(2-\(\dfrac{3}{4}\))\(^3\)/\(\dfrac{11}{16}\)
Điền dấu thích hợp (>,<,=) vào chỗ trống.
a,\(\dfrac{-6}{11}+\dfrac{5}{-11}....--1\) b.\(\dfrac{-5}{16}+\dfrac{-3}{16}...\dfrac{-1}{3}\)
c,\(\dfrac{2}{5}...\dfrac{3}{4}+\dfrac{-1}{6}\) d,\(\dfrac{5}{6}+\dfrac{-2}{3}...\dfrac{1}{12}+\dfrac{-4}{5}\)
Giúp mk nha :>>. Mk cảm ơn.
a. \(\dfrac{-6}{11}+\dfrac{5}{-11}< --1\)
b. \(\dfrac{-5}{16}+\dfrac{-3}{16}>-\dfrac{1}{3}\)
c. \(\dfrac{2}{5}>\dfrac{3}{4}+-\dfrac{1}{6}\)
d. \(\dfrac{5}{6}+\dfrac{-2}{3}>\dfrac{1}{12}+\dfrac{-4}{5}\)
1, A= \(\dfrac{-3}{4}.\left(0,125-1\dfrac{1}{2}\right):\dfrac{33}{16}-25\%\)
2, B= \(1\dfrac{13}{15}.0,75-\left(\dfrac{11}{20}+25\%\right):\dfrac{7}{3}\)
3, C= \(\dfrac{5}{16}:0,125-\left(2\dfrac{1}{4}-0,6\right).\dfrac{10}{11}\)
4, D= \(6\dfrac{5}{12}:2\dfrac{5}{4}+11\dfrac{1}{4}.\left(\dfrac{1}{3}-\dfrac{1}{5}\right)\)
\(1,A=-\dfrac{3}{4}.\left(0,125-1\dfrac{1}{2}\right):\dfrac{33}{16}-25\%\)
\(A=-\dfrac{3}{4}.\left(0,125-\dfrac{3}{2}\right):\dfrac{33}{16}-\dfrac{1}{4}\)
\(A=-\dfrac{3}{4}.\left(-\dfrac{11}{8}\right):\dfrac{33}{16}-\dfrac{1}{4}\)
\(A=\dfrac{33}{32}:\dfrac{33}{16}-\dfrac{1}{4}\)
\(A=\dfrac{33}{32}.\dfrac{16}{33}-\dfrac{1}{4}\)
\(A=\dfrac{1}{2}-\dfrac{1}{4}\)
\(A=\dfrac{2}{4}-\dfrac{1}{4}\)
\(A=\dfrac{1}{4}\)
\(D=6\dfrac{5}{12}:2\dfrac{5}{4}+11\dfrac{1}{4}.\left(\dfrac{1}{3}-\dfrac{1}{5}\right)\)
\(D=\dfrac{77}{12}:\dfrac{13}{4}+\dfrac{45}{4}.\dfrac{2}{15}\)
\(D=\dfrac{77}{39}+\dfrac{3}{2}\)
\(D=\dfrac{271}{78}\)
\(C=\dfrac{5}{16}:0,125-\left(2\dfrac{1}{4}-0,6\right).\dfrac{10}{11}\)
\(C=\dfrac{5}{16}:0,125-\left(\dfrac{9}{4}-0,6\right).\dfrac{10}{11}\)
\(C=\dfrac{5}{16}:0,125-\dfrac{33}{20}.\dfrac{10}{11}\)
\(C=\dfrac{5}{2}-\dfrac{3}{2}\)
\(C=1\)
Tính.
a) \(\dfrac{3}{11}+\dfrac{14}{11}\) b) \(\dfrac{1}{16}+\dfrac{3}{4}\) c) \(\dfrac{2}{20}+\dfrac{7}{10}\)
a) \(\dfrac{3}{11}+\dfrac{14}{11}=\dfrac{17}{11}\)
b) \(\dfrac{1}{16}+\dfrac{3}{4}=\dfrac{1}{16}+\dfrac{12}{16}=\dfrac{13}{16}\)
c) \(\dfrac{2}{20}+\dfrac{7}{10}=\dfrac{2}{20}+\dfrac{14}{20}=\dfrac{16}{20}=\dfrac{4}{5}\)
\(\)a) \(\dfrac{3}{5}+\dfrac{11}{20}\) b) \(\dfrac{5}{8}-\dfrac{4}{9}\) c) \(\dfrac{9}{16}\) x \(\dfrac{4}{3}\)
d) \(\dfrac{4}{7}:\dfrac{8}{11}\) e) \(\dfrac{3}{5}+\dfrac{4}{5}:\dfrac{2}{5}\)
a) \(\dfrac{3}{5}+\dfrac{11}{20}=\dfrac{12}{20}+\dfrac{11}{20}=\dfrac{23}{20}\)
b) \(\dfrac{5}{8}-\dfrac{4}{9}=\dfrac{45}{72}-\dfrac{32}{72}=\dfrac{13}{72}\)
c) \(\dfrac{9}{16}\times\dfrac{4}{3}=\dfrac{3}{4}\)
d) \(\dfrac{4}{7}:\dfrac{8}{11}=\dfrac{4}{7}\times\dfrac{11}{8}=\dfrac{11}{14}\)
e) \(\dfrac{3}{5}+\dfrac{4}{5}:\dfrac{2}{5}=\dfrac{3}{5}+\dfrac{4}{5}\times\dfrac{5}{2}=\dfrac{3}{5}+2=\dfrac{3}{5}+\dfrac{10}{5}=\dfrac{13}{5}\)
a)\(=\dfrac{12}{20}+\dfrac{11}{20}=\dfrac{23}{20}\)
b)\(=\dfrac{45}{72}-\dfrac{32}{72}=\dfrac{13}{72}\)
c)\(=\dfrac{9\times4}{16\times3}=\dfrac{3}{4}\)
d)\(=\dfrac{4}{7}\times\dfrac{11}{8}=\dfrac{11}{14}\)
e)\(=\dfrac{3}{5}+\dfrac{4}{2}=\dfrac{3}{5}+\dfrac{10}{5}=\dfrac{13}{5}\)
Bài 1 : Thực hiện phép tính
a , \(\left(\dfrac{1}{2}+\dfrac{16}{30}\right)-\left(1+\dfrac{1}{30}\right)\)
b , \(\dfrac{-5}{11}.\dfrac{4}{13}+\dfrac{-5}{11}.\dfrac{9}{13}+3\dfrac{5}{11}\)
c , \(3^2-12.\left(\dfrac{3}{4}-\dfrac{2}{3}\right)\)
Bài 2 : Cho đường thẳng xy , lấy điểm O thuộc đường thẳng xy . Trên tia Ox , lấy 2 điểm A , B sao cho OA = 3 cm , AB = 2 cm
a, Trong 3 điểm O , A , B điểm nào nằm giữa 2 điểm còn lại.
b, Có tất cả bao nhiêu tia ? nêu tên ?
c, Tính độ dài OB ?
Bài 3 : Tính A = \(\dfrac{9}{1.2}+\dfrac{9}{2.3}+\dfrac{9}{3.4}+...+\dfrac{9}{2021.2022}\)
bài 3:
\(A=\dfrac{9}{1\cdot2}+\dfrac{9}{2\cdot3}+\dfrac{9}{3\cdot4}+...+\dfrac{9}{2021\cdot2022}\)
\(=9\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{2021\cdot2022}\right)\)
\(=9\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2021}-\dfrac{1}{2022}\right)\)
\(=9\cdot\dfrac{2021}{2022}=\dfrac{6063}{674}\)
Bài 1:
a: \(\left(\dfrac{1}{2}+\dfrac{16}{30}\right)-\left(1+\dfrac{1}{30}\right)\)
\(=\dfrac{15+16}{30}-1-\dfrac{1}{30}\)
\(=\dfrac{30}{30}-1\)
=1-1
=0
b: \(\dfrac{-5}{11}\cdot\dfrac{4}{13}+\dfrac{-5}{11}\cdot\dfrac{9}{13}+3\dfrac{5}{11}\)
\(=-\dfrac{5}{11}\left(\dfrac{4}{13}+\dfrac{9}{13}\right)+3+\dfrac{5}{11}\)
\(=-\dfrac{5}{11}+3+\dfrac{5}{11}\)
=3
c: \(3^2-12\left(\dfrac{3}{4}-\dfrac{2}{3}\right)\)
\(=9-12\cdot\dfrac{9-8}{12}\)
=9-1
=8
tìm x
\([\dfrac{6:\dfrac{3}{5}-1\dfrac{1}{16}.\dfrac{16}{7}}{4\dfrac{1}{5}-\dfrac{10}{11}+5\dfrac{2}{11}}-\dfrac{\left(\dfrac{3}{20}+\dfrac{1}{2}-\dfrac{1}{5}\right).\dfrac{12}{49}}{3\dfrac{1}{3}+\dfrac{2}{9}}].x=2\dfrac{23}{96}\)
a.,\(\dfrac{4}{5}+5\dfrac{1}{2}\text{x }\left(4,5-2\right)=\dfrac{7}{10}\) b,125%x\(\dfrac{17}{4}:\left(1\dfrac{5}{16}-0,5\right)+2008\)
c,\(\dfrac{5}{11}+\left(\dfrac{16}{11}+1\right)\) d, \(\dfrac{3}{17}+\dfrac{11}{4}+\dfrac{5}{8}+\dfrac{14}{17}+\dfrac{3}{8}\)
`a)4/5+5 1/2 xx (4,5-2)+7/10`
`=4/5+11/2*2,5+7/10`
`=0,8+2,2+0,7`
`=3+0,7=3,7`
`b)125%xx 17/4:(1 5/16-0,5)+2008`
`=1,25xx4,25:13/16+2008`
`=85/13+2008`
`=2014 7/13`
`c)5/11+(16/11+1)`
`=5/11+1+5/11+1`
`=2+10/11=32/11`
`d)3/17+11/4+5/8+14/17+3/8`
`=3/17+14/17+5/8+3/8+11/4`
`=1+1+11/4`
`=19/4`
a)
\(\dfrac{4}{5}+5\dfrac{1}{2}x\left(4,5-2\right)=\dfrac{7}{10}\)
<=> \(\dfrac{11}{2}x\times2,5=\dfrac{7}{10}-\dfrac{4}{5}=\dfrac{-1}{10}\)
<=> \(\dfrac{55}{4}x=\dfrac{-1}{10}< =>x=\dfrac{-2}{275}\)
b) \(125\%\times\dfrac{17}{4}:\left(1\dfrac{5}{16}-0,5\right)+2008\)
= \(\dfrac{85}{16}:\left(\dfrac{21}{16}-\dfrac{1}{2}\right)+2008=\dfrac{85}{16}:\dfrac{13}{16}+2008=\dfrac{26189}{13}\)
c) \(\dfrac{5}{11}+\left(\dfrac{16}{11}+1\right)\)
= \(\dfrac{21}{11}+1=\dfrac{32}{11}\)
d) \(\left(\dfrac{3}{17}+\dfrac{14}{17}\right)+\left(\dfrac{5}{8}+\dfrac{3}{8}\right)+\dfrac{11}{4}\)
= 1 + 1 + \(\dfrac{11}{4}\) = \(\dfrac{19}{4}\)
Tính:
\(\dfrac{1}{2}-\dfrac{3}{8}\) \(\dfrac{4}{3}-\dfrac{8}{15}\) \(\dfrac{5}{6}-\dfrac{7}{12}\)
\(\dfrac{11}{4}-\dfrac{9}{8}\) \(\dfrac{17}{16}-\dfrac{3}{4}\) \(\dfrac{31}{36}-\dfrac{5}{6}\)
\(\dfrac{1}{2}-\dfrac{3}{8}=\dfrac{4}{2\times4}-\dfrac{3}{8}=\dfrac{4}{8}-\dfrac{3}{8}=\dfrac{1}{8}\)
\(\dfrac{4}{3}-\dfrac{8}{15}=\dfrac{4\times5}{3\times5}-\dfrac{8}{15}=\dfrac{20}{15}-\dfrac{8}{15}=\dfrac{12}{15}=\dfrac{4}{5}\)
\(\dfrac{5}{6}-\dfrac{7}{12}=\dfrac{5\times2}{6\times2}-\dfrac{7}{12}=\dfrac{10}{12}-\dfrac{7}{12}=\dfrac{3}{12}=\dfrac{1}{4}\)
\(\dfrac{11}{4}-\dfrac{9}{8}=\dfrac{11\times2}{4\times2}-\dfrac{9}{8}=\dfrac{22}{8}-\dfrac{9}{8}=\dfrac{13}{8}\)
\(\dfrac{17}{16}-\dfrac{3}{4}=\dfrac{17}{16}-\dfrac{3\times4}{4\times4}=\dfrac{17}{16}-\dfrac{12}{16}=\dfrac{5}{16}\)
\(\dfrac{31}{36}-\dfrac{5}{6}=\dfrac{31}{36}-\dfrac{5\times6}{6\times6}=\dfrac{31}{36}-\dfrac{30}{36}=\dfrac{1}{36}\)
Câu 1)
1) \(\dfrac{11}{24}\)−\(\dfrac{5}{41}\)+\(\dfrac{13}{24}\)+0,5−\(\dfrac{36}{41}\)=
2)12÷\(\left(\dfrac{3}{4}-\dfrac{5}{6}\right)^2\)=
3) (\(1+\dfrac{2}{3}-\dfrac{1}{4}\))\(\left(0,8-\dfrac{3}{4}\right)^2\) =
4)\(16\dfrac{2}{7}\)÷(\(\dfrac{-3}{5}\))+\(28\dfrac{2}{7}\)÷\(\dfrac{3}{5}\)
5)\(\left(2^2\div\dfrac{4}{3}-\dfrac{1}{2}\right)\times\dfrac{6}{5}-17\)
6)\(\left(\dfrac{1}{3}\right)^{50}\times\left(-9\right)^{25}-\dfrac{2}{3}\div4\)
1: \(\dfrac{11}{24}-\dfrac{5}{41}+\dfrac{13}{24}+0,5-\dfrac{36}{41}\)
\(=\left(\dfrac{11}{24}+\dfrac{13}{24}\right)-\left(\dfrac{5}{41}+\dfrac{36}{41}\right)+\dfrac{1}{2}\)
\(=1-1+\dfrac{1}{2}=\dfrac{1}{2}\)
2: \(12:\left(\dfrac{3}{4}-\dfrac{5}{6}\right)^2\)
\(=12:\left(\dfrac{9}{12}-\dfrac{10}{12}\right)^2\)
\(=12:\left(-\dfrac{1}{12}\right)^2=12:\dfrac{1}{144}=12\cdot144=1368\)
3: \(\left(1+\dfrac{2}{3}-\dfrac{1}{4}\right)\cdot\left(0,8-\dfrac{3}{4}\right)^2\)
\(=\dfrac{12+8-3}{12}\cdot\left(\dfrac{4}{5}-\dfrac{3}{4}\right)^2\)
\(=\dfrac{17}{12}\cdot\left(\dfrac{16-15}{20}\right)^2\)
\(=\dfrac{17}{12}\cdot\dfrac{1}{400}=\dfrac{17}{4800}\)
4: \(16\dfrac{2}{7}:\left(-\dfrac{3}{5}\right)+28\dfrac{2}{7}:\dfrac{3}{5}\)
\(=\dfrac{5}{3}\cdot\left(-16-\dfrac{2}{7}\right)+\dfrac{5}{3}\cdot\left(28+\dfrac{2}{7}\right)\)
\(=\dfrac{5}{3}\left(-16-\dfrac{2}{7}+28+\dfrac{2}{7}\right)\)
\(=12\cdot\dfrac{5}{3}=20\)
5: \(\left(2^2:\dfrac{4}{3}-\dfrac{1}{2}\right)\cdot\dfrac{6}{5}-17\)
\(=\left(4\cdot\dfrac{3}{4}-\dfrac{1}{2}\right)\cdot\dfrac{6}{5}-17\)
\(=\dfrac{5}{2}\cdot\dfrac{6}{5}-17=3-17=-14\)
6: \(\left(\dfrac{1}{3}\right)^{50}\cdot\left(-9\right)^{25}-\dfrac{2}{3}:4\)
\(=\left(\dfrac{1}{3}\right)^{50}\cdot\left(-1\right)\cdot3^{50}-\dfrac{2}{3\cdot4}\)
\(=-1-\dfrac{2}{12}=-1-\dfrac{1}{6}=-\dfrac{7}{6}\)