a: ĐKXĐ: \(x>=-\dfrac{1}{2}\)
\(\sqrt{18x+9}-\sqrt{8x+4}+\dfrac{1}{3}\cdot\sqrt{2x+1}=4\)
=>\(3\sqrt{2x+1}-2\cdot\sqrt{2x+1}+\dfrac{1}{3}\cdot\sqrt{2x+1}=4\)
=>\(\dfrac{4}{3}\sqrt{2x+1}=4\)
=>\(\sqrt{2x+1}=4:\dfrac{4}{3}=3\)
=>\(2x+1=3^2=9\)
=>2x=8
=>x=4(nhận)
b: ĐKXĐ: x>=1
\(\dfrac{2}{3}\cdot\sqrt{9x-9}-\dfrac{1}{4}\cdot\sqrt{16x-16}+27\cdot\sqrt{\dfrac{x-1}{81}}=4\)
=>\(\dfrac{2}{3}\cdot3\sqrt{x-1}-\dfrac{1}{4}\cdot4\sqrt{x-1}+27\cdot\dfrac{\sqrt{x-1}}{9}=4\)
=>\(2\sqrt{x-1}-\sqrt{x-1}+3\sqrt{x-1}=4\)
=>\(4\cdot\sqrt{x-1}=4\)
=>\(\sqrt{x-1}=1\)
=>x-1=1
=>x=1+1=2(nhận)
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