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Bài 8: \(sin\widehat{B}=\dfrac{AC}{BC}\) \(\Rightarrow\) \(\dfrac{AC}{BC}=sin30^o=\dfrac{1}{2}\) \(\Rightarrow\) BC=2AC.
Câu 1 :
\(\frac{4-x}{6-x}=\frac{x-3}{x-8}\)
\(\Leftrightarrow\left(4-x\right)\left(x-8\right)=\left(x-3\right)\left(6-x\right)\)
\(\Leftrightarrow4x-32-x^2+8x=6x-x^2-18+3x\)
\(\Leftrightarrow4x+8x-6x-3x=-18+32\)
\(\Leftrightarrow3x=14\)
\(\Leftrightarrow x=\frac{14}{3}\)
Câu 2 :
\(\frac{8-x}{4}=\frac{2x-3}{5}\)
\(\Leftrightarrow5.\left(8-x\right)=4.\left(2x-3\right)\)
\(\Leftrightarrow40-5x=8x-12\)
\(\Leftrightarrow-5x-8x=-12-40\)
\(\Leftrightarrow-13x=-52\)
\(\Leftrightarrow x=4\)
1\(\frac{1}{2}\)+2\(\frac{2}{3}\)+3\(\frac{3}{4}\)+4\(\frac{4}{5}\)+.......+50\(\frac{50}{51}\)+\(\frac{1}{2}\)+\(\frac{1}{3}\)+\(\frac{1}{4}\)+\(\frac{1}{5}\)+....+\(\frac{1}{51}\)
=(1\(\frac{1}{2}\)+\(\frac{1}{2}\))+(2\(\frac{2}{3}\)+\(\frac{1}{3}\))+(3\(\frac{3}{4}\)+\(\frac{1}{4}\))+.......+(50\(\frac{50}{51}\)+\(\frac{1}{51}\))
=2+3+4+.....+51
=1325
Vậy:1\(\frac{1}{2}\)+2\(\frac{2}{3}\)+3\(\frac{3}{4}\)+4\(\frac{4}{5}\)+.......+50\(\frac{50}{51}\)+\(\frac{1}{2}\)+\(\frac{1}{3}\)+\(\frac{1}{4}\)+\(\frac{1}{5}\)+....+\(\frac{1}{51}\)=1325
Học Tốt!
\(1\frac{1}{2}+2\frac{2}{3}+3\frac{3}{4}+4\frac{4}{5}+...+50\frac{50}{51}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{51}\)
\(=1+\frac{1}{2}+2+\frac{2}{3}+3+\frac{3}{4}+...+50+\frac{50}{51}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{51}\)
\(=\left(1+2+3+...+50\right)+\left(\frac{1}{2}+\frac{1}{2}\right)+\left(\frac{2}{3}+\frac{1}{3}\right)+...+\left(\frac{50}{51}+\frac{1}{51}\right)\)
\(=\frac{50.51}{2}+1+1+1+...+1\) ( có 50 số 1 )
\(=1275+50\)
\(=1325\)