a) \(\sqrt{8x+4}-\sqrt{\dfrac{10x+5}{5}}-2=\dfrac{1}{2}-\sqrt{18x+9}\) (ĐK: \(x\ge-\dfrac{1}{2}\))
\(\Leftrightarrow2\sqrt{2x+1}-\sqrt{\dfrac{5\cdot\left(2x+1\right)}{5}}-2=\dfrac{1}{2}-3\sqrt{2x+1}\)
\(\Leftrightarrow2\sqrt{2x+1}-\sqrt{2x+1}-2=\dfrac{1}{2}-3\sqrt{2x+1}\)
\(\Leftrightarrow\sqrt{2x+1}-2=\dfrac{1}{2}-3\sqrt{2x+1}\)
\(\Leftrightarrow\sqrt{2x+1}+3\sqrt{2x+1}=\dfrac{1}{2}+2\)
\(\Leftrightarrow4\sqrt{2x+1}=\dfrac{5}{2}\)
\(\Leftrightarrow\sqrt{2x+1}=\dfrac{5}{8}\)
\(\Leftrightarrow2x+1=\dfrac{25}{64}\)
\(\Leftrightarrow2x=-\dfrac{39}{64}\)
\(\Leftrightarrow x=-\dfrac{39}{128}\)
b) \(\sqrt{4x^2-4x+1}-3=2\)
\(\Leftrightarrow\sqrt{\left(2x-1\right)^2}=2+3\)
\(\Leftrightarrow\left|2x-1\right|=5\)
TH1: \(\left|2x-1\right|=2x-1\) khi \(x\ge\dfrac{1}{2}\)
Pt trở thành:
\(2x-1=5\) (ĐK: \(x\ge\dfrac{1}{2}\))
\(\Leftrightarrow2x=6\)
\(\Leftrightarrow x=3\left(tm\right)\)
TH2: \(\left|2x-1\right|=-\left(2x-1\right)\) khi \(x< \dfrac{1}{2}\)
Pt trở thành:
\(-\left(2x-1\right)=5\) (ĐK \(x< \dfrac{1}{2}\)
\(\Leftrightarrow2x-1=-5\)
\(\Leftrightarrow2x=-4\)
\(\Leftrightarrow x=-2\)
Vậy: ..
c: \(\Leftrightarrow\sqrt{4x^2-4x+5}=5-2x\)
=>5-2x>=0 và 4x^2-4x+5=(5-2x)^2
=>x<=5/2 và 4x^2-4x+5=4x^2-20x+25
=>x<=5/2 và -4x+5=-20x+25
=>x<=5/2 và 16x=20
=>x=5/4(nhận)
d: \(\Leftrightarrow x^2-4\sqrt{x-1}-9x+28=0\)
=>\(x^2-5x-4x+20-4\sqrt{x-1}+8=0\)
=>\(x\left(x-5\right)-4\left(x-5\right)-4\left(\sqrt{x-1}-2\right)=0\)
=>\(\left(x-5\right)\left(x-4\right)-4\left(\sqrt{x-1}-2\right)=0\)
=>\(\left(x-4\right)\cdot\left[\left(\sqrt{x-1}-2\right)\left(\sqrt{x-1}+2\right)\right]-4\left(\sqrt{x-1}-2\right)=0\)
=>\(\left(\sqrt{x-1}-2\right)\left[\left(x-4\right)\left(\sqrt{x-1}+2\right)-4\right]=0\)
=>căn(x-1)-2=0
=>căn(x-1)=2
=>x-1=4
=>x=5
