ĐKXĐ:\(x\ne1\)

\(\dfrac{1-5x}{x-1}\ge1\\ \Rightarrow1-5x\ge x-1\\ \Leftrightarrow x-1-1+5x\le0\\ \Leftrightarrow6x-2\le0\\ \Leftrightarrow x\le\dfrac{1}{3}\)

dòng thứ 3 bạn xem lại -3x^2 -2x +9 bạn mới là người sai

\(a,2x-5=-x+4\\ \Leftrightarrow3x=9\\ \Leftrightarrow x=3\\ b,\left(4x-10\right)\left(25+5x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}4x-10=0\\25+5x=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-5\end{matrix}\right.\\ c,\dfrac{x}{3}-\dfrac{2x+1}{2}=\dfrac{x}{6}-x\\ \Leftrightarrow\dfrac{2x}{6}-\dfrac{3\left(2x+1\right)}{6}-\dfrac{x}{6}+\dfrac{6x}{6}=0\\ \Leftrightarrow2x-6x-3-x+6x=0\\ \Leftrightarrow x-3=0\\ \Leftrightarrow x=3\)

d, ĐKXĐ:\(x\ne-2,x\ne3\)

\(1+\dfrac{x}{3-x}=\dfrac{5x}{\left(x+2\right)\left(3-x\right)}+\dfrac{2}{x+2}\\ \Leftrightarrow\dfrac{\left(x+2\right)\left(3-x\right)}{\left(x+2\right)\left(3-x\right)}+\dfrac{x\left(x+2\right)}{\left(x+2\right)\left(3-x\right)}-\dfrac{5x}{\left(x+2\right)\left(3-x\right)}-\dfrac{2\left(3-x\right)}{\left(x+2\right)\left(3-x\right)}=0\\ \Leftrightarrow\dfrac{-x^2+x+6}{\left(x+2\right)\left(3-x\right)}+\dfrac{x^2+2x}{\left(x+2\right)\left(3-x\right)}-\dfrac{5x}{\left(x+2\right)\left(3-x\right)}-\dfrac{6-2x}{\left(x+2\right)\left(3-x\right)}=0\)

\(\Leftrightarrow\dfrac{-x^2+x+6+x^2+2x-5x-6+2x}{\left(x+2\right)\left(3-x\right)}=0\\ \Rightarrow0=0\left(luôn.đúng\right)\)

\(3\left(x-2\right)^2+9\left(x-1\right)=3\left(x^2+x-3\right)\\ \Leftrightarrow3\left(x^2-4x+4\right)+9x-9=3x^2+3x-9\\ \Leftrightarrow3x^2-12x+12+9x-9-3x^2-3x+9=0\\ \Leftrightarrow-6x+12=0\\ \Leftrightarrow x=2\)

ĐKXĐ:\(17-x\ge0\Leftrightarrow x\le17\)

\(\left|2x-7\right|=17-x\\ \Leftrightarrow\left[{}\begin{matrix}2x-7=17-x\\2x-7=x-17\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}3x=24\\x=-10\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=8\left(tm\right)\\x=-10\left(tm\right)\end{matrix}\right.\)

C

C

a, \(\Delta'=\left(-2\right)^2-\left(1-2m\right)=4-1+2m=2m-3\)

Để pt có nghiệm thì \(\Delta'\ge0\Leftrightarrow2m-3\ge0\Leftrightarrow m\ge\dfrac{3}{2}\)

b, Theo Vi-ét:\(\left\{{}\begin{matrix}x_1+x_2=4\\x_1x_2=1-2m\end{matrix}\right.\)

\(x_1^2+x_2^2=6\\ \Leftrightarrow\left(x_1+x_2\right)^2-2x_1x_2=6\\ \Leftrightarrow4^2-2\left(1-2m\right)=6\\ \Leftrightarrow16-2+4m-6=0\\ \Leftrightarrow4m-8=0\\ \Leftrightarrow m=2\left(tm\right)\)

Ta có: \(\Delta=\left(-10\right)^2-4.3.2=100-24=76>0\)

Suy ra pt luôn có 2 nghiệm phân biệt

Theo Vi-ét:\(\left\{{}\begin{matrix}x_1+x_2=\dfrac{10}{3}\\x_1x_2=\dfrac{2}{3}\end{matrix}\right.\)

\(A=\dfrac{x_1-1}{x_2}+\dfrac{x_2-1}{x_1}-x_1^2x_2^2\\ =\dfrac{x_1\left(x_1-1\right)+x_2\left(x_2-1\right)}{x_1x_2}-\left(x_1x_2\right)^2\\ =\dfrac{x_1^2-x_1+x_2^2-x_2}{\dfrac{2}{3}}-\left(\dfrac{2}{3}\right)^2\\ =\dfrac{\left(x_1+x_2\right)^2-2x_1x_2-\left(x_1+x_2\right)}{\dfrac{2}{3}}-\dfrac{4}{9}\)

\(=\dfrac{\left(\dfrac{10}{3}\right)^2-2.\dfrac{2}{3}-\dfrac{10}{3}}{\dfrac{2}{3}}-\dfrac{4}{9}\\ =\dfrac{83}{9}\)

\(\dfrac{528}{264}=\dfrac{528:264}{264:264}=\dfrac{2}{1}=2\)