(\(\frac{1}{4.9}+\frac{1}{9.14}+...+\frac{1}{44.49}\)).\(\frac{1-3-5-...-49}{89}\)

= \(\frac{1}{5}.\left(\frac{5}{4.9}+\frac{5}{9.14}+...+\frac{5}{45.49}\right).\frac{1-3-5-...-49}{89}\)

\(=\frac{1}{5}.\left(\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+...+\frac{1}{44}-\frac{1}{49}\right).\frac{1-\frac{24.\left(49+3\right)}{2}}{89}\)

\(=\frac{1}{5}.\left(\frac{1}{4}-\frac{1}{49}\right).\left(-7\right)\)

\(=-\frac{9}{28}\)

Có chỗ ghi nhầm 44 thành 45. Tự sửa nhé

Bài 2/ a/

|2x + 3| = x + 2

Điều kiện \(x\ge-2\)

Với x < - 1,5 thì ta có

- 2x - 3 = x + 2

<=> 3x = - 5

<=> \(x=-\frac{5}{3}\)

Với \(x\ge-1,5\)thì ta có

2x + 3 = x + 2

<=> x = - 1

b/ A = |x - 2006| + |2007 - x| \(\ge\)|x - 2006 + 2007 - x| = 1

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

- #Ngân giải luôn nà. ^^ không gi lại đề nhá
\(=\left(\frac{5}{4\cdot9}+\frac{5}{9\cdot14}+\frac{5}{14\cdot19}+...+\frac{5}{44\cdot49}\right)\cdot\frac{1-\left(3+5+7+...+49\right)}{89}\)
\(=\frac{1}{5}\cdot\left(\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+\frac{1}{14}-\frac{1}{19}+...+\frac{1}{44}-\frac{1}{49}\right)\cdot\left(\frac{1-\frac{\left(52\cdot24\right)}{2}}{89}\right)\)
\(=\frac{1}{5}\cdot\left(\frac{1}{4}-\frac{1}{49}\right)\cdot\frac{1-624}{89}\)
\(=\frac{1}{5}\cdot\frac{45}{196}\cdot-\frac{623}{89}\)
\(=-\frac{9}{28}\)
#ĐÚng thì cho ngân ****+điểm

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

\(\left(\frac{1}{4.9}+\frac{1}{9.14}+\frac{1}{14.19}+....+\frac{1}{44.49}\right)\cdot\frac{1-3-5-7-....-49}{89}\)

  \(\text{Đặt }:\left(\frac{1}{4.9}+\frac{1}{9.14}+\frac{1}{14.19}+...+\frac{1}{44.49}\right)\)là \(A\)

            \(\frac{1-3-5-7-...-49}{89}\)là \(B\);ta có : 

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

\(A=\frac{1}{5}\cdot\left(\frac{1}{4}-\frac{1}{49}\right)=\frac{1}{5}\cdot\frac{45}{196}=\frac{9}{196}\)

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

Tổng của \(3+5+7+...+49\)là: 

\(\frac{\left(3+49\right).24}{2}=624\)

\(\Rightarrow\frac{1-624}{89}=\frac{-623}{89}=-7\)

\(\Rightarrow\left(\frac{1}{4.9}+\frac{1}{9.14}+...+\frac{1}{44.49}\right)\cdot\frac{1-3-5-7-...-49}{89}=A.B=\frac{9}{196}\cdot-7=-\frac{9}{28}\)

mk ko viết lại đề đâu

=\(\frac{1}{5}\left(\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+...+\frac{1}{44}-\frac{1}{49}\right)\)\(.\frac{1-\left(3+5+...+49\right)}{89}\)

=\(\frac{1}{5}\left(\frac{1}{4}-\frac{1}{49}\right).\frac{\left(1-\frac{\left(49+3\right).24}{2}\right)}{89}\)

=\(\frac{1}{5}.\frac{45}{196}.\frac{1-\left(\frac{52.24}{2}\right)}{89}\)

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

=\(\frac{-9}{28}\)

giúp mk vs mn

giúp mk vs

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

mk ko viết lại đề đâu

=\(\frac{1}{5}\left(\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+...+\frac{1}{44}-\frac{1}{49}\right)\)\(.\frac{1-\left(3+5+...+49\right)}{89}\)

=\(\frac{1}{5}\left(\frac{1}{4}-\frac{1}{49}\right).\frac{\left(1-\frac{\left(49+3\right).24}{2}\right)}{89}\)

=\(\frac{1}{5}.\frac{45}{196}.\frac{1-\left(\frac{52.24}{2}\right)}{89}\)

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

=\(\frac{-9}{28}\)

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