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\(a.\)
\(\left[6.\left(-\dfrac{1}{3}\right)^2-3\left(-\dfrac{1}{3}\right)+1\right]:\left(-\dfrac{1}{3}-1\right)\)
\(=\left[6.\dfrac{1}{9}+1+1\right]:\left(-\dfrac{4}{3}\right)\)
\(=\left(\dfrac{8}{3}\right):\left(-\dfrac{4}{3}\right)\)
\(=\left(\dfrac{8}{3}\right).\left(-\dfrac{3}{4}\right)\)
\(=-2\)
\(b.\)
\(\dfrac{\left(\dfrac{2}{3}\right)^3.\left(-\dfrac{3}{4}\right)^2.\left(-1\right)^{2003}}{\left(\dfrac{2}{5}\right)^2.\left(-\dfrac{5}{12}\right)^3}\)
\(=\dfrac{\dfrac{8}{27}.\dfrac{9}{16}.\left(-1\right)}{\dfrac{4}{25}.\left(-\dfrac{125}{1728}\right)}\)
\(=\dfrac{-\dfrac{1}{6}}{-\dfrac{5}{432}}\)
\(=\dfrac{72}{5}\)
Thực hiện các phép tính:
a) 9,6.212−(2.125−1512):149,6.212−(2.125−1512):14
b) 518−1,456:725+4,5.45518−1,456:725+4,5.45;
c) (12+0,8−113).(2,3+4725−1,28)(12+0,8−113).(2,3+4725−1,28)
d) (−5).12:[(−14)+12:(−2)]+113(−5).12:[(−14)+12:(−2)]+113.
Hướng dẫn làm bài:
a) 9,6.212−(2.125−1512):149,6.212−(2.125−1512):14
=9,6.52−(250−1712)×4=9,6.52−(250−1712)×4
=4,8.5−(1000−173)=4,8.5−(1000−173)
=24−1000+173=24−1000+173
=−976+173=−976+173
=−97013=−97013
b) 518−1,456:725+4,5.45518−1,456:725+4,5.45;
=518−1,456×257+92.45=518−1,456×257+92.45
=518−0,208×25+185=518−0,208×25+185
=518−5,2+185=518−5,2+185
=25−468+32490=25−468+32490
=−11990=−11990
c) (12+0,8−113).(2,3+4725−1,28)(12+0,8−113).(2,3+4725−1,28)
=(12+45−43).(2310+10725−3225)=(12+45−43).(2310+10725−3225)
=(15+24−4030).(2310+10725−3225)=(15+24−4030).(2310+10725−3225)
=(15+24−4030).(115+214−6450)=(15+24−4030).(115+214−6450)
=−130.26550=−130.26550
=−53300=−53300
d) (−5).12:[(−14)+12:(−2)]+113(−5).12:[(−14)+12:(−2)]+113
=−60:[14+12×(−12)]+1.13=−60:[14+12×(−12)]+1.13
=−60:[−14−14]+113=−60:[−14−14]+113
=−60:(12)+113=−60:(12)+113
=120+113=120+113
=12113
a) \(9,6.2\dfrac{1}{2}-\left(2.125-1\dfrac{5}{12}\right):\dfrac{1}{4}\)
\(=9,6.\dfrac{5}{2}-\left(250-\dfrac{17}{12}\right).4\)
\(=4,8.5-\left(1000-\dfrac{17}{3}\right)\)
\(=24-1000+\dfrac{17}{3}\)
\(=-976+\dfrac{17}{3}=-970\dfrac{1}{3}\)
b) \(\dfrac{5}{18}-1,456:\dfrac{7}{25}+4,5.\dfrac{4}{5}\)
\(=\dfrac{5}{18}-1,456.\dfrac{25}{7}+\dfrac{9}{2}.\dfrac{4}{5}\)
\(=\dfrac{5}{18}-0,208.25+\dfrac{18}{5}\)
\(=\dfrac{5}{18}-5,2+\dfrac{18}{5}\)
\(=-\dfrac{119}{90}\)
c) \(\left(\dfrac{1}{2}+0,8-1\dfrac{1}{3}\right).\left(2,3+4\dfrac{7}{25}-1,28\right)\)
\(=\left(\dfrac{1}{2}+\dfrac{4}{5}-\dfrac{4}{3}\right).\left(\dfrac{23}{10}+\dfrac{107}{25}-\dfrac{32}{25}\right)\)
\(=-\dfrac{1}{30}.\dfrac{265}{50}=-\dfrac{53}{300}\)
d) \(\left(-5\right).12:\left[\left(-\dfrac{1}{4}\right)+\dfrac{1}{2}:\left(-2\right)\right]+1\dfrac{1}{3}\)
\(=-60:\left[\dfrac{1}{4}+\dfrac{1}{2}.\dfrac{-1}{2}\right]+1.\dfrac{1}{3}\)
\(=-60:\left[-\dfrac{1}{4}-\dfrac{1}{4}\right]+1\dfrac{1}{3}\)
\(=-60:\left(\dfrac{1}{2}\right)+1\dfrac{1}{3}\)
\(=121\dfrac{1}{3}\)
A = \(\left(-2\right).\left(-1\dfrac{1}{2}\right).\left(-1\dfrac{1}{3}\right).\left(-1\dfrac{1}{4}\right)...\left(-1\dfrac{1}{214}\right)\)
= \(\left(-2\right).\left(-\dfrac{3}{2}\right).\left(-\dfrac{4}{3}\right).\left(-\dfrac{5}{4}\right)...\left(-\dfrac{215}{214}\right)\)
= \(\dfrac{\left(-2\right).\left(-3\right).\left(-4\right).\left(-5\right)...\left(-215\right)}{1.2.3.4...214}\)
= \(\dfrac{2.3.4.5...215}{1.2.3.4...214}\)
= \(\dfrac{215}{1}=215\)
B = \(\left(-1\dfrac{1}{2}\right).\left(-1\dfrac{1}{3}\right).\left(-1\dfrac{1}{4}\right)....\left(-1\dfrac{1}{299}\right)\)
= \(\left(-\dfrac{3}{2}\right).\left(-\dfrac{4}{3}\right).\left(-\dfrac{5}{4}\right)...\left(-\dfrac{300}{299}\right)\)
= \(\dfrac{\left(-3\right).\left(-4\right).\left(-5\right)...\left(-300\right)}{2.3.4...299}\)
= \(\dfrac{3.4.5...300}{2.3.4.5...299}\)
= \(\dfrac{300}{2}=150\)
\(5\dfrac{1}{3}+2,5\left(\dfrac{1}{3}-\dfrac{5}{12}\right)\)
\(=\dfrac{16}{3}+2,5.\dfrac{-1}{12}\)
\(=\dfrac{16}{3}+\dfrac{-5}{24}=\dfrac{41}{8}\)
\(\sqrt{12^2-44}-\left(1\dfrac{3}{4}-2\right)^2+\left(-1,5\right)^5:\left(\dfrac{-3}{2}\right)^4\)
\(=\sqrt{144-44}-\left(\dfrac{7}{4}-2\right)^2+\left(-1,5\right)^5:\left(\dfrac{3}{-2}\right)^4\)
\(=\sqrt{100}-\left(\dfrac{-1}{4}\right)^2+\left(1,5\right)^5:\left(\dfrac{3}{-2}\right)^4\)
\(=10-\dfrac{1}{16}+\dfrac{243}{32}:\dfrac{81}{16}\)
\(=\dfrac{183}{16}\)
mình tính cái này hay nhầm lẫn lắm nên bấm máy trước khi chép
(Ko chép lại đề nữa nhé, đánh đề bài xoắn cả tay)
\(P=\dfrac{\left(17,005-4,505\right)^2+125,075}{\left\{\left[0,1936:0,88+3,53\right]^2-7,5625\right\}:0,52}\)
\(=\dfrac{\left(12,5\right)^2+125,075}{\left\{\left[3,75\right]^2-7,5625\right\}:0,52}\)
\(=\dfrac{156,25+125,075}{\left\{14,0625-7,5625\right\}:0,52}\)
\(=\dfrac{281,325}{6,5:0,52}\)
\(=\dfrac{281,325}{12,5}\)
\(=22,506\)
\(Q=\dfrac{\left(\dfrac{53}{4}-\dfrac{59}{27}-\dfrac{65}{6}\right).\dfrac{1016}{5}+\dfrac{187}{4}}{\left(\dfrac{10}{7}+\dfrac{10}{3}\right):\left(\dfrac{37}{3}-\dfrac{100}{7}\right)}\)
\(=\dfrac{\left(\dfrac{1431}{108}-\dfrac{236}{108}-\dfrac{1170}{108}\right).\dfrac{1016}{5}+\dfrac{187}{4}}{\left(\dfrac{30}{21}+\dfrac{70}{21}\right):\left(\dfrac{259}{21}-\dfrac{300}{21}\right)}\)
\(=\dfrac{\dfrac{25}{108}.\dfrac{1016}{5}+\dfrac{187}{4}}{\dfrac{100}{21}:\dfrac{-1}{21}}\)
\(=\dfrac{\dfrac{1270}{27}+\dfrac{187}{4}}{-100}\)
\(=\dfrac{\dfrac{5080}{108}+\dfrac{5049}{108}}{-100}\)
\(=\dfrac{10129}{108}.\left(-\dfrac{1}{100}\right)\)
\(=-\dfrac{10129}{10800}\)
\(A=1+\dfrac{1}{2}\left(1+2\right)+\dfrac{1}{3}\left(1+2+3\right)+...+\dfrac{1}{12}\left(1+2+3+...+12\right)\)\(A=1+\dfrac{2+1}{2}+\dfrac{3+1}{2}+...+\dfrac{12+1}{2}\)
\(A=1+\dfrac{1}{2}\left(3+4+5+...+13\right)\)
\(A=1+\dfrac{1}{2}\left(1+2+3+...+13\right)-\dfrac{3}{2}\)
\(A=1+\dfrac{1}{2}.\dfrac{13\left(13+1\right)}{2}-\dfrac{3}{2}\)
\(A=1+\dfrac{91}{2}-\dfrac{3}{2}=45\)
Lời giải:
Ta có công thức tính tổng $n$ số hạng đầu tiên là: \(1+2+3+...+n=\frac{n(n+1)}{2}\)
\(\Rightarrow \frac{1+2+..+n}{n}=\frac{n(n+1)}{2n}=\frac{n+1}{2}\)
Áp dụng vào bài toán đã cho:
\(B=1+\frac{1}{2}(1+2)+\frac{1}{3}(1+2+3)+...+\frac{1}{12}(1+2+3+...+12)\)
\(B=1+\frac{2+1}{2}+\frac{3+1}{2}+\frac{4+1}{2}+...+\frac{12+1}{2}\)
\(B=1+\frac{1}{2}(3+4+5+...+13)\)
\(B=1+\frac{1}{2}(1+2+3+...+13)-\frac{3}{2}=1+\frac{1}{2}.\frac{13(13+1)}{2}-\frac{3}{2}\)
\(B=1+\frac{91}{2}-\frac{3}{2}=45\)