\(\sqrt{\left(2x\right)^2+2.2x.5+5^2}+\sqrt{x^2+2.x.3+3^2}=10x-20\)
\(\Leftrightarrow\sqrt{\left(2x+5\right)^2}+\sqrt{\left(x+3\right)^2}=10x-20\)
\(\Leftrightarrow2x+5+x+3=10x-20\)
\(\Leftrightarrow7x=28\Leftrightarrow x=4\)
so sánh \(\sqrt{12}+\sqrt{20}+\sqrt{30}+\sqrt{42}\)và \(\sqrt{401}\)mong ad giúp mình youtube của mình là long vh nhớ đăng ký nhé
\(\Leftrightarrow\left|2x+5\right|+\left|x+3\right|+10x-20\) + Nếu x<-3 \(\Leftrightarrow-2x-5-x-3=10x-20\) \(\Leftrightarrow-3x-8=10x-20\) \(\Leftrightarrow x=\dfrac{12}{13}\left(loại\right)\) + Nếu -3\(^{\le x\le-\dfrac{5}{2}}\) \(\Leftrightarrow-2x-5+x+3=10x-20\) \(\Leftrightarrow-x-2=10x-20\) \(\Leftrightarrow x=\dfrac{18}{11}\left(loại\right)\) + Nếu x lớn hơn \(\dfrac{-5}{2}\) \(\Leftrightarrow2x+5+x+3=10x-20\) \(\Leftrightarrow3x+8+10x-20\) \(\Leftrightarrow x=4\left(tm\right)\) Vậy x=4