\(35-\left|2x-1\right|=14\\ \left|2x-1\right|=21\\ \left(2x-1\right)^2=21^2\\ \left(2x-1+21\right)\left(2x-1-21\right)=0\\ 4\left(x+10\right)\left(x-11\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-10\\x=11\end{matrix}\right.\)

\(\left|x\right|-3+6=16\\ \left|x\right|=13\Rightarrow x=\pm13\)

ĐKXĐ: x khác +-2

\(\dfrac{x-2}{x+2}-\dfrac{3}{x-2}=\dfrac{2\left(x-11\right)}{x^2-4}\\ \dfrac{x^2-4x+4-3x-6}{x^2-4}=\dfrac{2x-22}{x^2-4}\\ \Rightarrow x^2-12x-2=2x-22\\ x^2-14x+20=0\\ x^2-14x+49=49-20\\ \left(x-7\right)^2=29\\ \Rightarrow\left[{}\begin{matrix}x-7=29\\x-7=-29\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=36\\x=-22\end{matrix}\right.\)

Cò lùi là cò ko tiến cò ko tiến là tiền ko có 

Tick mk nha bạn 

\(\dfrac{-7x^2+4}{x^3+1}=\dfrac{5}{x^2-x+1}-\dfrac{1}{x+1}\\ \dfrac{-7x^2+4}{x^3+1}=\dfrac{5\left(x+1\right)-\left(x^2-x+1\right)}{x^3+1}\\ \Rightarrow-7x^2+4=-x^2+6x-4\\ 6x^2+6x-8=0\\ x^2+x-\dfrac{4}{3}=0\\ x^2+x+\dfrac{1}{4}=\dfrac{4}{3}+\dfrac{1}{4}\\ \left(x+\dfrac{1}{2}\right)^2=\dfrac{19}{12}\\ \Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\sqrt{\dfrac{19}{12}}\\x+\dfrac{1}{2}=-\sqrt{\dfrac{19}{12}}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\sqrt{\dfrac{19}{12}}-\dfrac{1}{2}\\x=-\sqrt{\dfrac{19}{12}}-\dfrac{1}{2}\end{matrix}\right.\)

\(\left(x+2\right)\left(x^2-3x+5\right)=\left(x+2\right)x^2\\\left(x+2\right)\left(x^2-3x+5\right)-\left(x+2\right)x^2=0\\ \left(x+2\right)\left(5-3x\right)=0\\ \Rightarrow\left[{}\begin{matrix}x+2=0\\5-3x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{5}{3}\end{matrix}\right.\)

đơn giản vkl

7+6=13 :))

c)

\(S=-x^2+4x-9\\ S=-\left(x^2-4x+4\right)-5\\ S=-\left(x-2\right)^2-5\le-5\)

đẳng thức xảy ra khi x=2

vậy MAX S=-5 tại x=2

a)

\(Q=2x-2-3x^2\\ Q=-3\left(x-\dfrac{1}{3}\right)^2+\dfrac{3.\left(-3\right)\left(-2\right)-2^2}{4.\left(-3\right)}\\ Q=-3\left(x-\dfrac{1}{3}\right)^2-\dfrac{14}{12}\le-\dfrac{14}{12}\)

đẳng thức xảy ra khi x=1/3

vậy MAX Q=-14/12 tại x=1/3

\(x^3+y^3+z^3=3xyz\\ \Leftrightarrow\left(x+y+z\right)\left[\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\right]=0\\ \Rightarrow x=y=z\)

tháy vào M ta có:

\(M=\left(1+\dfrac{x}{y}\right)\left(1+\dfrac{y}{z}\right)\left(1+\dfrac{z}{x}\right)\\ M=\left(1+1\right)\left(1+1\right)\left(1+1\right)=\left(1+1\right)^3=8\)