Question

алгебра, на картинке

Answer 1

$1); ; frac{(3x^2+x-2)(x^2-x)}{(3x-2)(x^2-x+1)}leq 0\\a); ; 3x^2+x-2=0; ,; D=25; ; ,; ; x_1=frac{2}{3}; ,; ; x_2=-1\\3x^2+x-2=3(x-frac{2}{3})(x+1)\\b); ; x^2-x+1=0; ,; ; D=-3 textless 0; ; to ; ; x^2-x+1 textgreater 0\\c); ; frac{3(x-frac{2}{3})(x+1)cdot xcdot (x-1)}{3cdot (x-frac{2}{3})} leq 0; ,; ; xne frac{2}{3}\\ (x+1)cdot xcdot (x-1) leq 0\\---[-1, ]+++[, 0, ]---[, 1, ]+++\\underline {xin (-infty ,-1, ]cup [, 0,frac{2}{3})cup (frac{2}{3},1, ]}$

$2); ; (x-1)(x^2+1)(x^3-1)(x^4+1) textless 0\\a); ; x^2+1 textgreater 0; ,; ; ; x^4+1 textgreater 0; ; to \\b); ; (x-1)(x^3-1) textless 0\\(x-1)(x-1)(x^2+x+1) textless 0\\(x-1)^2(underbrace {x^2+x+1}_{ textgreater 0}) textless 0\\x^2+x+1=0; ,; ; D=-3 textless 0; ; to ; ; x^2+x+1 textgreater 0\\c); ; (x-1)^2 textless 0; ; Rightarrow ; ; underline {xin varnothing }; ,; ; t.k.; ; (x-1)^2 geq 0; .$

Answer 2

спасибо <3