Cả 2 dãy mỗi dãy nhóm 2 số lại

a ) 2 - 3 + 4 - 5 + 6 - 7 + 8 - ... + 209 + 210

( 2 - 3 ) + ( 4 - 5 ) + ( 6 - 7 ) + ( 8 - 9 ) + ... + ( 207 - 208 ) + ( 209 + 210 )

= - 1 + ( -1 ) + ( - 1 ) + ( - 1 ) + ... + ( - 1 ) + 219

Bài này hơi kì kì sao ấy !

Đây là toán lớp 6 mà lại là tính hợp lí, thuộc vào chương số nguyên nên mấy bé lớp 5,4,3,2,1 đừng dại gì mà giải nhé 

A=2+3-4-5+6+7-8-9+...-100-101+102

=(2+3-4-5)+(6+7-8-9)+...+(98+99-100-101)+102

=(-4)+(-4)+...+(-4)+102

=-4*25+102

=102-100=2

\(-\dfrac{2}{5}.\dfrac{4}{7}+\dfrac{-3}{5}.\dfrac{2}{7}+\dfrac{-3}{5}\)

= \(\dfrac{-8}{35}+\dfrac{-6}{35}+\dfrac{-3}{5}\) 

=\(\dfrac{-8-6-21}{35}\) 

=\(\dfrac{-35}{35}\) 

=-1

`(-2)/5.4/7+(-3)/5.2/7+(-3)/5`

`=(-8)/3 +(-6)/35+(-3)/5`

`=(-8)-6-21/35`

`=(-35)/35`

`=(-1)`

`@Nae`

(-2)/5 x 4/7 + (-3)/5 x 2/7 + (-3)/5

`=(-8)/3 +(-6)/35+(-3)/5`

`=(-8)-6-21/35`

`=(-35)/35`

`=(-1)`

`@Nae`

\(a,-\frac{4}{7}+\left(-\frac{2}{3}+\frac{3}{7}\right)=-\frac{4}{7}-\frac{2}{3}+\frac{3}{7}=-\frac{4}{7}+\frac{3}{7}-\frac{2}{3}=-\frac{1}{7}-\frac{2}{3}=-\frac{17}{21}\)

\(b,-\frac{3}{10}\cdot\frac{5}{12}+-\frac{3}{10}\cdot\frac{7}{12}+5\cdot\frac{3}{10}=\frac{3}{10}\cdot-\frac{5}{12}+\frac{3}{10}\cdot-\frac{7}{12}+5\cdot\frac{3}{10}=\frac{3}{10}\cdot\left(-\frac{5}{12}-\frac{7}{12}+5\right)\)

\(=\frac{3}{10}\cdot4=\frac{6}{5}\)

ti ck nha

đúng đó

xin đó

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a, -4/7 + (-2/3 + 3/7 )

= -4/7 + (-2/3) + 3/7

= -4/7 + 3/7 + (-2/3)

= -1/7 + -14/21

= -3/21 + -14/21

= -17/21

b, -3/10 . 5/12 + -3/10 . 7/12 + 5. - 3/10

-3/10.(5/12 + 7/12 + 60/12)

-3/10.6

-18/10= -9/5

Đoàn Lê Thu Trang

Đặt x = 2k ( k thuộc N )

Số số hạng của dãy số đó là :

\(\frac{2k-2}{2}+1=k-1+1=k\) ( số hạng )

Do đó :

\(\frac{k\times\left(2k+2\right)}{2}=2450\)

\(k\times k+1=2450=49\times50\)

\(\Rightarrow k=49\)

Vậy x chỉ có thể bằng : 

49 x 2 = 98

b) 1+2-3-4+5+6-7-8+...-499-500+501+502

= (1+2-3-4)+(5+6-7-8)+...+(497+498-499-500)+501+502

= (-4)+(-4)+...+(-4)+501+502

= (-4.125)+501+502

= (-500)+501+502

= 503

\(\frac{-4}{7}+\left(\frac{-2}{3}+\frac{4}{7}\right)\)

\(=\frac{-4}{7}+\frac{-2}{3}+\frac{4}{7}\)

\(=\left(\frac{-4}{7}+\frac{4}{7}\right)+\frac{-2}{3}\)

\(=0+\frac{-2}{3}=\frac{-2}{3}\)

~ Hok tốt ~

\(\frac{-3}{10}.\frac{5}{12}+\frac{-3}{10}.\frac{7}{12}+5\frac{3}{10}\)

\(=\frac{-3}{10}.\frac{5}{12}+\frac{-3}{10}.\frac{7}{12}+\frac{53}{10}\)

\(=\frac{-3}{10}.\left(\frac{5}{12}+\frac{7}{12}\right)+\frac{53}{10}\)

\(=\frac{-3}{10}.1+\frac{53}{10}\)

\(=\frac{-3}{10}+\frac{53}{10}=\frac{50}{10}=5\)

~ Hok tốt ~

\(\dfrac{2}{7}-\dfrac{3}{11}+\dfrac{5}{7}-\dfrac{8}{11}\\ =\left(\dfrac{2}{7}+\dfrac{5}{7}\right)-\left(\dfrac{3}{11}+\dfrac{8}{11}\right)\\ =\dfrac{7}{7}-\dfrac{11}{11}\\ =1-1\\ =0\)

= (2/7-5/7)+(3/11-8/11)

= 1 + (-1)

= 0

2/7-3/11+5/7-8/11

=0

\(a.\dfrac{-3}{8}-\dfrac{13}{65}+\dfrac{3}{8}=\left(\dfrac{3}{8}-\dfrac{3}{8}\right)-\dfrac{13}{65}=-\dfrac{13}{65}\)

\(b.\left(\dfrac{-13}{7}-\dfrac{4}{9}\right)-\left(-\dfrac{10}{7}-\dfrac{4}{9}\right)=\dfrac{-13}{7}-\dfrac{4}{9}+\dfrac{10}{7}+\dfrac{4}{9}\\ =\left(\dfrac{-13}{7}+\dfrac{10}{7}\right)+\left(\dfrac{4}{9}-\dfrac{4}{9}\right)=-\dfrac{3}{7}\)

\(c.17\dfrac{1}{3}\cdot\left(\dfrac{-3}{7}\right)+3\dfrac{2}{3}\cdot\left(\dfrac{-3}{7}\right)=\dfrac{-3}{7}\cdot\left(17\dfrac{1}{3}+3\dfrac{2}{3}\right)\\ =\dfrac{-3}{7}\cdot\left(\dfrac{52}{3}+\dfrac{11}{3}\right)=\dfrac{-3}{7}\cdot21=-9\)

a)

\(\frac{5}{6}-\frac{57}{12}\)

=\(-\frac{47}{12}\)

b)

\(=\frac{3}{7}\left(\frac{9}{11}+\frac{2}{11}\right)\)

\(=\frac{3}{7}.1=\frac{3}{7}\)