\(4.\left(-1\right)-1-8+\left(-1\right).\dfrac{1}{2}+6+2+2.\left(-1\right)^2.\dfrac{1}{2}+2\)
Thực hiện phép tính:
1, \(\left(\dfrac{-1}{2}\right)^2.\left|+8\right|-\left(-\dfrac{1}{2}\right)^3:\left|-\dfrac{1}{16}\right|\)
2, \(\left|-0,25\right|-\left(-\dfrac{3}{2}\right)^2:\dfrac{1}{4}+\dfrac{3}{4}.2017^0\)
3, \(\left|\dfrac{2}{3}-\dfrac{5}{6}\right|.\left(3,6:2\dfrac{2}{5}\right)^3\)
4, \(\left|\left(-0,5\right)^2+\dfrac{7}{2}\right|.10-\left(\dfrac{29}{30}-\dfrac{7}{15}\right):\left(-\dfrac{2017}{2018}\right)^0\)
5, \(\dfrac{8}{3}+\left(3-\dfrac{1}{2}\right)^2-\left|\dfrac{-7}{3}\right|\)
`1//([-1]/2)^2 . |+8|-(-1/2)^3:|-1/16|=1/4 .8+1/8 .16=2+2=4`
`2//|-0,25|-(-3/2)^2:1/4+3/4 .2017^0=0,25-2,25.4+0,75.1=0,25-9+0,75=-8,75+0,75-8`
`3//|2/3-5/6|.(3,6:2 2/5)^3=|-1/6|.(3/2)^3=1/6 . 27/8=9/16`
`4//|(-0,5)^2+7/2|.10-(29/30-7/15):(-2017/2018)^0=|1/4+7/2|.10-1/2:1=|15/4|.10-1/2=15/4 .10-1/2=75/2-1/2=37`
`5// 8/3+(3-1/2)^2-|[-7]/3|=8/3+(5/2)^2-7/3=8/3+25/4-7/3=107/12-7/3=79/12`
Tính rồi so A và B :
\(A=\left(0,25\right)^{-1}.\left(1\dfrac{1}{4}\right)^2+25\left[\left(\dfrac{4}{3}\right)^{-2}:\left(1,25\right)^3\right]:\left(\dfrac{-2}{3}\right)^{-3}\)
\(B=\left(0,2\right)^{-3}.\left[\left(\dfrac{-1}{5}\right)^{-2}\right]^{-1}+\left[\left(\dfrac{1}{2}\right)^{-3}\right]^{-2}:\left(\dfrac{1}{8}\right)^{-1}-\left(2^{-3}\right)^{-2}:\dfrac{1}{2^6}\)
\(A=4.\dfrac{25}{16}+25.\left[\dfrac{9}{16}:\dfrac{125}{64}\right]:\dfrac{-27}{8}\)
\(=\dfrac{25}{16}+25.\dfrac{36}{125}:\dfrac{-27}{8}=-\dfrac{137}{240}\left(1\right)\)
\(B=125.\left[\dfrac{1}{25}+\dfrac{1}{64}:8\right]-64.\dfrac{1}{64}\)
\(=125.\dfrac{89}{1600}:8-64.\dfrac{1}{64}=\dfrac{-67}{512}\left(2\right)\)
Vì (2) > (1) => B > A
Tính các tích sau:
P\(_1\) =\(\left(1+\dfrac{2}{4}\right)\left(1+\dfrac{2}{10}\right)\left(1+\dfrac{2}{18}\right)....\left(1+\dfrac{2}{n^2+3n}\right)\)
P\(_2\) =\(\left(1+\dfrac{1}{3}\right)\left(1+\dfrac{1}{8}\right)\left(1+\dfrac{1}{15}\right)....\left(1+\dfrac{2}{n^2+2n}\right)\)
P\(_3\) = \(\left(1-\dfrac{1}{1+2}\right)\left(1-\dfrac{1}{1+2+3}\right)\left(1-\dfrac{1}{1+2+3+4}\right).....\left(1-\dfrac{1}{1+2+3+4+...+n}\right)\)
P\(_4\) = \(\dfrac{2^4+4}{4^4+4}.\dfrac{6^4+4}{8^4+4}.\dfrac{8^4+4}{10^4+4}....\dfrac{18^4+4}{20^4+4}\)
a, \(\dfrac{1}{24}-\left\{\dfrac{1}{4}-\left(\dfrac{1}{2}-\dfrac{7}{ }8\right)\right\}\)
b,\(\left(\dfrac{5}{7}-\dfrac{7}{5}\right)-\left\{\dfrac{1}{2}\left(\dfrac{2}{7}-\dfrac{1}{10}\right)\right\}\)
c,\(3-\left(\dfrac{-6}{7}\right)^6+\left(\dfrac{1}{2}\right)^2:2\)
d,\(\left(5^{-5}\right)^{-1}.\left(\dfrac{1}{2}\right)^2.\dfrac{1}{10^5}\)
Các bạn trả lời giúp mk nha. Mk đang cần gấp. Chều nay mk kiểm tra rồi
Tính giá trị biểu thức:
\(e,\dfrac{18}{37}+\dfrac{8}{24}+\dfrac{19}{37}-1\dfrac{23}{24}+\dfrac{2}{3}\)
\(f,\left(-2\right)^3.\left(\dfrac{3}{4}-0,25\right):\left(2\dfrac{1}{4}-1\dfrac{1}{6}\right)\)
\(g,\left(\dfrac{2}{5}\right)^2+5\dfrac{1}{2}.\left(4,5-2\right)+\dfrac{2^3}{\left(-4\right)}\)
\(h,\dfrac{4}{9}.19\dfrac{1}{3}-\dfrac{4}{9}.39\dfrac{1}{3}\)
\(i,\left(-\dfrac{1}{2}\right)^2:\dfrac{1}{4}-2\left(-\dfrac{1}{2}\right)^2\)
\(j,125\%.\left(\dfrac{-1}{2}\right)^2:\left(1\dfrac{5}{16}-1,5\right)+2008^0\)
\(k,\left(-2\right)^3.\dfrac{-1}{24}+\left(\dfrac{4}{3}-1\dfrac{5}{6}\right):\dfrac{5}{12}\)
e: \(=\left(\dfrac{18}{37}+\dfrac{19}{37}\right)+\left(\dfrac{8}{24}+\dfrac{2}{3}\right)-\dfrac{47}{24}=2-\dfrac{47}{24}=\dfrac{1}{24}\)
f: \(=-8\cdot\dfrac{1}{2}:\left(\dfrac{9}{4}-\dfrac{7}{6}\right)\)
\(=-4:\dfrac{13}{12}=\dfrac{-48}{13}\)
g: \(=\dfrac{4}{25}+\dfrac{11}{2}\cdot\dfrac{5}{2}-\dfrac{8}{4}=\dfrac{4}{25}+\dfrac{55}{4}-2=\dfrac{1191}{100}\)
tính
a) \(\left[\dfrac{0.8\div\left(\dfrac{4}{5}\cdot1025\right)}{0.64-1}+\dfrac{\left(1.08-\dfrac{2}{25}\right)\div\dfrac{4}{7}}{\left(6\dfrac{5}{7}-3\dfrac{1}{4}\right)\cdot2\dfrac{2}{17}}+\left(1.2\cdot0.5\right)\div\dfrac{4}{5}\right]\)
b) \(\left(0.2\right)^{-3}\left[\left(-\dfrac{1}{5}\right)^{-2}\right]^{-1}+\left[\left(\dfrac{1}{2}\right)^{-3}\right]^{-2}\div\left(2^{-3}\right)^{-1}-\left(0.175\right)^{-2}\)
c) \(2+\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{1+\dfrac{1}{2}}}}\)
d) \(\dfrac{1}{90}-\dfrac{1}{72}-\dfrac{1}{56}-\dfrac{1}{42}-\dfrac{1}{3}\)
e) \(\left(\dfrac{1}{3}\right)^{-1}-\left(-\dfrac{6}{7}\right)^0+\left(\dfrac{1}{2}\right)^2\div2\)
f) \(\dfrac{\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}}{\dfrac{2}{3}-\dfrac{2}{7}-\dfrac{2}{13}}\cdot\dfrac{\dfrac{3}{4}-\dfrac{3}{16}-\dfrac{3}{64}-\dfrac{3}{256}}{1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}}+\dfrac{5}{8}\)
g) \(\dfrac{1}{-\left(2017\right)\left(-2015\right)}+\dfrac{1}{\left(-2015\right)\left(-2013\right)}+...+\dfrac{1}{\left(-3\right)\cdot\left(-1\right)}\)
h) \(\left(1-\dfrac{1}{1\cdot2}\right)+\left(1-\dfrac{1}{2\cdot3}+...+\left(1-\dfrac{1}{2017\cdot2018}\right)\right)\)
c)
Ta có :\(2+\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{1+\dfrac{1}{2}}}}\)
\(=2+\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{\dfrac{3}{2}}}}\) \(=2+\dfrac{1}{1+\dfrac{1}{2+\dfrac{2}{3}}}\) \(=2+\dfrac{1}{1+\dfrac{1}{\dfrac{8}{3}}}\) \(=2+\dfrac{1}{1+\dfrac{3}{8}}\) \(=2+\dfrac{1}{\dfrac{11}{8}}\) \(=2+\dfrac{8}{11}\) \(=\dfrac{30}{11}\)
d) \(\left(\dfrac{1}{3}\right)^{-1}-\left(-\dfrac{6}{7}\right)^0+\left(\dfrac{1}{2}\right)^2:2\)
\(=3-1+\left(\dfrac{1}{2}\right)^2:2\)
\(=3-1+\dfrac{1}{4}:2\)
\(=3-1+\dfrac{1}{8}\)
\(=\dfrac{17}{8}\)
f) \(\dfrac{\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}}{\dfrac{2}{3}-\dfrac{2}{7}-\dfrac{2}{13}}.\dfrac{\dfrac{3}{4}-\dfrac{3}{16}-\dfrac{3}{64}-\dfrac{3}{256}}{1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}}+\dfrac{5}{8}\)
\(=\dfrac{1\left(\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}\right)}{2\left(\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}\right)}.\dfrac{\dfrac{3}{4}-\dfrac{3}{16}-\dfrac{3}{64}-\dfrac{3}{256}}{1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}}+\dfrac{5}{8}\)
\(=\dfrac{1}{2}.\dfrac{\dfrac{3}{4}-\dfrac{3}{16}-\dfrac{3}{64}-\dfrac{3}{256}}{1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}}+\dfrac{5}{8}\)
\(=\dfrac{1}{2}.\dfrac{\dfrac{3}{4}\left(1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}\right)}{1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}}+\dfrac{5}{8}\)
\(=\dfrac{1}{2}.\dfrac{3}{4}+\dfrac{5}{8}\)
\(=\dfrac{3}{8}+\dfrac{5}{8}\)
\(=1\)
a, \(\left(18\dfrac{1}{3}:\sqrt{225}+8\dfrac{2}{3}.\sqrt{\dfrac{49}{4}}\right)\): \(\left[\left(12\dfrac{1}{3}+8\dfrac{6}{7}\right)-\dfrac{\left(\sqrt{7}\right)^2}{\left(3\sqrt{2}\right)^2}\right]\): \(\dfrac{1704}{445}\)
b, \(\dfrac{1}{1.2}\)+\(\dfrac{1}{2.3}\)+...+\(\dfrac{1}{99.100}\)
c, \(\left(1-\dfrac{1}{2}\right)\)x\(\left(1-\dfrac{1}{3}\right)\)x.....x\(\left(1-\dfrac{1}{n+1}\right)\) (n ϵ N)
d, -66 x \(\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{11}\right)\) + 124 x -37 + 63 x -124
e, \(\dfrac{7}{4}\) x \(\left(\dfrac{33}{12}+\dfrac{3333}{2020}+\dfrac{333333}{303030}+\dfrac{33333333}{42424242}\right)\)
a: \(\left(18\dfrac{1}{3}:\sqrt{225}+8\dfrac{2}{3}\cdot\sqrt{\dfrac{49}{4}}\right):\left[\left(12\dfrac{1}{3}+8\dfrac{6}{7}\right)-\dfrac{\left(\sqrt{7}\right)^2}{\left(3\sqrt{2}\right)^2}\right]:\dfrac{1704}{445}\)
\(=\left(\dfrac{55}{3}:15+\dfrac{26}{3}\cdot\dfrac{7}{4}\right):\left[\left(12+\dfrac{1}{3}+8+\dfrac{6}{7}\right)-\dfrac{7}{18}\right]\cdot\dfrac{445}{1704}\)
\(=\left(\dfrac{55}{45}+\dfrac{91}{6}\right):\left[20+\dfrac{101}{126}\right]\cdot\dfrac{445}{1704}\)
\(=\dfrac{295}{18}:\dfrac{2621}{126}\cdot\dfrac{445}{1704}\)
\(=\dfrac{295}{18}\cdot\dfrac{126}{2621}\cdot\dfrac{445}{1704}\simeq0,21\)
b: \(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{99\cdot100}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(=1-\dfrac{1}{100}=\dfrac{99}{100}\)
c: \(\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\cdot...\cdot\left(1-\dfrac{1}{n+1}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot...\cdot\dfrac{n}{n+1}\)
\(=\dfrac{1}{n+1}\)
d: \(-66\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{11}\right)+124\cdot\left(-37\right)+63\cdot\left(-124\right)\)
\(=-66\cdot\dfrac{33-22+6}{66}+124\left(-37-63\right)\)
\(=-17-12400=-12417\)
e: \(\dfrac{7}{4}\left(\dfrac{33}{12}+\dfrac{3333}{2020}+\dfrac{333333}{303030}+\dfrac{33333333}{42424242}\right)\)
\(=\dfrac{7}{4}\left(\dfrac{33}{12}+\dfrac{33}{20}+\dfrac{33}{30}+\dfrac{33}{42}\right)\)
\(=\dfrac{7}{4}\cdot33\cdot\left(\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}\right)\)
\(=33\cdot\dfrac{7}{4}\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}\right)\)
\(=33\cdot\dfrac{7}{4}\cdot\left(\dfrac{1}{3}-\dfrac{1}{7}\right)\)
\(=33\cdot\dfrac{7}{4}\cdot\dfrac{4}{21}=\dfrac{33\cdot1}{3}=11\)
BT2: Tính nhanh
7) \(\left(-\dfrac{1}{2}\right)-\left(-\dfrac{3}{5}\right)+\left(-\dfrac{1}{9}\right)+\dfrac{1}{71}-\left(-\dfrac{2}{7}\right)+\dfrac{4}{35}-\dfrac{7}{18}\)
8)\(\left(3-\dfrac{1}{4}+\dfrac{2}{3}\right)-\left(5-\dfrac{1}{3}-\dfrac{6}{5}\right)-\left(6-\dfrac{7}{4}+\dfrac{3}{2}\right)\)
a) \(8\left(x+\dfrac{1}{x}\right)^2+4\left(x^2+\dfrac{1}{x^2}\right)^2-4\left(x^2+\dfrac{1}{x^2}\right)\left(x+\dfrac{1}{x}\right)^2=\left(x+4\right)^2\)
b) \(\left(6x+7\right)^2\left(3x+4\right)\left(x+1\right)=6\)
c) \(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)=840\)