giúp mk vs mn

giúp mk vs

Đặt \(A=\frac{1}{4\cdot9}+\frac{1}{9\cdot14}+\cdots+\frac{1}{44\cdot49}\)

\(=\frac15\left(\frac{5}{4\cdot9}+\frac{5}{9\cdot14}+\cdots+\frac{5}{44\cdot49}\right)\)

\(=\frac15\left(\frac14-\frac19+\frac19-\frac{1}{14}+\frac{1}{14}-\frac{1}{19}+\cdots+\frac{1}{44}-\frac{1}{49}\right)\)

\(=\frac15\left(\frac14-\frac{1}{49}\right)=\frac15\cdot\frac{45}{196}=\frac{9}{196}\)

Đặt B=3+5+7+...+49

Số số hạng của dãy số là: \(\frac{49-3}{2}+1=\frac{46}{2}+1=24\) (số)

Tổng của dãy số là \(\left(49+3\right)\cdot\frac{24}{2}=52\cdot\frac{24}{2}=26\cdot24=624\)

\(M=\left(\frac{1}{4\cdot9}+\frac{1}{9\cdot14}+\cdots+\frac{1}{44\cdot49}\right)\cdot\frac{1-3-5-\cdots-49}{89}\)

\(=\frac{9}{196}\cdot\frac{1-624}{89}=\frac{9}{196}\cdot\frac{-623}{89}=\frac{9}{196}\cdot\left(-7\right)=-\frac{9}{28}\)

\(S=\left(\frac{1}{4\cdot9}+\frac{1}{9\cdot14}+\frac{1}{14\cdot19}+...+\frac{1}{44\cdot49}\right)\cdot\frac{1-3-5-7-...-49}{89}\\ S=\frac{1}{5}\left(\frac{5}{4\cdot9}+\frac{5}{9\cdot14}+\frac{5}{14\cdot19}+...+\frac{5}{44\cdot49}\right)\cdot\frac{1-\left(3+5+7+...+49\right)}{89}\\ S=\frac{1}{5}\left(\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+\frac{1}{14}-\frac{1}{19}+....+\frac{1}{44}-\frac{1}{49}\right)\cdot\frac{1-\frac{\left(49+3\right)\cdot24}{2}}{89}\\ S=\frac{1}{5}\left(\frac{1}{4}-\frac{1}{49}\right)\cdot\left(-7\right)\\ S=\frac{1}{5}\cdot\frac{45}{196}\cdot\left(-7\right)\\ S=\frac{-9}{28}\)

ta có

1/5(5/36+5/126+...+5/44*49)1-3-5-7-9-...-49/89

=1/5(1/4-1/9+1/9-1/14+...+1/44-1/49)-623/89

=1/5*-7(1/4-1/49)

=-7/5*45/196

=-9/128

bạn ơi 9/28 chứ không phải 9/128 đâu

Đặt A = \(\frac{1}{4.9}+\frac{1}{9.14}+\frac{1}{14.19}+...+\frac{1}{44.49}\)

\(\Rightarrow\) A = \(\frac{1}{5}\left(\frac{5}{4.9}+\frac{5}{9.14}+\frac{5}{14.19}+...+\frac{1}{44.49}\right)\)

\(\Rightarrow\) A = \(\frac{1}{5}.\left(\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+\frac{1}{14}-\frac{1}{19}+...+\frac{1}{44}-\frac{1}{49}\right)\)

\(\Rightarrow\) A = \(\frac{1}{5}\left(\frac{1}{4}-\frac{1}{49}\right)\)

\(\Rightarrow\)A = \(\frac{1}{5}.\frac{45}{196}=\frac{9}{196}\)

Đặt B = \(\frac{1-3-5-7-9-...-49}{89}\)

\(\Rightarrow\)B = \(\frac{1-\left(3+5+7+9+...+49\right)}{89}\)

\(\Rightarrow\)B = \(\frac{1-624}{89}=-7\)

Vậy M =\(\frac{9}{196}.-7=-\frac{9}{28}\)

cho mình hỏi vì sao

1-3-5-7-...-49= 1-(3+5+7+9)

1/Tính

\(\left(\frac{3}{7}\right)^{20}:\left(\frac{9}{49}\right)^5\)

\(=\left(\frac{3}{7}\right)^{20}:\left(\frac{3^2}{7^2}\right)^5\)

\(=\left(\frac{3}{7}\right)^{20}:\left(\frac{3}{7}\right)^{10}\)

\(=\left(\frac{3}{7}\right)^{10}\)

2/ Ta có:A+B+C = 180 độ ( tổng 3 góc tam giác)

Và : \(A.\frac{1}{2}=B.\frac{1}{3}=C.\frac{2}{5}\)

hay \(\frac{A}{\frac{2}{1}}=\frac{B}{\frac{3}{1}}=\frac{C}{\frac{5}{2}}\)

Áp dụng tính chất dãy tỉ số bằng nhau:

\(\frac{A}{\frac{2}{1}}=\frac{B}{\frac{3}{1}}=\frac{C}{\frac{5}{2}}=\frac{A+B+C}{\frac{2}{1}+\frac{3}{1}+\frac{5}{2}}=\frac{180}{\frac{15}{2}}=24\)

=> \(A=24.\frac{2}{1}=48\)độ

     \(B=24.\frac{3}{1}=72\)độ

      \(C=24.\frac{5}{2}=60\)độ