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\(a.\)
\(\left[6.\left(-\dfrac{1}{3}\right)^2-3\left(-\dfrac{1}{3}\right)+1\right]:\left(-\dfrac{1}{3}-1\right)\)
\(=\left[6.\dfrac{1}{9}+1+1\right]:\left(-\dfrac{4}{3}\right)\)
\(=\left(\dfrac{8}{3}\right):\left(-\dfrac{4}{3}\right)\)
\(=\left(\dfrac{8}{3}\right).\left(-\dfrac{3}{4}\right)\)
\(=-2\)
\(b.\)
\(\dfrac{\left(\dfrac{2}{3}\right)^3.\left(-\dfrac{3}{4}\right)^2.\left(-1\right)^{2003}}{\left(\dfrac{2}{5}\right)^2.\left(-\dfrac{5}{12}\right)^3}\)
\(=\dfrac{\dfrac{8}{27}.\dfrac{9}{16}.\left(-1\right)}{\dfrac{4}{25}.\left(-\dfrac{125}{1728}\right)}\)
\(=\dfrac{-\dfrac{1}{6}}{-\dfrac{5}{432}}\)
\(=\dfrac{72}{5}\)
\(P=\left(0,5-\dfrac{3}{5}\right):\left(-3\right)+\dfrac{1}{3}-\left(-\dfrac{1}{6}\right):\left(-2\right)\)
\(=\left(-\dfrac{1}{2}-\dfrac{3}{5}\right):\left(-3\right)+\dfrac{1}{3}-\left(-\dfrac{1}{6}\right).\left(-\dfrac{1}{2}\right)\)
\(=\left(\dfrac{-5-6}{10}\right):\left(-3\right)+\dfrac{1}{3}-\dfrac{1}{12}\)
\(=-\dfrac{11}{10}:\left(-3\right)+\dfrac{1}{4}\)
\(=-\dfrac{11}{10}.\left(-\dfrac{1}{3}\right)+\dfrac{1}{4}=\dfrac{11}{30}+\dfrac{1}{4}=\dfrac{37}{60}\)
Vậy \(P=\dfrac{37}{60}\)
\(Q=\left(\dfrac{2}{25}-1,008\right):\dfrac{4}{7}:\left[\left(3\dfrac{1}{4}-6\dfrac{5}{9}\right):2\dfrac{2}{17}\right]\)
\(=\left(\dfrac{2}{25}-\dfrac{126}{125}\right):\dfrac{4}{7}:\left[\left(\dfrac{13}{4}-\dfrac{59}{9}\right).\dfrac{36}{17}\right]\)
\(=-\dfrac{116}{125}.\dfrac{7}{4}:\left(-\dfrac{119}{36}.\dfrac{36}{17}\right)\)
\(=\dfrac{-29.7}{125}:\left(-7\right)=\dfrac{29}{125}\)
Vậy \(Q=\dfrac{29}{125}\)
\(\frac{\left(\frac{2}{3}\right)^3.\left(\frac{-3}{4}\right)^2.\left(-1\right)^5}{\left(\frac{2}{5}\right)^2.\left(\frac{-5}{12}\right)^3}=\frac{\frac{2^3}{3^3}.\frac{3^2}{4^2}.\left(-1\right)}{\frac{2^2}{5^2}.\frac{\left(-5\right)^3}{12^3}}=\)\(\frac{\frac{1}{6}.\left(-1\right)}{\frac{2^2}{5^2}.\frac{5^3}{2^6.3^3}.\left(-1\right)}=\frac{\frac{1}{2.3}}{\frac{5}{2^4.3^3}}=\frac{2^3.3^2}{5}=\frac{72}{5}\)
A=\(\dfrac{5}{4}\).(5-\(\dfrac{4}{3}\)).(\(-\dfrac{1}{11}\))
= \(\dfrac{5}{4}\).\(\dfrac{11}{3}\).(\(-\dfrac{1}{11}\))
=\(\dfrac{5}{4}\).[\(\dfrac{11}{3}.\left(-\dfrac{1}{11}\right)\text{]}\)
=\(\dfrac{5}{4}.\dfrac{1}{3}\)
=\(\dfrac{5}{12}\) (1)
B=\(\dfrac{3}{4}:\left(-12\right).\left(-\dfrac{2}{3}\right)\) =\(\dfrac{3}{4}:\text{[}\left(-12\right).\left(-\dfrac{2}{3}\right)\text{]}\)
=\(\dfrac{3}{4}:8\) =\(\dfrac{3}{4}.\dfrac{1}{8}\)=\(\dfrac{3}{32}\)(2)
C=\(\dfrac{5}{4}:\left(-15\right).\left(-\dfrac{2}{5}\right)\) =\(\dfrac{5}{4}:\text{[}\left(-15\right).\left(-\dfrac{2}{5}\right)\text{]}\)
=\(\dfrac{5}{4}:6=\dfrac{5}{4}.\dfrac{1}{6}=\dfrac{5}{24}\left(3\right)\)
D=(-3).\(\left(\dfrac{2}{3}-\dfrac{5}{4}\right):\left(-7\right)\) =(-3).\(\left(-\dfrac{7}{12}\right)\):(-7)=\(\dfrac{7}{4}:\left(-7\right)\)=\(\dfrac{7}{4}\).\(\left(\dfrac{-1}{7}\right)\)=\(\dfrac{-1}{4}\) (4)
Từ (1),(2),(3)và(4)=>Ta có thể sắp xếp các kết quả trên theo thứ tự tăng dần là:
(Bạn tự làm nhé! mình bận đi học rồi
)
(Ko chép lại đề nữa nhé, đánh đề bài xoắn cả tay)
\(P=\dfrac{\left(17,005-4,505\right)^2+125,075}{\left\{\left[0,1936:0,88+3,53\right]^2-7,5625\right\}:0,52}\)
\(=\dfrac{\left(12,5\right)^2+125,075}{\left\{\left[3,75\right]^2-7,5625\right\}:0,52}\)
\(=\dfrac{156,25+125,075}{\left\{14,0625-7,5625\right\}:0,52}\)
\(=\dfrac{281,325}{6,5:0,52}\)
\(=\dfrac{281,325}{12,5}\)
\(=22,506\)
\(Q=\dfrac{\left(\dfrac{53}{4}-\dfrac{59}{27}-\dfrac{65}{6}\right).\dfrac{1016}{5}+\dfrac{187}{4}}{\left(\dfrac{10}{7}+\dfrac{10}{3}\right):\left(\dfrac{37}{3}-\dfrac{100}{7}\right)}\)
\(=\dfrac{\left(\dfrac{1431}{108}-\dfrac{236}{108}-\dfrac{1170}{108}\right).\dfrac{1016}{5}+\dfrac{187}{4}}{\left(\dfrac{30}{21}+\dfrac{70}{21}\right):\left(\dfrac{259}{21}-\dfrac{300}{21}\right)}\)
\(=\dfrac{\dfrac{25}{108}.\dfrac{1016}{5}+\dfrac{187}{4}}{\dfrac{100}{21}:\dfrac{-1}{21}}\)
\(=\dfrac{\dfrac{1270}{27}+\dfrac{187}{4}}{-100}\)
\(=\dfrac{\dfrac{5080}{108}+\dfrac{5049}{108}}{-100}\)
\(=\dfrac{10129}{108}.\left(-\dfrac{1}{100}\right)\)
\(=-\dfrac{10129}{10800}\)
Vì A= \(\frac{605}{36}\)
B=\(\frac{-1}{24}\)
C=\(\frac{-1}{30}\)
D= \(\frac{-1}{4}\)
tức là : A= \(\frac{6050}{360}\)
B=\(\frac{-15}{360}\)
C=\(\frac{-12}{360}\)
D=\(\frac{-90}{360}\)
nÊN được sắp xếp theo thứ tự tăng dần là B < C < D < A
D=
\(a,A=\dfrac{7}{35}+\left(-1\dfrac{3}{4}+\dfrac{12}{7}\right)-\left(\dfrac{1}{4}-\dfrac{2}{7}-\dfrac{12}{35}\right)-\dfrac{3}{7}\)\(A=\dfrac{7}{35}-\dfrac{7}{4}+\dfrac{12}{7}-\dfrac{1}{4}+\dfrac{2}{7}+\dfrac{13}{35}-\dfrac{3}{7}\\ A=\left(\dfrac{7}{35}+\dfrac{13}{35}\right)-\left(\dfrac{7}{4}-\dfrac{1}{4}\right)+\left(\dfrac{12}{7}+\dfrac{2}{7}-\dfrac{3}{7}\right)\)
\(A=\dfrac{4}{7}-\dfrac{3}{2}+\dfrac{11}{7}\\ A=\left(\dfrac{4}{7}+\dfrac{11}{7}\right)-\dfrac{3}{2}\\ A=\dfrac{15}{7}-\dfrac{3}{2}=\dfrac{9}{14}\)
Giải:
\(B=\dfrac{\left(\dfrac{2}{3}\right)^3.\left(-\dfrac{3}{4}\right)^2.\left(-1\right)^{2003}}{\left(\dfrac{2}{5}\right)^2.\left(-\dfrac{5}{12}\right)^3}\)
\(\Leftrightarrow B=\dfrac{\left(\dfrac{2}{3}\right)^3.\left(\dfrac{3}{4}\right)^2.\left(-1\right)}{\left(\dfrac{2}{5}\right)^2.\left(-\dfrac{5}{12}\right)^3}\)
\(\Leftrightarrow B=\dfrac{\left(\dfrac{2}{3}\right)^3.\left(\dfrac{3}{4}\right)^2}{\left(\dfrac{2}{5}\right)^2.\left(\dfrac{5}{12}\right)^3}\)
\(\Leftrightarrow B=\dfrac{\dfrac{2^3.3^2}{3^3.4^2}}{\dfrac{2^2.5^3}{5^2.12^3}}\)
\(\Leftrightarrow B=\dfrac{2^3.3^2.5^2.12^3}{3^3.4^2.2^2.5^3}\)
\(\Leftrightarrow B=\dfrac{2^3.3^2.5^2.2^6.3^3}{3^3.2^4.2^2.5^3}\)
\(\Leftrightarrow B=\dfrac{2^9.3^5.5^2}{3^3.2^6.5^3}\)
\(\Leftrightarrow B=\dfrac{2^3.3^2}{5}\)
\(\Leftrightarrow B=\dfrac{72}{5}\)
Vậy ...
\(B=\dfrac{\left(\dfrac{2}{3}\right)^3.\left(-\dfrac{3}{4}\right)^2.\left(-1\right)^{2003}}{\left(\dfrac{2}{5}\right)^2.\left(-\dfrac{5}{12}\right)^3}=\dfrac{\dfrac{8}{27}.\dfrac{9}{16}.\left(-1\right)}{\dfrac{4}{25}.\left(-\dfrac{125}{1728}\right)}=\dfrac{\dfrac{-1}{6}}{\dfrac{-5}{432}}=\dfrac{72}{5}\)