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\(\text{A=}\left(\dfrac{7}{8}-0,25\right):\left(\dfrac{5}{6}-0,75\right)^2\)
\(A=\left(\dfrac{7}{8}-\dfrac{1}{4}\right):\left(\dfrac{5}{6}-\dfrac{3}{4}\right)^2\)
\(A=\left(\dfrac{7}{8}+\dfrac{-1}{4}\right):\dfrac{1}{144}\)
\(A=\dfrac{5}{8}.144=90\)
\(\text{B=}\dfrac{3}{4}.1\dfrac{4}{9}-\left(\dfrac{-3}{4}\right)\)
\(B=\dfrac{3}{4}.\dfrac{13}{9}+\dfrac{3}{4}\)
\(B=\dfrac{13}{12}+\dfrac{3}{4}\)
\(B=\dfrac{11}{6}\)
\(a,\left(31\dfrac{6}{13}+5\dfrac{9}{41}\right)-36\dfrac{6}{13}\\ =\left(31-36\right)+\left(\dfrac{6}{13}-\dfrac{6}{13}\right)+5\dfrac{9}{41}\\ =-5+0+5\dfrac{9}{41}\\ =\left(-5+5\right)+\dfrac{9}{41}=\dfrac{9}{41}\)
\(b,\dfrac{5}{3}+\left(-\dfrac{2}{7}\right)-\left(-1,2\right)\\ =\dfrac{5}{3}-\dfrac{2}{7}+\dfrac{6}{5}\\ =\dfrac{5.35-2.15+6.21}{105}=\dfrac{271}{105}\\ c,0,25+\dfrac{3}{5}-\left(\dfrac{1}{8}-\dfrac{2}{5}+1\dfrac{1}{4}\right)=\dfrac{1}{4}+\dfrac{3}{5}-\dfrac{1}{8}+\dfrac{2}{5}-1\dfrac{1}{4}\\ =\left(-1\dfrac{1}{4}+\dfrac{1}{4}\right)+\left(\dfrac{3}{5}+\dfrac{2}{5}\right)-\dfrac{1}{8}=-1+1-\dfrac{1}{8}=-\dfrac{1}{8}\)
a) (31 6/13 + 5 9/41) - 36 6/13
= 409/13 + 214/41 - 474/13
= (409/13 - 474/13) + 214/41
= -5 + 214/41
= 9/41
b) 5/3 + (-2/7) - (-1,2)
= 5/3 - 2/7 + 6/5
= 29/21 + 6/5
= 271/105
c) 0,25 + 3/5 - (1/8 - 2/5 + 1 1/4)
= 1/4 + 3/5 - 1/8 + 2/5 - 5/4
= (1/4 - 5/4) + (3/5 + 2/5) - 1/8
= -1 + 1 - 1/8
= -1/8
Bài 1:
a) Ta có:
\(-3\dfrac{1}{5}=-2,8\)
\(\dfrac{37}{10}=3,7\)
Vì \(-3,25< -2,8< 3,7\)
Nên \(-3,25< -3\dfrac{1}{5}< \dfrac{37}{10}\)
b) Ta có:
\(\dfrac{-567}{568}>-1\)
\(\dfrac{-568}{567}< -1\)
\(\Leftrightarrow\dfrac{-567}{568}>-1>\dfrac{-568}{567}\)
\(\Leftrightarrow\dfrac{-567}{568}>\dfrac{-568}{567}\)
c) \(-0,625=-\dfrac{5}{8}\)
Vì \(8>7\)
\(\Leftrightarrow\dfrac{1}{8}< \dfrac{1}{7}\)
\(\Leftrightarrow\dfrac{-5}{8}>\dfrac{-5}{7}\)
\(\Leftrightarrow-0,625>\dfrac{-5}{7}\)
Vậy ...
Giải:
a) \(3,6-\dfrac{-5}{6}.\left(-2,4\right).\dfrac{3}{5}\)
\(=\dfrac{18}{5}-\dfrac{-5}{6}.\left(-\dfrac{12}{5}\right).\dfrac{3}{5}\)
\(=\dfrac{18}{5}-\dfrac{6}{5}\)
\(=\dfrac{12}{5}\)
Vậy ...
b) \(\dfrac{1}{4}-0,5-6\dfrac{1}{2}+\dfrac{5}{8}\)
\(=\dfrac{1}{4}-\dfrac{1}{2}-6\dfrac{1}{2}+\dfrac{5}{8}\)
\(=-\dfrac{49}{8}\)
Vậy ...
c) \(1,1-\left(-1,2\right)-\left|-1,3\right|-2\dfrac{3}{4}\)
\(=1,1+1,2-1,3-2,75\)
\(=-\dfrac{7}{4}=-1,75\)
Vậy ...
\(A=\left(7-\dfrac{3}{4}+\dfrac{1}{3}\right)-\left(6+\dfrac{5}{4}-\dfrac{4}{3}\right)-\left(5-\dfrac{7}{4}+\dfrac{5}{3}\right)\)
\(\Rightarrow A=7-\dfrac{3}{4}+\dfrac{1}{3}-6-\dfrac{5}{4}+\dfrac{4}{3}-5+\dfrac{7}{4}-\dfrac{5}{3}\)
\(\Rightarrow A=\left(7-6-5\right)-\dfrac{3}{4}-\dfrac{5}{4}+\dfrac{7}{4}+\dfrac{1}{3}+\dfrac{4}{3}-\dfrac{5}{3}\)
\(\Rightarrow A=-4-\dfrac{1}{4}+0\)
\(\Rightarrow A=-\dfrac{17}{4}\)
Cách 2 tính trong dấu ngoặc đơn của A, bạn tự tính nhé.
\(M=\left(\dfrac{\dfrac{2}{5}-\dfrac{2}{9}+\dfrac{2}{11}}{\dfrac{7}{5}-\dfrac{7}{9}+\dfrac{7}{11}}-\dfrac{\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}}{\dfrac{7}{6}-\dfrac{7}{8}+\dfrac{7}{10}}\right)\times\dfrac{2022}{2021}\)
\(M=\left(\dfrac{\dfrac{178}{495}}{\dfrac{623}{495}}-\dfrac{\dfrac{17}{60}}{\dfrac{119}{120}}\right)\times\dfrac{2022}{2021}\)
\(M=\left(\dfrac{2}{7}-\dfrac{2}{7}\right)\times\dfrac{2022}{2021}\)
\(M=0\times\dfrac{2022}{2021}\)
M=0
Ta lần lượt có:
\(\dfrac{1}{0,25}=\dfrac{100}{25}=4;\) \(\left(1\dfrac{1}{4}\right)^2=\left(\dfrac{5}{4}\right)^2=\dfrac{25}{16};\)
\(\dfrac{1}{\left(\dfrac{4}{3}\right)^2}=\dfrac{1}{\dfrac{16}{9}}=\dfrac{9}{16};\) \(\left(\dfrac{5}{4}\right)^3=\dfrac{125}{64}\) và \(\dfrac{1}{\left(-\dfrac{2}{3}\right)^3}=\dfrac{1}{-\dfrac{8}{27}}=-\dfrac{27}{8}\)
Vậy \(P=4.\dfrac{25}{16}+25.\left[\dfrac{9}{16}:\dfrac{125}{64}\right]:\left(-\dfrac{27}{8}\right)\)
\(=\dfrac{25}{4}-25.\dfrac{9}{16}.\dfrac{64}{125}.\dfrac{8}{27}\)
\(=\dfrac{25}{4}-\dfrac{32}{15}\)
\(=\dfrac{375-128}{60}=\dfrac{247}{60}\)
P=4.(25/16)+25.(9/16:25/16):(-27/8)
=25/4 +25.(9/25):(-27/8)
=25/4 +9:(-27/8)
=25/4 +(-8/3)
=43/12
\(a,=\dfrac{5}{3}-\dfrac{2}{7}+\dfrac{6}{5}=\dfrac{271}{105}\\ b,=\dfrac{1}{4}+\dfrac{3}{5}-\dfrac{1}{8}+\dfrac{2}{5}-\dfrac{5}{4}=1-1-\dfrac{1}{8}=-\dfrac{1}{8}\)
1a) \(\dfrac{5}{3}-\dfrac{2}{7}+1,2=\dfrac{5}{3}-\dfrac{2}{7}+\dfrac{6}{5}=\dfrac{175}{105}-\dfrac{30}{105}+\dfrac{126}{105}=\dfrac{107-30+126}{105}=\dfrac{203}{205}\)
\(D=\left(-8\right).\left(\dfrac{3}{4}-\dfrac{1}{4}\right):\left(\dfrac{9}{4}-\dfrac{7}{6}\right)=\left(-8\right)\dfrac{1}{2}:\dfrac{13}{12}=\left(-4\right).\dfrac{12}{13}=-\dfrac{48}{13}\)
\(\left(3\dfrac{1}{2}+0,25\right)-\left(\dfrac{1}{4}-3,25+\dfrac{1}{2}\right)\)
\(=\left(\dfrac{7}{2}+\dfrac{1}{4}\right)-\left(\dfrac{1}{4}-\dfrac{13}{4}+\dfrac{1}{2}\right)\)
\(=\dfrac{7}{2}+\dfrac{1}{4}-\dfrac{1}{4}+\dfrac{13}{4}-\dfrac{1}{2}\)
\(=\left(\dfrac{7}{2}-\dfrac{1}{2}\right)+\dfrac{13}{4}\)
\(=3+\dfrac{13}{4}\)
\(=\dfrac{12}{4}+\dfrac{13}{4}\)
\(=\dfrac{25}{4}\)
\(\left(3\dfrac{1}{2}+0,25\right)-\left(\dfrac{1}{4}-3,25+\dfrac{1}{2}\right)\)
\(=3,5+0,25-0,25+3,25-0,5\)
\(=6,75-0,5=6,25\)=25/4