

Trịnh Minh Hoàng
Giới thiệu về bản thân



































\(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{10\cdot11}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{11}\\ =1-\dfrac{1}{11}\\ =\dfrac{11}{11}-\dfrac{1}{11}=\dfrac{10}{11}\)
\(5^2\cdot x-2^4\cdot x=3^4-6\cdot3^2\\ \Rightarrow25x-16x=81-6\cdot9\\ \Rightarrow9x=81-54\\ \Rightarrow9x=27\\ \Rightarrow x=27:9\\ \Rightarrow x=3\)
\(2\cdot\left(x-3\right)-\left(4\cdot x-1\right)=0\\ \Rightarrow2x-6-4x+1=0\\ \Rightarrow\left(2x-4x\right)+\left(-6+1\right)=0\\ \Rightarrow-2x-5=0\\ \Rightarrow-2x=5\\ \Rightarrow x=-\dfrac{5}{2}\)
Kết quả là: `46`
\(\dfrac{392-x}{32}+\dfrac{390-x}{34}+\dfrac{388-x}{36}+\dfrac{386-x}{38}+\dfrac{384-x}{40}=-5\\\Rightarrow\dfrac{392-x}{32}+1+\dfrac{390-x}{34}+1+\dfrac{388-x}{36}+1+\dfrac{386-x}{38}+1+\dfrac{384-x}{40}+1=0\\ \Rightarrow\dfrac{424-x}{32}+\dfrac{424-x}{34}+\dfrac{424-x}{36}+\dfrac{424-x}{38}+\dfrac{424-x}{40}=0\\ \Rightarrow\left(424-x\right)\left(\dfrac{1}{32}+\dfrac{1}{34}+\dfrac{1}{36}+\dfrac{1}{38}+\dfrac{1}{40}\right)=0\\ \Rightarrow424-x=0\\ \Rightarrow x=424\)
\(a,x^2=25\\ \Rightarrow x^2=5^2\\ \Rightarrow x=5\)
\(b,6\cdot x^2=150\\ \Rightarrow x^2=150:6\\ \Rightarrow x^2=25\\ \Rightarrow x^2=5^2\\ \Rightarrow x=5\)
\(a,2^n+2^{n+4}=272\\ \Rightarrow2^n+2^n.2^4=272\\ \Rightarrow2^n+2^n.16=272\\ \Rightarrow2^n.17=272\\ \Rightarrow2^n=16\\ \Rightarrow2^n=2^4\\ \Rightarrow n=4\)
\(b,5^{n+2}-5^n=600\\ \Rightarrow5^n.5^2-5^n=600\\ \Rightarrow5^n\left(25-1\right)=600\\ \Rightarrow5^n.24=600\\ \Rightarrow5^n=25\\ \Rightarrow5^n=5^2\\ \Rightarrow n=2\)