\(\dfrac{7}{5}+\dfrac{4}{7}-\dfrac{9}{10}=\dfrac{49-20}{35}-\dfrac{9}{10}=\dfrac{19}{35}-\dfrac{9}{10}=\dfrac{190-315}{350}=\dfrac{-125}{350}\)

\(\dfrac{2}{1}+\dfrac{3}{4}\text{×}\dfrac{8}{5}=\dfrac{8+3}{4}\text{×}\dfrac{8}{5}=\dfrac{11\text{×}8}{4\text{×}5}=\dfrac{88}{20}\)

mấy câu kia áp dụng là dc!

\(a,=\frac{7-1}{1.3.7}+\frac{9-3}{3.7.9}+\frac{13-7}{7.9.13}+\frac{15-9}{9.13.15}\)\(+\frac{19-13}{13.15.19}\)

\(=\frac{1}{1.3}-\frac{1}{3.7}+\frac{1}{3.7}-\frac{1}{7.9}+\frac{1}{7.9}-\frac{1}{9.13}+\frac{1}{9.13}-\frac{1}{13.15}+\frac{1}{13.15}-\frac{1}{15.19}\)

\(=\frac{1}{1.3}-\frac{1}{15.19}=\frac{95}{285}-\frac{1}{285}=\frac{94}{285}\)

\(b,=\frac{1}{6}.\left(\frac{6}{1.3.7}+\frac{6}{3.7.9}+\frac{6}{7.9.13}+\frac{6}{9.13.15}+\frac{6}{13.15.19}\right)\)

làm giống như trên

\(c,=\frac{1}{8}.\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{48.49.50}\right)\)

\(=\frac{1}{16}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{48.49.50}\right)\)

\(=\frac{1}{16}.\left(\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+...+\frac{50-48}{48.49.50}\right)\)

\(=\frac{1}{16}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{48.49}-\frac{1}{49.50}\right)\)

\(=\frac{1}{16}.\left(\frac{1}{2}-\frac{1}{2450}\right)=\frac{1}{16}.\left(\frac{1225}{2450}-\frac{1}{2450}\right)=\frac{153}{4900}\)

\(d,=\frac{5}{7}.\left(\frac{7}{1.5.8}+\frac{7}{5.8.12}+\frac{7}{8.12.15}+...+\frac{7}{33.36.40}\right)\)

\(=\frac{5}{7}.\left(\frac{8-1}{1.5.8}+\frac{12-5}{5.8.12}+\frac{15-8}{8.12.15}+...+\frac{40-33}{33.36.40}\right)\)

\(=\frac{5}{7}.\left(\frac{1}{1.5}-\frac{1}{5.8}+\frac{1}{5.8}-\frac{1}{8.12}+\frac{1}{8.12}-\frac{1}{12.15}+...+\frac{1}{33.36}-\frac{1}{36.40}\right)\)

\(=\frac{5}{7}.\left(\frac{1}{5}-\frac{1}{1440}\right)=\frac{5}{7}.\left(\frac{288}{1440}-\frac{1}{1440}\right)=\frac{41}{288}\)

P/S: . là nhân nha

sxasxsxxsxsxssxsxsxsxsx232332321322

Cái này dễ thế mà ko bít lm

a) \(\left(\frac{4}{13}.\frac{6}{5}+\frac{4}{13}.\frac{2}{5}\right).\left(2x+1\right)^2=\frac{10}{13}\)

\(\left(\frac{4}{13}.\frac{8}{5}\right).\left(2x+1\right)^2=\frac{10}{13}\)

\(\frac{32}{65}.\left(2x+1\right)^2=\frac{10}{13}\)

\(\left(2x+1\right)^2=\frac{10}{13}\div\frac{32}{65}\)

\(\left(2x+1\right)^2=\frac{25}{16}\)

\(\Rightarrow2x+1\in\left\{\frac{5}{4};-\frac{5}{4}\right\}\)

\(\hept{\begin{cases}2x+1=\frac{5}{4}\\2x+1=-\frac{5}{4}\end{cases}\Rightarrow\hept{\begin{cases}2x=\frac{1}{4}\\2x=-\frac{9}{4}\end{cases}\Rightarrow}\hept{\begin{cases}x=\frac{1}{8}\\x=-\frac{9}{8}\end{cases}}}\)

Vậy \(x\in\left\{\frac{1}{8};-\frac{9}{8}\right\}\)

\(x^3-\frac{9}{16}.x=0\)

\(x\left(x^2-\frac{9}{16}\right)=0\)

\(\hept{\begin{cases}x=0\\x^2-\frac{9}{16}=0\end{cases}\Rightarrow\hept{\begin{cases}x=0\\x^2=\frac{9}{16}\end{cases}\Rightarrow}\hept{\begin{cases}x=0\\x=\pm\frac{3}{4}\end{cases}}}\)

Vậy \(x\in\left\{0;\frac{3}{4};-\frac{3}{4}\right\}\)

Là B bạn nha

Vì dãy số tự nhiên không giới hạn nha bạn

Câu B nha

TL: Câu B

không biết làm bài này hả

 = 0 nha em

bằng 0

nhìn th

a ) 13/20

B)

C..........................................................

minh dang tính

lấy máy tính mà bấm

trả lời kiểu gì vậy hả bạn lê thế trung

a) 1/7 . 5/6 + 1/7 . 1/6 + -8/7

= 1/7 . ( 5/6 + 1/6 ) + -8/7

= 1/7 . 1 + -8/7

= 1/7 + -8/7 = -1

b) 3/5 . -4/9 . 5/3 . 18/7

= ( 3/5 . 5/3 ) . ( -4/9 . 18/7 ) 

= 1 . -8/7 = -8/7

c) 1/2 + -3/4 . 16/9

= 1/2 + -4/3 = -5/6

d) ( 1/7 + 5/14 ) . -28/3

= 1/2 . -28/3 = -14/3