Алгебра, вопрос задал viktoriyaorlov2 , 9 лет назад

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Ответил NNNLLL54
0
 frac{x^3-x^2-x+1}{x+1} + frac{4}{x^2-2x+1}  leq 5\\ frac{x^2(x-1)-(x-1)}{x+1} +  frac{4}{(x-1)^2} leq 5\\ frac{(x-1)(x^2-1)}{x+1} +frac{4}{(x-1)^2} leq 5\\ frac{(x-1)(x-1)(x+1)}{x+1} + frac{4}{(x-1)^2} -5 leq 0\\(x-1)^2+frac{4}{(x-1)^2}-5 leq 0\\t=(x-1)^2; ,; ;  frac{t^2-5t+4}{t}  leq 0; ,; ;  frac{(t-1)(t-4)}{t}  leq 0\\Znaki:; ; ---(0)+++(1)---(4)+++

tin (-infty ,0)cup [, 1,4, ]; ; to

  left { {{(x-1)^2 textless  0; net; reshenij} atop {1<=(x-1)^2<=4}} right.  

 left { {{(x-1)^2 geq 1} atop {(x-1)^2 leq 4}} right. ;  left { {{(x-1)^2-1 geq 0} atop {(x-1)^2-4 leq 0}} right. ;  left { {{(x-1-1)(x-1+1) geq 0} atop {(x-1-2)(x-1+2) leq 0}} right. ;  left { {{(x-2)x geq 0} atop {(x-3)(x+1) leq 0}} right. \\ left { {{x leq 0; ili; x geq 2} atop {-1 leq x leq 3}} right. ; ; Rightarrow ; ; xin [-1;0, ]cup [, 2;3, ]\\ODZ:; ; xne -1; ;; xne 1\\Otvet:; ; (-1;0, ]cup [, 2;3, ]

2); ;  frac{sqrt2cosx-1}{sqrt{-5sinx}} =0; ; to ; ;  left { {{sqrt2cosx-1=0} atop {sinx textless  0}} right. \\ left { {{cosx=frac{1}{sqrt2}=frac{sqrt2}{2}} atop {-pi +2pi n textless  x textless  2pi n,; nin Z}} right. ; ;  left { {{x=pm frac{pi}{4}+2pi k,; kin Z} atop {-pi +2pi n textless  x textless  2pi n,; nin Z}} right. \\x=-frac{pi}{4}+2pi n,; nin Z\\xin [, frac{3pi }{2};3pi , ]; ; to ; ; ; x=frac{7pi}{4}
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