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Câu 1:
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x.\left(x+1\right):2}=\frac{1991}{1993}.\)
\(\frac{1}{2.3:2}+\frac{1}{3.4:2}+\frac{1}{4.5:2}+...+\frac{1}{x.\left(x+1\right):2}=\frac{1991}{1993}\)
\(\frac{1}{2.3}.2+\frac{1}{3.4}.2+\frac{1}{4.5}.2+...+\frac{1}{x.\left(x+1\right)}.2=\frac{1991}{1993}\)
\(2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x.\left(x+1\right)}\right)=\frac{1991}{1993}\)
\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{1991}{3986}\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{1991}{3986}\)
...
e tự tính nốt nha
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\left(x+1\right)\div2}=\frac{1991}{1993}\)
\(\Leftrightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{1991}{1993}\)
\(\Leftrightarrow\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{1991}{1993}\div2\)
\(\Leftrightarrow\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{1991}{3986}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{1991}{3986}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{1991}{3986}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{1991}{3986}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{1993}\)
\(\Leftrightarrow x+1=1993\)
\(\Leftrightarrow x=1993-1\)
\(\Leftrightarrow x=1992\)
Vậy x = 1992
\(\left(\frac{1}{3^2}-1\right).\left(\frac{1}{4^2}-1\right).\left(\frac{1}{5^2}-1\right)...\left(\frac{1}{50^2}-1\right)\)
\(=\frac{-8}{3^2}.\frac{-15}{4^2}.\frac{-24}{25}...\frac{-2499}{50^2}\)
\(=\frac{8}{3^2}.\frac{15}{4^2}.\frac{24}{5^2}...\frac{2499}{50^2}\) (vì có 48 thừa số âm nên kết quả là dương)
\(=\frac{2.4}{3.3}.\frac{3.5}{4.4}.\frac{4.6}{5.5}...\frac{49.51}{50.50}\)
\(=\frac{2.3.4...49}{3.4.5...50}.\frac{4.5.6...51}{3.4.5...50}\)
\(=\frac{2}{50}.\frac{51}{3}\)
\(=\frac{1}{25}.17=\frac{17}{25}\)
\(a;\dfrac{2}{3}x-50\%x-\left(-\dfrac{4}{5}\right):1\dfrac{3}{5}=-0,12+1\dfrac{3}{25}\\ \dfrac{1}{6}x+\dfrac{1}{2}=1\Rightarrow\dfrac{1}{6}x=1-\dfrac{1}{2}=\dfrac{1}{2}\\ \Rightarrow x=\dfrac{1}{2}:\dfrac{1}{6}=3\\ b;\left(-1\dfrac{1}{6}+\dfrac{2}{3}-\dfrac{3}{4}\right):x+\left(-1\dfrac{11}{12}\right)\cdot1\dfrac{21}{23}=-6\dfrac{1}{3}\\ -\dfrac{5}{4}:x-\dfrac{11}{3}=-\dfrac{19}{3}\\ -\dfrac{5}{4}:x=-\dfrac{19}{3}+\dfrac{11}{3}=-\dfrac{8}{3}\\ x=-\dfrac{5}{4}:\left(-\dfrac{8}{3}\right)=\dfrac{15}{32}\\ c;50\%x-\dfrac{1}{3}x-\left(-\dfrac{2}{3}\right)^2\cdot\left(-1\dfrac{1}{8}\right)=-119\dfrac{3}{4}+120\dfrac{5}{6}\\ \dfrac{1}{6}x+\dfrac{1}{2}=\dfrac{13}{12}\Rightarrow\dfrac{1}{6}x=\dfrac{13}{12}-\dfrac{1}{2}=\dfrac{7}{12}\\ x=\dfrac{7}{12}:\dfrac{1}{6}=\dfrac{7}{2}\)
1.\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...........\frac{49}{50}=\frac{1}{50}\)
\(P=\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right)...\left(1-\frac{1}{50^2}\right)\)
\(\Rightarrow P=\left(\frac{4}{4}-\frac{1}{4}\right)\left(\frac{9}{9}-\frac{1}{9}\right)\left(\frac{16}{16}-\frac{1}{16}\right)...\left(\frac{2500}{2500}-\frac{1}{2500}\right)\)
\(\Rightarrow P=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}...\frac{2499}{2500}\)
\(\Rightarrow P=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}...\frac{49.51}{50.50}\)
\(\Rightarrow P=\frac{\left(1.2.3...49\right)\left(3.4.5...51\right)}{\left(2.3.4...50\right)\left(2.3.4...50\right)}\)
\(\Rightarrow P=\frac{51}{50.2}=\frac{51}{100}>\frac{50}{100}=\frac{1}{2}\)
Vậy \(P>\frac{1}{2}\)
Ta có:
\(P=\left(1-\frac{1}{2^2}\right).\left(1-\frac{1}{3^2}\right).\left(1-\frac{1}{4^2}\right).....\left(1-\frac{1}{50^2}\right)\)
\(P=\left(1-\frac{1}{4}\right).\left(1-\frac{1}{9}\right).\left(1-\frac{1}{16}\right).....\left(1-\frac{1}{2500}\right)\)
\(P=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}.....\frac{2499}{2500}\)
\(P=\frac{3.8.15.....2499}{4.9.16.....2500}\)
Tới chỗ này rồi tiếp tục rút gọn
Kết quả cuối cùng là: \(P>\frac{1}{2}\)
Xin lỗi nha, tớ ko có giỏi ở phần rút gọn.