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Popular >

(sqrt((x+3))-sqrt(3)+1)/((2sqrt(3))x)<= 0.1

Solution

\frac{( x + 3 ) - 3 + 1}{( 2 3 ) x} \leq 0.1

x \geq \frac{5 1 + 20 3 + 10 3 - 5}{6}

Solution steps

\frac{x + 3 - 3 + 1}{2 3 x} \leq 0.1

Multiply both sides by

23

\frac{( x + 3 - 3 + 1 ) \cdot 2 3}{2 3 x} \leq 0.1 \cdot 23

Simplify

\frac{x + 3 - 3 + 1}{x} \leq 3 \cdot 0.2

Multiply both sides by

10

\frac{10 ( x + 3 - 3 + 1 )}{x} \leq 23

Multiply both sides by

x

10 x + 3 - 103 + 10 \leq 23 x

Move

103

to the right side

10 x + 3 + 10 \leq 23 x + 103

Move

10

to the right side

10 x + 3 \leq 23 x + 103 - 10

Divide both sides by

10

x + 3 \leq \frac{3 x + 5 3 - 5}{5}

f (x) \leq g (x)

then

g (x) \geq 0

\frac{3 x + 5 3 - 5}{5} \geq 0 : x \geq \frac{- 5 3 + 5}{3}

x + 3 \leq \frac{3 x + 5 3 - 5}{5} : x \leq - \frac{5 - 10 3 + 5 1 + 20 3}{6} or x \geq \frac{5 1 + 20 3 + 10 3 - 5}{6}

Find singularity points

Find non-negative values for radicals:

x \geq - 3

Combine the intervals

x \geq \frac{- 5 3 + 5}{3} and (x \leq - \frac{5 - 10 3 + 5 1 + 20 3}{6} or x \geq \frac{5 1 + 20 3 + 10 3 - 5}{6}) and x \geq - 3

Merge Overlapping Intervals

x \geq \frac{5 1 + 20 3 + 10 3 - 5}{6}

Verify Solutions:

x \geq \frac{5 1 + 20 3 + 10 3 - 5}{6}

True

The solution is

x \geq \frac{5 1 + 20 3 + 10 3 - 5}{6}

\frac{x - 3}{6} = 2 - \frac{5 ( x + 3 )}{12}

- 7 x > 10

lo g_{2} (\frac{x}{4})

10 x + 3 < 8

(1 \frac{1}{2} x)^{2}