Gọi vận tốc thực của ca nô là: x (km/h) (a>0)

⇒ vận tốc thực của ca nô khi xuôi dòng là: x+4 (km/h)

    vận tốc thực của ca nô khi ngược dòng là: x-4 (km/h) (a>4)

Khi gặp bè cách A 8km thì ca nô cách B 16km

Theo bài ra, ta có pt:

\(\dfrac{24}{x+4}+\dfrac{16}{x-4}=\dfrac{8}{4}=2\)

\(\Leftrightarrow24\left(x-4\right)+16\left(x+4\right)=2\left(x^2-16\right)\)

\(\Leftrightarrow24x-96+16x+64=2\left(x^2-16\right)\)

\(\Leftrightarrow20x-16=x^2-16\)

\(\Leftrightarrow x^2-20x=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=0\left(L\right)\\x=20\left(TM\right)\end{matrix}\right.\)

⇒ Vận tốc thực của ca nô là: \(20\) (km/h)

\(\dfrac{-15}{35}=\dfrac{x}{42}=\dfrac{-30}{y}=\dfrac{60}{z}\)

\(\dfrac{-3}{7}=\dfrac{x}{42}=\dfrac{-30}{y}=\dfrac{60}{z}\)

\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{\left(-3\right)\times42}{7}=-18\\y=\dfrac{\left(-30\right)\times7}{\left(-3\right)}=70\\z=\dfrac{7\times60}{\left(-3\right)}=-140\end{matrix}\right.\)

\(=\dfrac{98}{12}=\dfrac{49}{6}\)

\(MSC:143\)

\(\dfrac{11}{13}=\dfrac{11\times143:13}{13\times143:13}=\dfrac{121}{143}\)

\(\dfrac{9}{11}=\dfrac{9\times143:11}{11\times143:11}=\dfrac{117}{143}\)

\(\left(1-\dfrac{1}{3}\right)\times\left(1-\dfrac{3}{7}\right)\times\left(1-\dfrac{3}{10}\right)\times\left(1-\dfrac{3}{97}\right)\times\left(1-\dfrac{3}{100}\right)\)

\(=\dfrac{2}{3}\times\dfrac{4}{7}\times\dfrac{7}{10}\times...\times\dfrac{94}{97}\times\dfrac{97}{100}\)

\(=\dfrac{2\times4\times7\times...\times94\times97}{3\times7\times10\times...\times97\times100}\)

\(=\dfrac{2\times4}{3\times100}=\dfrac{8}{300}=\dfrac{2}{75}\)

\(1,\dfrac{x+2}{x-2}-\dfrac{x-2}{x+2}=\dfrac{4}{x^2-4}\) \(\left(đk:x\ne\pm2\right)\)

\(\Leftrightarrow\left(x+2\right)^2-\left(x-2\right)^2=4\)

\(\Leftrightarrow x^2+4x+4-x^2+4x-4=4\)

\(\Leftrightarrow8x=4\)

\(\Leftrightarrow x=\dfrac{1}{2}\)

\(A=1\times3+3\times5+5\times7+...+99\times101\)

\(=1\left(1+2\right)+3\left(3+2\right)+5\left(5+2\right)+...+99\left(99+2\right)\)

\(=\left(1^2+3^2+5^2+...+99^2\right)+2\left(1+3+5+...+99\right)\)

Ta có:

\(1^2+2^2+3^2+...+n^2=\dfrac{n\left(n+1\right)\left(2n+1\right)}{6}\)

⇒ \(A=\left(1^2+2^2+3^2+...+100^2\right)-2^2\left(1^2+2^2+3^2+...+50^2\right)+2\left(1+3+5+...+99\right)\)

\(=\dfrac{100.101.201}{6}+\dfrac{4.50.51.101}{6}+\dfrac{\left(99+1\right).\left[\left(99-1\right):2+1\right]}{2}\)

\(=338350-171700+5000\)

\(=166650+5000=171650\)

\(\Leftrightarrow\left(x+3\right)^2\left(x-3\right)^2=6\left(x+3\right)\)

\(\Leftrightarrow\left(x+3\right)\left[\left(x+3\right)\left(x-3\right)^2-6\right]=0\)

Vì \(\left[\left(x+3\right)\left(x-3\right)^2-6\right]\ne0\)

\(\Rightarrow x+3=0\)

\(\Rightarrow x=-3\)

Vì nước tiểu thải ra các chất đọc hại giúp cho ta ít bị bệnh hơn