\(\dfrac{-4}{x}=\dfrac{x}{-49}\\ \Rightarrow x^2=\left(-4\right)\left(-49\right)\\ \Rightarrow x^2=196\\ \Rightarrow x=\pm14\)

\(\dfrac{3.6}{x-3}=\dfrac{5}{3}\\ \Rightarrow5\left(x-3\right)=3.3.6\\ \Rightarrow5\left(x-3\right)=54\\ \Rightarrow x-3=\dfrac{54}{5}\\ \Rightarrow x=\dfrac{54}{5}+3\\ \Rightarrow x=\dfrac{69}{15}\)

\(\left(2x+1\right):2=12:3\\ \left(2x+1\right):2=4\\2x+1=2\\ 2x=1\\ x=\dfrac{1}{2} \)

\(\left(2x-14\right):3=12:9\\ \left(2x-14\right):3=\dfrac{4}{3}\\ 2x-14=4\\ 2x=16\\ x=8\)

CÂU NÀY RẤT DỄ. ANH ĐÃ BIẾT KẾT QUẢ TỪ KHI MỚI NHÌN ĐẦU BÀI: KẾT QUẢ LÀ .Z.O.L.......L.O.Z..............................FDGR...................HAPPY........BEAUTYFULLY.>>>>>,<<<<<<<<< .THẰNG NÀO KO HIỂU CHỨNG TỎ NGU . THANKS

a>(8x^2y+10xy6^2-6xy):2xy=4xy+5y-3

b>(3x^2-4x).(2x-6)=6x^3-26x^2+24x

a, \(x^2+y^2-2x+6y-30\)

\(=x^2-2x+1+y^2+6y+9-40\)

\(=\left(x-1\right)^2+\left(y+3\right)^2-40\ge-40\)

\(min=-40\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-3\end{matrix}\right.\)

a)x^2+y^2-2x+6y-30=(x-1)^2+(y+3)^2-40\(\ge\) -40

dấu = xảy ra khi x=1,y=-3

b, \(4-2x^2\le4\)

\(max=4\Leftrightarrow x=0\)

c, \(-x^2+10x-5=-\left(x^2-10x+25\right)+20=-\left(x-5\right)^2+20\le2\text{​​}0\)

\(max=20\Leftrightarrow x=5\)