September 2018 1 24 Report

rozwiąż nierówności:

2x-5>2(x+3)

(x-2)(2x+5)>3x+4+2x[kwadrat]

\frac{x+2}{3}-\frac{x+4}{4}>\frac{x+1}{12}

\frac{x-5}{7}<\frac{3x+2}{21}

4,24,2(x+5)+1,8\leq-2,4

\frac{5,6x-4}8}\leq0,3x+4,5

-1\frac{2}{3}(x+6)\geq\frac{1}{6}(x+3)+0,5

(3-x)(x+1)\geq\frac{5-2x^{2}}{2}

pilne!! :)


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