sqrt{5}\right)}{\left(2-\sqrt{5}\right)\left(2+\sqrt{5}\right)}\)
=\(\frac{4+2\sqrt{5}-4+2\sqrt{5}}{2^2-\sqrt{5}^2}\)
=\(\frac{4\sqrt{5}}{4-5}\)
=\(\frac{4\sqrt{5}}{-1}\)
\(-4\sqrt{5}\)
a/ \(\Rightarrow\sqrt{\left(3x+1\right)^2}=1\Rightarrow3x+1=1\Rightarrow3x=0\Rightarrow x=0\)
b/ \(\Rightarrow\frac{3}{2}\sqrt{3x}-\sqrt{3x}-\frac{1}{2}\sqrt{3x}=5\)
\(\Rightarrow\sqrt{3x}\left(\frac{3}{2}-1-\frac{1}{2}\right)=5\)
\(\Rightarrow\sqrt{3x}.0=5\Rightarrow0=5\) (vô lí)
Vậy pt vô nghiệm
1: Đặt \(a=9x^2-6x\)
=>\(45x^2-30x=5\left(9x^2-6x\right)=5a\)
\(\sqrt{9x^2-6x+2}+\sqrt{45x^2-30x+9}=\sqrt{6x-9x^2+8}\)
=>\(\sqrt{a+2}+\sqrt{5a+9}=\sqrt{-a+8}\)
=>\(\sqrt{a+2}-1+\sqrt{5a+9}-2=\sqrt{-a+8}-3\)
=>\(\frac{a+2-1}{\sqrt{a+2}+1}+\frac{5a+9-4}{\sqrt{5a+9}+2}=\frac{-a+8-9}{\sqrt{-a+8}+3}\)
=>\(\left(a+1\right)\left(\frac{1}{\sqrt{a+2}+1}+\frac{5}{\sqrt{5a+9}+2}+\frac{1}{\sqrt{a+8}+3}\right)=0\)
=>a+1=0
=>a=-1
=>\(9x^2-6x=-1\)
=>\(9x^2-6x+1=0\)
=>\(\left(3x-1\right)^2=0\)
=>3x-1=0
=>3x=1
=>x=1/3
2: Đặt \(x^2-2x=a\)
\(\sqrt{2x^2-4x+3}+\sqrt{3x^2-6x+7}=2-x^2+2x\)
=>\(\sqrt{2\left(x^2-2x\right)+3}+\sqrt{3\left(x^2-2x\right)+7}=-\left(x^2-2x\right)+2\)
=>\(\sqrt{2a+3}+\sqrt{3a+7}=-a+2\)
=>\(\sqrt{2a+3}-1+\sqrt{3a+7}-2=-a+2-3\)
=>\(\frac{2a+2}{\sqrt{2a+3}+1}+\frac{3a+7-4}{\sqrt{3a+7}+2}=-a-1\)
=>\(\left(a+1\right)\left(\frac{2}{\sqrt{2a+3}+1}+\frac{3}{\sqrt{3a+7}+2}+1\right)=0\)
=>a+1=0
=>\(x^2-2x+1=0\)
=>\(\left(x-1\right)^2=0\)
=>x-1=0
=>x=1
\(3x+2-\sqrt{9x^2+6x+1}\)
\(=3x+2-\sqrt{\left(3x+1\right)^2}\)
\(=3x+2-\left(3x+1\right)\)(Chú ý \(x>\frac{1}{3}\))
\(=3x+2-3x-1=1\)