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A = \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(=1-\frac{1}{50}=\frac{49}{50}\)
B = \(\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}.\frac{5^2}{4.6}=\frac{\left(2.3.4.5\right).\left(2.3.4.5\right)}{\left(1.2.3.4\right).\left(3.4.5.6\right)}=\frac{5.2}{1.6}=\frac{5}{3}\)
C = \(\frac{3}{5.7}+\frac{3}{7.9}+...+\frac{3}{59.61}=\frac{3}{2}.\left(\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{59.61}\right)\)
\(=\frac{3}{2}.\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{59}-\frac{1}{61}\right)=\frac{3}{2}.\left(\frac{1}{5}-\frac{1}{61}\right)=\frac{3}{2}.\frac{56}{305}=\frac{74}{305}\)
Bài làm:
1) \(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
\(A=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{50-49}{49.50}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(A=1-\frac{1}{50}=\frac{49}{50}\)
2) \(B=\frac{2^2.3^2.4^2.5^2}{1.2.3^2.4^2.5.6}=\frac{2.5}{6}=\frac{5}{3}\)
3) \(C=\frac{3}{5.7}+\frac{3}{7.9}+...+\frac{3}{59.61}\)
\(C=\frac{3}{2}\left(\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{59.61}\right)\)
\(C=\frac{3}{2}\left(\frac{7-5}{5.7}+\frac{9-7}{7.9}+...+\frac{61-59}{59.61}\right)\)
\(C=\frac{3}{2}\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{59}-\frac{1}{61}\right)\)
\(C=\frac{3}{2}\left(\frac{1}{5}-\frac{1}{61}\right)\)
\(C=\frac{3}{2}.\frac{56}{305}=\frac{84}{305}\)
\(=-2.\frac{2}{3}.\frac{1}{3}:\left(\frac{-1}{6}+0,5\right)-\left(-2009^0\right)-\left(-2\right)^2\)
\(=\frac{4}{3}.\frac{1}{3}:\left(\frac{-1}{6}+\frac{1}{2}\right)-1.4\)
\(=\frac{4}{3}.\frac{1}{3}+4\)
\(=4+4\)
\(=8\)
1) \(\frac{2}{3}+x=-\frac{4}{5}\)
\(x=\left(-\frac{4}{5}\right)-\frac{2}{3}\)
\(x=-1\frac{7}{15}\)
Vậy \(x=-1\frac{7}{15}\)
2) \(\frac{2}{5}-x=-\frac{1}{3}\)
\(x=\frac{2}{5}-\left(-\frac{1}{3}\right)\)
\(x=\frac{11}{15}\)
Vậy \(x=\frac{11}{15}\)
3) \(1-\frac{x}{3}=1\frac{1}{2}\)
\(\frac{x}{3}=1-1\frac{1}{2}\)
\(\frac{x}{3}=-\frac{1}{2}\)
\(\Rightarrow x=\frac{\left(-1\right)\cdot3}{2}\)
\(x=-1\frac{1}{2}\)
4) \(1-\left(\frac{2x}{3}+2\right)=-1\)
\(\frac{2x}{3}+2=1-\left(-1\right)\)
\(\frac{2x}{3}+2=2\)
\(\frac{2x}{3}=2-2\)
\(\frac{2x}{3}=0\)
\(\Rightarrow x=0\)
Vậy \(x=0\)
Mik lười quá bạn tham khảo câu 3 tại đây nhé:
Câu hỏi của nguyen linh nhi - Toán lớp 6 - Học toán với OnlineMath
\(S=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+...+\frac{1}{37\cdot38\cdot39}\)
\(2S=\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+...+\frac{1}{37\cdot38}-\frac{1}{38\cdot39}\)
\(2S=\frac{1}{2}-\frac{1}{38\cdot39}\)
\(S=\frac{1}{4}-\frac{1}{2\cdot38\cdot39}< \frac{1}{4}\)
1. \(\frac{25}{100}x+x-\frac{1}{5}x=\frac{1}{5}\)
\(\Leftrightarrow\frac{1}{4}x+x-\frac{1}{5}x=\frac{1}{5}\)
\(\Leftrightarrow\left(\frac{1}{4}+1-\frac{1}{5}\right)x=\frac{1}{5}\)
\(\Leftrightarrow\frac{21}{20}x=\frac{1}{5}\)
\(\Leftrightarrow x=\frac{1}{5}:\frac{21}{20}\)
\(\Leftrightarrow x=\frac{4}{21}\)
\(1)\) \(\frac{\frac{3}{41}-\frac{12}{47}+\frac{27}{53}}{\frac{4}{41}-\frac{16}{47}+\frac{36}{53}}=\frac{3\left(\frac{1}{41}-\frac{4}{47}+\frac{9}{53}\right)}{4\left(\frac{1}{41}-\frac{4}{47}+\frac{9}{53}\right)}=\frac{3}{4}\)
\(2)\) Đặt \(A=4+2^2+2^4+...+2^{20}\)
\(4A=2^4+2^4+2^6+...+2^{22}\)
\(4A-A=\left(2^4+2^4+2^6+...+2^{22}\right)+\left(2^2+2^2+2^4+...+2^{20}\right)\)
\(3A=2^4+2^{22}-2^2-2^2\)
\(3A=2^{22}+2^4-2^3\)
\(A=\frac{2^{22}+2^4-2^3}{3}\)
\(3)\) \(\frac{5^2}{1.6}+\frac{5^2}{6.11}+\frac{5^2}{11.16}+...+\frac{5^2}{26.31}\) ( bạn ghi đầy đủ ra nhé ở đây mk viết "..." cho nhanh )
\(=\)\(5\left(\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}+...+\frac{5}{26.31}\right)\)
\(=\)\(5\left(\frac{1}{1}-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+...+\frac{1}{26}-\frac{1}{31}\right)\)
\(=\)\(5\left(1-\frac{1}{31}\right)\)
\(=\)\(5.\frac{30}{31}\)
\(=\)\(\frac{150}{31}\)
Chúc bạn học tốt ~
Ta có:
\(\frac{3\left(\frac{1}{41}-\frac{4}{47}+\frac{9}{53}\right)}{4\left(\frac{1}{41}-\frac{4}{47}+\frac{9}{53}\right)}=\frac{3}{4}\)
\(1,\frac{3737.43-4343.37}{1^2+2^3+...+27^2}=\frac{101.43.37-101.43.37}{..........}=0\)