Question

tìm x biết

a,\(\sqrt{3x}=4\)

b,\(\sqrt{3x}-\dfrac{1}{2}\sqrt{3x}+\dfrac{3}{4}\sqrt{3x}+5=5\sqrt{3x}\)

c,\(\sqrt{\left(1-2x\right)^2}=2\)

Answer 1

a)\(\sqrt{3x}=4\)

\(\Leftrightarrow3x=16\)

\(\Leftrightarrow x=\dfrac{16}{3}\)

Answer 2

c)\(\sqrt{\left(1-2x\right)^2}=2\)

\(\Leftrightarrow1-2x=2\)

\(\Leftrightarrow-2x=1\)

\(\Leftrightarrow x=\dfrac{-1}{2}\)

Answer 3

b) \(\sqrt{3x}-\dfrac{1}{2}\sqrt{3x}+\dfrac{3}{4}\sqrt{3x}+5=5\sqrt{3x}\)

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

\(\Leftrightarrow\dfrac{-15}{4}\sqrt{3x}=-5\)

\(\Leftrightarrow\sqrt{3x}=\dfrac{4}{3}\)

\(\Leftrightarrow3x=\dfrac{16}{9}\Leftrightarrow x=\dfrac{16}{27}\)

Answer 4

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