Số số hạng là

( 19 - 1 ) : 2 + 1 = 10 ( số )

Tổng dãy số là

( 19 + 1 ) x 10 : 2 = 100

X là

( 724 - 100 ) : 3 = 208

Đáp số : 208

Ủng hộ nhé ! 

X=208

dó là 208

Tính dãy số 1 + 3 + 5 + .. + 19 trước

Số số hạng là: ( 19 - 1 ) : 2 + 1 = 10 (số)

Tổng là: ( 19 + 1 ) x 10 : 2 = 100

Ráp lại có: 100 + X x 3 = 724

                           X x 3  = 724 - 100

                           X x 3  = 624

                          X        = 624 : 3

                          X        = 208

Khoảng cách của số liền sau và số liền trước của dãy số 1+3+5+7+9+....+19 là:

                    3-1=5-3=7-5=9-7=2

Số số hạng của dãy số 1+3+5+7+9+....+19 là:

                    (19-1):2+1=10 (số số hạng)

Tổng của dãy số trên là:

                     (19+1)x10:2=100

Chuyển thành bài toán tìm x,ta có:

                      100+xX3=724

                       xX3=724-100

                        xX3=624

                         x=624:3

                        x=208

                                 Đ/s+208

(1+19)+(3+17)+(5+15)+(7+13)+(9+11)x3=724

20+20+20+20+20x3=724

20X5+xX3=724

100+xX3=724

xX3=724-100

xX3=624

x=624:3

x=208

1+3+5+7+9+...+19+X*3=724

100+X*3=724

       X*3=724-100

       X*3=624

          X=624:3

          X=208

TK mk nha Duy mk dg am diem 

1+3+5+7+9+...+19 +X x 3

=(1+19)+(3+17)+...+(9+11) + X x3

 = 20+20+...+20+ X x 3

=20x5+ X x3=100+X x3=724

                               X x 3= 724-100=624

                              X= 624/3=208

1 + 3 + 5 + 7 + 9 + ....+ 19 + X x 3 = 724

ta có số số hạng vế trái

( 19 - 1 ) : 2 + 1 = 10 (số hạng )

tổng vế trái là

( 19 + 1 ) x 10 : 2 = 100

khi đó ta có : 100 + X x 3 = 724

                               X x 3 = 724 - 100

                               X x 3 = 624

                               X       = 624 : 3

                               X       = 208

vậy X = 208

x= 208 nhé hihi

k cho mình nha

Chọn A

1) Ta có: \(\frac{-4}{7}-\frac{11}{19}+\frac{13}{19}\cdot\frac{-3}{7}+\frac{2}{19}:\frac{-7}{4}\)

\(=\frac{-4}{7}-\frac{11}{19}-\frac{39}{133}-\frac{8}{133}\)

\(=\frac{-76}{133}-\frac{77}{133}-\frac{39}{133}-\frac{8}{133}\)

\(=\frac{-200}{133}\)

2) Ta có: \(\left(\frac{-4}{9}+\frac{3}{5}\right):\frac{1}{\frac{1}{5}}+\left(\frac{1}{5}-\frac{5}{9}\right):\frac{1}{\frac{1}{5}}\)

\(=\left(\frac{-4}{9}+\frac{3}{5}\right)\cdot\frac{1}{5}+\left(\frac{1}{5}-\frac{5}{9}\right)\cdot\frac{1}{5}\)

\(=\frac{1}{5}\left(\frac{-4}{9}+\frac{3}{5}+\frac{1}{5}-\frac{5}{9}\right)\)

\(=\frac{1}{5}\left(-1+\frac{4}{5}\right)\)

\(=\frac{1}{5}\cdot\frac{-1}{5}=\frac{-1}{25}\)

3) Ta có: \(\frac{4}{5}-\left(-\frac{2}{7}\right)-\frac{7}{10}\)

\(=\frac{4}{5}+\frac{2}{7}-\frac{7}{10}\)

\(=\frac{56}{70}+\frac{20}{70}-\frac{49}{70}\)

\(=\frac{27}{70}\)

4) Ta có: \(\frac{2}{7}-\left(-\frac{13}{15}+\frac{4}{9}\right)-\left(\frac{5}{9}-\frac{2}{15}\right)\)

\(=\frac{2}{7}+\frac{13}{15}-\frac{4}{9}-\frac{5}{9}+\frac{2}{15}\)

\(=\frac{2}{7}+1-1=\frac{2}{7}\)

a: \(\left(-\dfrac{5}{6}+\dfrac{2}{5}\right):\dfrac{3}{8}+\left(\dfrac{4}{5}-\dfrac{11}{30}\right):\dfrac{3}{8}\)

\(=\left(-\dfrac{5}{6}+\dfrac{2}{5}\right)\cdot\dfrac{8}{3}+\left(\dfrac{4}{5}-\dfrac{11}{30}\right)\cdot\dfrac{8}{3}\)

\(=\dfrac{8}{3}\left(-\dfrac{5}{6}+\dfrac{2}{5}+\dfrac{4}{5}-\dfrac{11}{30}\right)\)

\(=\dfrac{8}{3}\cdot\dfrac{-25+36-11}{30}\)

=0

b: \(\left(-\dfrac{3}{4}+\dfrac{2}{5}\right):\dfrac{3}{7}+\left(\dfrac{3}{5}+\dfrac{-1}{4}\right):\dfrac{3}{7}\)

\(=\left(-\dfrac{3}{4}+\dfrac{2}{5}\right)\cdot\dfrac{7}{3}+\left(\dfrac{3}{5}-\dfrac{1}{4}\right)\cdot\dfrac{7}{3}\)

\(=\dfrac{7}{3}\left(-\dfrac{3}{4}+\dfrac{2}{5}+\dfrac{3}{5}-\dfrac{1}{4}\right)\)

\(=\dfrac{7}{3}\cdot0=0\)

c: \(\dfrac{-13}{18}\cdot\dfrac{5}{8}+\dfrac{-5}{18}\cdot\dfrac{2}{9}+\dfrac{-13}{18}\cdot\dfrac{3}{8}+\dfrac{-5}{18}\cdot\dfrac{7}{9}\)

\(=\left(-\dfrac{13}{18}\cdot\dfrac{5}{8}+\dfrac{-13}{18}\cdot\dfrac{3}{8}\right)+\left(-\dfrac{5}{18}\cdot\dfrac{2}{9}+\dfrac{-5}{18}\cdot\dfrac{7}{9}\right)\)

\(=-\dfrac{13}{18}\left(\dfrac{5}{8}+\dfrac{3}{8}\right)+\dfrac{-5}{18}\left(\dfrac{2}{9}+\dfrac{7}{9}\right)\)

\(=-\dfrac{13}{18}-\dfrac{5}{18}=-\dfrac{18}{18}=-1\)

d: Sửa đề: \(\dfrac{-11}{19}\cdot\dfrac{4}{9}+\dfrac{-8}{19}\cdot\dfrac{3}{7}+\dfrac{-11}{19}\cdot\dfrac{5}{9}+\dfrac{-8}{19}\cdot\dfrac{4}{7}\)

\(=\left(-\dfrac{11}{19}\cdot\dfrac{4}{9}+\dfrac{-11}{19}\cdot\dfrac{5}{9}\right)+\left(\dfrac{-8}{19}\cdot\dfrac{3}{7}+\dfrac{-8}{19}\cdot\dfrac{4}{7}\right)\)

\(=-\dfrac{11}{19}\left(\dfrac{4}{9}+\dfrac{5}{9}\right)+\dfrac{-8}{19}\left(\dfrac{3}{7}+\dfrac{4}{7}\right)\)

\(=-\dfrac{11}{19}-\dfrac{8}{19}=-\dfrac{19}{19}=-1\)

\(a.\left(-\dfrac{5}{6}+\dfrac{2}{5}\right):\dfrac{3}{8}+\left(\dfrac{4}{5}-\dfrac{11}{30}\right):\dfrac{3}{8}\)

\(=\left(-\dfrac{13}{30}\right):\dfrac{3}{8}+\dfrac{13}{30}:\dfrac{3}{8}\)

\(=\left[\left(-\dfrac{13}{30}+\dfrac{13}{30}\right)\right]:\dfrac{3}{8}\)

\(=0:\dfrac{3}{8}=0\)

\(b.\left(-\dfrac{3}{4}+\dfrac{2}{5}\right):\dfrac{3}{7}+\left(\dfrac{3}{5}+-\dfrac{1}{4}\right):\dfrac{3}{7}\)

\(=\left(-\dfrac{7}{20}\right):\dfrac{3}{7}+\dfrac{7}{20}:\dfrac{3}{7}\)

\(=\left[\left(-\dfrac{7}{20}+\dfrac{7}{20}\right)\right]:\dfrac{3}{7}=0:\dfrac{3}{7}=0\)

\(c.-\dfrac{13}{18}.\dfrac{5}{8}+-\dfrac{5}{18}.\dfrac{2}{9}+-\dfrac{13}{18}.\dfrac{3}{8}+-\dfrac{5}{18}.\dfrac{7}{9}\)

\(=\left(\dfrac{5}{8}+\dfrac{3}{8}\right).-\dfrac{13}{18}+\left(\dfrac{2}{9}+\dfrac{7}{9}\right).-\dfrac{5}{18}\)

\(=1.-\dfrac{13}{18}+1.-\dfrac{5}{18}=-\dfrac{13}{18}+-\dfrac{5}{18}=-1\)

\(d.-\dfrac{11}{19}.\dfrac{4}{9}+\dfrac{-8}{19}.\dfrac{3}{7}+-\dfrac{11}{19}.\dfrac{5}{9}+-\dfrac{9}{19}.\dfrac{4}{7}\)

\(=\left(\dfrac{4}{9}+\dfrac{5}{9}\right).-\dfrac{11}{19}+-\dfrac{24}{133}+-\dfrac{36}{133}\)

\(=-\dfrac{11}{19}+-\dfrac{60}{133}=-\dfrac{137}{133}\)

$x\times\dfrac{3}{7}=1-\dfrac{5}{8}$

$=>x\times\dfrac{3}{7}=\dfrac{3}{8}$

$=>x=\dfrac{3}{8}:\dfrac{3}{7}$

$=>x=\dfrac{7}{8}$