Kết quả là: `46`

\(\dfrac{392-x}{32}+\dfrac{390-x}{34}+\dfrac{388-x}{36}+\dfrac{386-x}{38}+\dfrac{384-x}{40}=-5\\\Rightarrow\dfrac{392-x}{32}+1+\dfrac{390-x}{34}+1+\dfrac{388-x}{36}+1+\dfrac{386-x}{38}+1+\dfrac{384-x}{40}+1=0\\ \Rightarrow\dfrac{424-x}{32}+\dfrac{424-x}{34}+\dfrac{424-x}{36}+\dfrac{424-x}{38}+\dfrac{424-x}{40}=0\\ \Rightarrow\left(424-x\right)\left(\dfrac{1}{32}+\dfrac{1}{34}+\dfrac{1}{36}+\dfrac{1}{38}+\dfrac{1}{40}\right)=0\\ \Rightarrow424-x=0\\ \Rightarrow x=424\)

\(a,x^2=25\\ \Rightarrow x^2=5^2\\ \Rightarrow x=5\)

\(b,6\cdot x^2=150\\ \Rightarrow x^2=150:6\\ \Rightarrow x^2=25\\ \Rightarrow x^2=5^2\\ \Rightarrow x=5\)

\(a,2^n+2^{n+4}=272\\ \Rightarrow2^n+2^n.2^4=272\\ \Rightarrow2^n+2^n.16=272\\ \Rightarrow2^n.17=272\\ \Rightarrow2^n=16\\ \Rightarrow2^n=2^4\\ \Rightarrow n=4\)

\(b,5^{n+2}-5^n=600\\ \Rightarrow5^n.5^2-5^n=600\\ \Rightarrow5^n\left(25-1\right)=600\\ \Rightarrow5^n.24=600\\ \Rightarrow5^n=25\\ \Rightarrow5^n=5^2\\ \Rightarrow n=2\)

\(a,27\cdot3^{n-1}=81\\ \Rightarrow3^{n-1}=81:27\\ \Rightarrow3^{n-1}=3\\ \Rightarrow n-1=1\\ \Rightarrow n=2\)

\(b,5\cdot5^{6-n}=625\\ \Rightarrow5^{6-n}=625:5\\ \Rightarrow5^{6-n}=125\\ \Rightarrow5^{6-n}=5^3\\ \Rightarrow6-n=3\\ \Rightarrow n=3\)

\(\left(\dfrac{-5}{7}\right).\left(\dfrac{2}{5}-x\right)+\dfrac{-1}{3}=\dfrac{1}{5}+\dfrac{-3}{10}\\ \Rightarrow-\dfrac{5}{7}\cdot\dfrac{2}{5}+\dfrac{5}{7}\cdot x-\dfrac{1}{3}=\dfrac{1}{5}-\dfrac{3}{10}\\ \Rightarrow-\dfrac{2}{7}+\dfrac{5}{7}x-\dfrac{1}{3}=-\dfrac{1}{10}\\ \Rightarrow-\dfrac{13}{21}+\dfrac{5}{7}x=-\dfrac{1}{10}\\ \Rightarrow\left(\dfrac{5}{7}x\right)=-\dfrac{1}{10}+\dfrac{13}{21}\\ \Rightarrow\left(\dfrac{5}{7}x\right)=\dfrac{109}{210}\\ \Rightarrow x=\dfrac{109}{150}\)

\(\left(3x-21\right)\cdot\left(43-x\right)=0\\ \Rightarrow\left[{}\begin{matrix}3x-21=0\\43-x=0\end{matrix}\right.\\\Rightarrow\left[{}\begin{matrix}x=7\\x=43\end{matrix}\right. \)

Vậy: \(x=7;x=43\)

1435; 3192; 2307; 1435; 3192; 2307; 1435; 3192; 2307; 1435