\(a,x\in\left\{\pm3;\pm2;\pm1;4;0\right\}\\ b,x\in\left\{\pm9;\pm8;\pm7;\pm6;\pm5;\pm4;\pm3;\pm2;\pm1;12;0\right\}\\ c,x\in\left\{\pm14;\pm13;\pm12;\pm11;\pm10;\pm9;\pm8;\pm7;\pm6;\pm5;\pm4;\pm3;\pm2;\pm1;0\right\}\)

do mik có hợp hay chơi vs người ta thân ko nếu ko thì sẽ tạch còn nhờ vapf vẻ bề ngoài nữa =))

\(\rightarrow\) =))??? 

tính nhẩm chx chắc đúng ặ :v 

\(=-\dfrac{5}{7}\left(\dfrac{2}{11}+\dfrac{9}{11}\right)+\dfrac{12}{7}\\ =-\dfrac{5}{7}\cdot1+\dfrac{12}{7}\\ =-\dfrac{5}{7}+\dfrac{12}{7}\\ =1\)

=))) 2 câu đấy khó qớ 

-.- còn TH x=-3 nx ạ 

\(c,\dfrac{1}{2}x^2-4\dfrac{1}{2}=0\\ \Rightarrow\dfrac{1}{2}x^2-\dfrac{9}{2}=0\\ \Rightarrow\dfrac{1}{2}x^2=\dfrac{9}{2}\\ \Rightarrow x^2=9\\ \Rightarrow x^2=\left(\pm3\right)^2\\ \Rightarrow x=\left\{-3;3\right\}\)

\(d,16-4\left(x+\dfrac{1}{2}\right)^2=7\\ \Rightarrow4\left(x+\dfrac{1}{2}\right)^2=9\\ \Rightarrow\left(x+\dfrac{1}{2}\right)^2=\dfrac{9}{4}\\ \Rightarrow\left(x+\dfrac{1}{2}\right)^2=\left(\pm\dfrac{3}{2}\right)^2\\ \Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{3}{2}\\x+\dfrac{1}{2}=-\dfrac{3}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)

Gọi số hs cần tìm là `x-3` 

`x-3 \vdots 10;12;18`

`=>x-3 in BC(12;10;18)={0;180;360;540;...}`

`=>x-3=360`

`=>x=363`