 павликпавлик2 
                                                20.08.2020 18:49
                                                
                                                павликпавлик2 
                                                20.08.2020 18:49
                                            ![\sqrt{1-x}=x+1 \\ \\ \left \{ {{1-x=(x+1)^2} \atop {1-x \geq 0,x+1 \geq 0}} \right. \\ \\ \left \{ {{1-x=x^2+2x+1} \atop {x \leq 1,x \geq -1}} \right. \\ \\ \left \{ {{x^2+3x=0} \atop {x E[-1; 1]}} \right. \\ \\ \left \{ {{x*(x+3)=0} \atop {x E[-1; 1]}} \right. \\ \\ \left \{ {{x=0, x=-3} \atop {x E[-1; 1]}} \right. \\ x=0](/tpl/images/0753/2718/d77a9.png) 
                                                 lvalerika 
                                                21.10.2020 12:06
                                                
                                                lvalerika 
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                                                mindrehumor11111 
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                                                nail4iklatyopo 
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                                                рипрмот 
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                                                oleca3 
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                                                pro100miss 
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                                                irinar1 
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