\(\dfrac{2}{3}-4.\left(\dfrac{1}{2}\right)+\left(\dfrac{3}{4}\right)\)=
Những câu hỏi liên quan
\(a,\left(\dfrac{4}{9}+\dfrac{1}{3}\right)^2\)
\(b,\left(\dfrac{1}{2}-\dfrac{3}{5}\right)^3\)
c,\(\left(\dfrac{-10}{3}\right)^5.\left(\dfrac{-6}{4}\right)^4\)
\(\left(\dfrac{3}{4}\right)^3:\left(\dfrac{3}{4}\right)^2:\left(\dfrac{-3}{2}\right)^3\)
a: \(\left(\dfrac{4}{9}+\dfrac{1}{3}\right)^2=\dfrac{49}{81}\)
b: \(\left(\dfrac{1}{2}-\dfrac{3}{5}\right)^3=-\dfrac{1}{1000}\)
c: \(\left(-\dfrac{10}{3}\right)^5\cdot\left(-\dfrac{6}{4}\right)^4=-\dfrac{6250}{3}\)
d: \(\left(\dfrac{3}{4}\right)^3:\left(\dfrac{3}{4}\right)^2:\left(-\dfrac{3}{2}\right)^3=-\dfrac{2}{9}\)
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3dfrac{3}{3}.dfrac{1}{3}-dfrac{3}{4}.dfrac{1}{3}left[dfrac{11}{3}right]-left(dfrac{-1}{2}right)^2-4dfrac{1}{2}left(dfrac{3}{2}-dfrac{5}{4}+dfrac{1}{3}right):left(dfrac{4}{3}+2dfrac{3}{2}-dfrac{3}{4}right)5dfrac{5}{27}+dfrac{7}{23}+0,5+dfrac{-5}{27}+dfrac{16}{23}2dfrac{5}{4}+left(-2018right)^0-left[dfrac{-1}{4}right]dfrac{19}{11}.dfrac{6}{5}+dfrac{6^2}{11}.dfrac{6}{5}-left(dfrac{1}{2}right)^0
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\(3\dfrac{3}{3}.\dfrac{1}{3}-\dfrac{3}{4}.\dfrac{1}{3}\)
\(\left[\dfrac{11}{3}\right]-\left(\dfrac{-1}{2}\right)^2-4\dfrac{1}{2}\)
\(\left(\dfrac{3}{2}-\dfrac{5}{4}+\dfrac{1}{3}\right):\left(\dfrac{4}{3}+2\dfrac{3}{2}-\dfrac{3}{4}\right)\)
\(5\dfrac{5}{27}+\dfrac{7}{23}+0,5+\dfrac{-5}{27}+\dfrac{16}{23}\)
\(2\dfrac{5}{4}+\left(-2018\right)^0-\left[\dfrac{-1}{4}\right]\)
\(\dfrac{19}{11}.\dfrac{6}{5}+\dfrac{6^2}{11}.\dfrac{6}{5}-\left(\dfrac{1}{2}\right)^0\)
Bài 8:
b)\(\left(\dfrac{-4}{3}\right)+\left(\dfrac{-2}{5}\right)+\left(\dfrac{-3}{2}\right)\)
c) \(\dfrac{4}{5}-\left(\dfrac{-2}{7}\right)-\dfrac{-7}{10}\)
d) \(\dfrac{2}{3}-\left[\left(\dfrac{-7}{4}\right)-\left(\dfrac{1}{2}+\dfrac{3}{8}\right)\right]\)
\(b,=-\dfrac{40}{30}-\dfrac{12}{30}-\dfrac{45}{30}=-\dfrac{97}{30}\\ c,=\left(\dfrac{4}{5}+\dfrac{7}{10}\right)+\dfrac{2}{7}=\dfrac{3}{2}+\dfrac{2}{7}=\dfrac{25}{14}\\ d,=\dfrac{2}{3}+\dfrac{7}{4}+\dfrac{1}{2}+\dfrac{3}{8}\\ =\left(\dfrac{2}{3}+\dfrac{1}{2}\right)+\left(\dfrac{7}{4}+\dfrac{3}{8}\right)=\dfrac{7}{6}+\dfrac{17}{8}=\dfrac{79}{24}\)
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c: \(\dfrac{4}{5}-\dfrac{-2}{7}-\dfrac{-7}{10}\)
\(=\dfrac{56}{70}+\dfrac{20}{70}+\dfrac{49}{70}\)
\(=\dfrac{125}{70}=\dfrac{25}{14}\)
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\(\left(3-x\right)^3=-\dfrac{27}{64};\left(x-5\right)^3=\dfrac{1}{-27};\left(x-\dfrac{1}{2}\right)^3=\dfrac{27}{8};\left(2x-1\right)^2=\dfrac{1}{4};\left(2-3x\right)^2=\dfrac{9}{4};\left(1-\dfrac{2}{3}\right)^2=\dfrac{4}{9}\)
\(\left(3-x\right)^3=-\dfrac{27}{64}\)
\(\left(3-x\right)^3=\left(\dfrac{-3}{4}\right)^3\)
\(=>3-x=\dfrac{-3}{4}\)
\(x=3-\dfrac{-3}{4}=\dfrac{12}{4}+\dfrac{3}{4}\)
\(x=\dfrac{15}{4}\)
________
\(\left(x-5\right)^3=\dfrac{1}{-27}\)
\(\left(x-5\right)^3=\left(\dfrac{-1}{3}\right)^3\)
\(=>x-5=\dfrac{-1}{3}\)
\(x=\dfrac{-1}{3}+5=\dfrac{-1}{3}+\dfrac{15}{3}\)
\(x=\dfrac{14}{3}\)
_____________
\(\left(x-\dfrac{1}{2}\right)^3=\dfrac{27}{8}\)
\(\left(x-\dfrac{1}{2}\right)^3=\left(\dfrac{3}{2}\right)^3\)
\(=>x-\dfrac{1}{2}=\dfrac{3}{2}\)
\(x=\dfrac{3}{2}+\dfrac{1}{2}\)
\(x=2\)
________
\(\left(2x-1\right)^2=\dfrac{1}{4}\)
\(\left(2x-1\right)^2=\left(\dfrac{1}{2}\right)^2\) hoặc \(\left(2x-1\right)^2=\left(\dfrac{-1}{2}\right)^2\)
\(=>2x-1=\dfrac{1}{2}\) \(2x-1=\dfrac{-1}{2}\)
\(2x=\dfrac{1}{2}+1=\dfrac{1}{2}+\dfrac{2}{2}\) \(2x=\dfrac{-1}{2}+1=\dfrac{-1}{2}+\dfrac{2}{2}\)
\(2x=\dfrac{3}{2}\) \(2x=\dfrac{1}{2}\)
\(x=\dfrac{3}{2}:2=\dfrac{3}{2}.\dfrac{1}{2}\) \(x=\dfrac{1}{2}:2=\dfrac{1}{2}.\dfrac{1}{2}\)
\(x=\dfrac{3}{4}\) \(x=\dfrac{1}{4}\)
____________
\(\left(2-3x\right)^2=\dfrac{9}{4}\)
\(\left(2-3x\right)^2=\left(\dfrac{3}{2}\right)^2\) hoặc \(\left(2-3x\right)^2=\left(\dfrac{-3}{2}\right)^2\)
\(=>2-3x=\dfrac{3}{2}\) \(2-3x=\dfrac{-3}{2}\)
\(3x=2-\dfrac{3}{2}=\dfrac{4}{2}-\dfrac{3}{2}\) \(3x=2-\dfrac{-3}{2}=\dfrac{4}{2}+\dfrac{3}{2}\)
\(3x=\dfrac{1}{2}\) \(3x=\dfrac{7}{2}\)
\(x=\dfrac{1}{2}.\dfrac{1}{3}\) \(x=\dfrac{7}{2}.\dfrac{1}{3}\)
\(x=\dfrac{1}{6}\) \(x=\dfrac{7}{6}\)
______________
\(\left(1-\dfrac{2}{3}\right)^2=\dfrac{4}{9}\) -> Kiểm tra đề câu này
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(3-x)3=(-\(\dfrac{3}{4}\))3
3-x=-\(\dfrac{3}{4}\)
x=3-(-\(\dfrac{3}{4}\))
x=\(\dfrac{15}{4}\)
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Rút gọn bt: A = \(\dfrac{\left(1^4+4\right)\left(5^4+4\right)\left(9^4+4\right)...\left(21^4+4\right)}{\left(3^4+4\right)\left(7^4+4\right)\left(11^4+4\right)...\left(23^4+4\right)}\)
B = \(\left(\dfrac{n-1}{1}+\dfrac{n-2}{2}+\dfrac{n-3}{3}+..+\dfrac{2}{n-2}+\dfrac{1}{n-1}\right):\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{n}\right)\)
A = \(\dfrac{\left(1^4+4\right)\left(5^4+4\right)\left(9^4+4\right)...\left(21^4+4\right)}{\left(3^4+4\right)\left(7^4+4\right)\left(11^4+4\right)...\left(23^4+4\right)}\)
Xét: n4 + 4 = (n2+2)2 - 4n2 = (n2-2n+2)(n2+2n+2) = [(n-1)2+1][(x+1)2+1] nên: A = \(\dfrac{\left(0^2+1\right)\left(2^2+1\right)}{\left(2^2+1\right)\left(4^2+1\right)}.\dfrac{\left(4^2+1\right)\left(6^2+1\right)}{\left(6^2+1\right)\left(8^2+1\right)}.....\dfrac{\left(20^2+1\right)\left(22^2+1\right)}{\left(22^2+1\right)\left(24^2+1\right)}=\dfrac{1}{24^2+1}=\dfrac{1}{577}\)
B = \(\left(\dfrac{n-1}{1}+\dfrac{n-2}{2}+...+\dfrac{2}{n-2}+\dfrac{1}{n-1}\right):\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{n}\right)\)
Đặt C = \(\dfrac{n-1}{1}+\dfrac{n-2}{2}+...+\dfrac{n-\left(n-2\right)}{n-2}+\dfrac{n-\left(n-1\right)}{n-1}\)
= \(\dfrac{n}{1}+\dfrac{n}{2}+...+\dfrac{n}{n-2}+\dfrac{n}{n-1}-1-1-...-1\)
= \(n+\dfrac{n}{2}+\dfrac{n}{3}+...+\dfrac{n}{n-1}-\left(n-1\right)\)
= \(\dfrac{n}{2}+\dfrac{n}{3}+...+\dfrac{n}{n-1}+\dfrac{n}{n}\)
= \(n\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{n}\right)\)
Vậy ...
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Thực hiện phép tính:1, left(dfrac{-1}{2}right)^2.left|+8right|-left(-dfrac{1}{2}right)^3:left|-dfrac{1}{16}right|2, left|-0,25right|-left(-dfrac{3}{2}right)^2:dfrac{1}{4}+dfrac{3}{4}.2017^03, left|dfrac{2}{3}-dfrac{5}{6}right|.left(3,6:2dfrac{2}{5}right)^34, left|left(-0,5right)^2+dfrac{7}{2}right|.10-left(dfrac{29}{30}-dfrac{7}{15}right):left(-dfrac{2017}{2018}right)^05, dfrac{8}{3}+left(3-dfrac{1}{2}right)^2-left|dfrac{-7}{3}right|
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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`
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-2/5 : \(1\dfrac{1}{3}\)- ( 1/2 )\(^2\)
\(\left(\dfrac{1}{2}-\dfrac{2}{3}+\dfrac{5}{6}\right).\left(\dfrac{-3}{2}\right)^2\)
\(\left(1+\dfrac{2}{3}-\dfrac{1}{4}\right).\left(\dfrac{4}{5}-\dfrac{3}{4}\right)^2\)
Cậu ơi, câu đầu tớ vẫn chưa hiểu ý câu hỏi là như nào hết, cậu có thể viết lại cho tớ đko ạ? 
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Tìm x :
1) \(\left(-0,75x+\dfrac{5}{2}\right).\dfrac{4}{7}-\left(-\dfrac{1}{3}\right)=-\dfrac{5}{6}\)
2) \(\left(4x-9\right)\left(2,5+\dfrac{-7}{3}x\right)=0\)
3) \(\left|x-\dfrac{3}{4}\right|-\dfrac{1}{2}=0\)
4)\(\left(\dfrac{3}{5}-\dfrac{2}{3}x\right)^3=\dfrac{-64}{125}\)
3: \(\left|x-\dfrac{3}{4}\right|-\dfrac{1}{2}=0\)
\(\Leftrightarrow\left|x-\dfrac{3}{4}\right|=\dfrac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{3}{4}=\dfrac{1}{2}\\x-\dfrac{3}{4}=-\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{4}\\x=\dfrac{1}{4}\end{matrix}\right.\)
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tính
a) left[dfrac{0.8divleft(dfrac{4}{5}cdot1025right)}{0.64-1}+dfrac{left(1.08-dfrac{2}{25}right)divdfrac{4}{7}}{left(6dfrac{5}{7}-3dfrac{1}{4}right)cdot2dfrac{2}{17}}+left(1.2cdot0.5right)divdfrac{4}{5}right]
b) left(0.2right)^{-3}left[left(-dfrac{1}{5}right)^{-2}right]^{-1}+left[left(dfrac{1}{2}right)^{-3}right]^{-2}divleft(2^{-3}right)^{-1}-left(0.175right)^{-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) l...
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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 :\(2+\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{1+\dfrac{1}{2}}}}\)
\(=2+\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{\dfrac{3}{2}}}}\) \(=2+\dfrac{1}{1+\dfrac{1}{2+\dfrac{2}{3}}}\) \(=2+\dfrac{1}{1+\dfrac{1}{\dfrac{8}{3}}}\) \(=2+\dfrac{1}{1+\dfrac{3}{8}}\) \(=2+\dfrac{1}{\dfrac{11}{8}}\) \(=2+\dfrac{8}{11}\) \(=\dfrac{30}{11}\)
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d) \(\left(\dfrac{1}{3}\right)^{-1}-\left(-\dfrac{6}{7}\right)^0+\left(\dfrac{1}{2}\right)^2:2\)
\(=3-1+\left(\dfrac{1}{2}\right)^2:2\)
\(=3-1+\dfrac{1}{4}:2\)
\(=3-1+\dfrac{1}{8}\)
\(=\dfrac{17}{8}\)
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f) \(\dfrac{\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}}{\dfrac{2}{3}-\dfrac{2}{7}-\dfrac{2}{13}}.\dfrac{\dfrac{3}{4}-\dfrac{3}{16}-\dfrac{3}{64}-\dfrac{3}{256}}{1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}}+\dfrac{5}{8}\)
\(=\dfrac{1\left(\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}\right)}{2\left(\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}\right)}.\dfrac{\dfrac{3}{4}-\dfrac{3}{16}-\dfrac{3}{64}-\dfrac{3}{256}}{1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}}+\dfrac{5}{8}\)
\(=\dfrac{1}{2}.\dfrac{\dfrac{3}{4}-\dfrac{3}{16}-\dfrac{3}{64}-\dfrac{3}{256}}{1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}}+\dfrac{5}{8}\)
\(=\dfrac{1}{2}.\dfrac{\dfrac{3}{4}\left(1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}\right)}{1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}}+\dfrac{5}{8}\)
\(=\dfrac{1}{2}.\dfrac{3}{4}+\dfrac{5}{8}\)
\(=\dfrac{3}{8}+\dfrac{5}{8}\)
\(=1\)
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Tính:
a) (-0,4)2 - (0,4)3 . (-3)
b) \(\left(1\dfrac{3}{4}\right)^3\) - \(\left(1\dfrac{3}{4}\right)^2\) + (-1,031)0
c) \(\left(\dfrac{2}{3}\right)^3\) - 4. \(\left(-1\dfrac{3}{4}\right)^2\) + \(\left(-\dfrac{2}{3}\right)^3\)
a) = 0,16 - 0,064 . (-3)
= 0,16 + 0,192
= 0,352
b) = (7/4)³ - (7/4)² + 1
= 343/64 - 49/16 + 1
= 147/64 + 1
= 211/64
c) = 8/27 - 4.(-7/4)² - 8/27
= -4.49/16
= -49/4
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