Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\dfrac{7}{5}+\dfrac{4}{7}-\dfrac{9}{10}=\dfrac{49-20}{35}-\dfrac{9}{10}=\dfrac{19}{35}-\dfrac{9}{10}=\dfrac{190-315}{350}=\dfrac{-125}{350}\)
\(\dfrac{2}{1}+\dfrac{3}{4}\text{×}\dfrac{8}{5}=\dfrac{8+3}{4}\text{×}\dfrac{8}{5}=\dfrac{11\text{×}8}{4\text{×}5}=\dfrac{88}{20}\)
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
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 \(x\in\left\{0;\frac{3}{4};-\frac{3}{4}\right\}\)
a ) 13/20
B)
C..........................................................
minh dang tính
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
a,4/5x2/9+1/5x2/9
=2/9*(4/5+1/5)
=2/9*1
=2/9
b)7/8:1/5-3/8:1/5
=(7/8-3/8):1/5
=1/2:1/5
=5/2
c)6/11x8/9+5/11x8/9
=(6/11+5/11)*8/9
=1*8/9
=8/9
d)11/12:2/3-3/12:2/3
=(11/12-3/12):2/3
=2/3:2/3
=1
MK sẽ giúp bn
a)\(\frac{4}{5}x\frac{2}{9}+\frac{1}{5}x\frac{2}{9}\)
\(=\left(\frac{4}{5}+\frac{1}{5}\right)x\frac{2}{9}\)
\(=1x\frac{2}{9}\)
\(=\frac{2}{9}\)
b)\(\frac{7}{8}:\frac{1}{5}-\frac{3}{8}:\frac{1}{5}\)
\(=\left(\frac{7}{8}-\frac{3}{8}\right):\frac{1}{5}\)
\(=\frac{5}{8}:\frac{1}{5}\)
\(=\frac{25}{8}\)
c)\(\frac{6}{11}x\frac{8}{9}+\frac{5}{11}x\frac{8}{9}\)
\(=\left(\frac{6}{11}+\frac{5}{11}\right)x\frac{8}{9}\)
\(=\frac{8}{9}x1\)
\(=\frac{8}{9}\)
d)\(\frac{11}{12}:\frac{2}{3}-\frac{3}{12}:\frac{2}{3}\)
\(=\left(\frac{11}{12}-\frac{3}{12}\right):\frac{2}{3}\)
\(=\frac{2}{3}:\frac{2}{3}\)
\(=1\)