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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) \(\left(\dfrac{1}{3}\right)^{-1}-\left(-\dfrac{6}{7}\right)^0+\left(\dfrac{1}{2}\right)^2:2\)
\(=3-1+\left(\dfrac{1}{2}\right)^2:2\)
\(=3-1+\dfrac{1}{4}:2\)
\(=3-1+\dfrac{1}{8}\)
\(=\dfrac{17}{8}\)
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 ...
\(B=\dfrac{\left(\dfrac{2}{3}\right)^3.\left(-\dfrac{3}{4}\right)^2.\left(-1\right)^{2003}}{\left(\dfrac{2}{5}\right)^2.\left(-\dfrac{5}{12}\right)^3}=\dfrac{\dfrac{8}{27}.\dfrac{9}{16}.\left(-1\right)}{\dfrac{4}{25}.\left(-\dfrac{125}{1728}\right)}=\dfrac{\dfrac{-1}{6}}{\dfrac{-5}{432}}=\dfrac{72}{5}\)
(Ko chép lại đề nữa nhé, đánh đề bài xoắn cả tay)
\(P=\dfrac{\left(17,005-4,505\right)^2+125,075}{\left\{\left[0,1936:0,88+3,53\right]^2-7,5625\right\}:0,52}\)
\(=\dfrac{\left(12,5\right)^2+125,075}{\left\{\left[3,75\right]^2-7,5625\right\}:0,52}\)
\(=\dfrac{156,25+125,075}{\left\{14,0625-7,5625\right\}:0,52}\)
\(=\dfrac{281,325}{6,5:0,52}\)
\(=\dfrac{281,325}{12,5}\)
\(=22,506\)
\(Q=\dfrac{\left(\dfrac{53}{4}-\dfrac{59}{27}-\dfrac{65}{6}\right).\dfrac{1016}{5}+\dfrac{187}{4}}{\left(\dfrac{10}{7}+\dfrac{10}{3}\right):\left(\dfrac{37}{3}-\dfrac{100}{7}\right)}\)
\(=\dfrac{\left(\dfrac{1431}{108}-\dfrac{236}{108}-\dfrac{1170}{108}\right).\dfrac{1016}{5}+\dfrac{187}{4}}{\left(\dfrac{30}{21}+\dfrac{70}{21}\right):\left(\dfrac{259}{21}-\dfrac{300}{21}\right)}\)
\(=\dfrac{\dfrac{25}{108}.\dfrac{1016}{5}+\dfrac{187}{4}}{\dfrac{100}{21}:\dfrac{-1}{21}}\)
\(=\dfrac{\dfrac{1270}{27}+\dfrac{187}{4}}{-100}\)
\(=\dfrac{\dfrac{5080}{108}+\dfrac{5049}{108}}{-100}\)
\(=\dfrac{10129}{108}.\left(-\dfrac{1}{100}\right)\)
\(=-\dfrac{10129}{10800}\)
5) \(\left(-2\right)^2+\sqrt{36}-\sqrt{9}+\sqrt{25}\)
=\(4+6-3+5\)
=\(12\)
2) \(\dfrac{11}{25}.\left(-24,8\right)-\dfrac{11}{25}.75,2\)
=\(\dfrac{11}{25}.\left(-24,8-75,2\right)\)
=\(\dfrac{11}{25}.\left(-100\right)\)
=\(-44\)
câu 1 \(A=\dfrac{3^2}{5^2}.5^2-\dfrac{9^3}{4^3}:\dfrac{3^3}{4^3}+\dfrac{1}{2}\)
\(A=\dfrac{3^2}{5^2}.5^2-\dfrac{\left(3^2\right)^3}{4^3}.\dfrac{4^3}{3^3}+\dfrac{1}{2}\)
\(A=\dfrac{3^2}{5^2}.5^2-\dfrac{3^6}{4^3}.\dfrac{4^3}{3^3}+\dfrac{1}{2}=3^2-3^3+\dfrac{1}{2}=-18+\dfrac{1}{2}=-\dfrac{35}{2}\)
\(B=\left[\dfrac{4}{11}+\dfrac{7}{22}.2\right]^{2010}-\left(\dfrac{1}{2^2}.\dfrac{4^4}{8^2}\right)^{2009}\)
\(B=\left[\dfrac{4}{11}+\dfrac{7}{11}\right]^{2010}-\left(\dfrac{1}{2^2}.\dfrac{\left(2^2\right)^4}{\left(2^3\right)^2}\right)^{2009}\)
\(B=1^{2010}-\left(\dfrac{1}{2^2}.\dfrac{2^8}{2^6}\right)^{2009}\)
\(B=1^{2010}-\left(\dfrac{2^8}{2^8}\right)^{2009}\)
\(B=1^{2010}-1^{2009}=1-1=0\)
câu 2
a) \(2x-\dfrac{5}{4}=\dfrac{20}{15}\)
\(\Leftrightarrow2x=\dfrac{4}{3}+\dfrac{5}{4}\)
\(\Leftrightarrow2x=\dfrac{31}{12}\)
\(\Leftrightarrow x=\dfrac{31}{24}\)
b) \(\left(x+\dfrac{1}{3}\right)^3=\left(-\dfrac{1}{2}\right)^3\)
\(\Leftrightarrow x+\dfrac{1}{3}=-\dfrac{1}{2}\)
\(\Leftrightarrow x=-\dfrac{1}{2}-\dfrac{1}{3}\)
\(\Leftrightarrow x=-\dfrac{5}{6}\)
a) = 4. 5/4 + 25. [ 3/2 : (5/4)2] : 27/8
= 5 + 25. 12/5: 27/8
=5 +160/9
=205/9
b) = 8+ 3- 1+2.8
=11-1+2.8
=10+2.8
=10+ 16
= 26
c)= 3+1+1/4:2
= 4+ 0,125
=4,125
e)\(16\dfrac{2}{7}:\left(-\dfrac{3}{5}\right)+28\dfrac{2}{7}:\left(-\dfrac{3}{5}\right)\)
=\(\left(16\dfrac{2}{7}+28\dfrac{2}{7}\right):\left(-\dfrac{3}{5}\right)\)
=\(\dfrac{312}{7}\)\(:\left(-\dfrac{3}{5}\right)\)
=\(-\dfrac{516}{7}\)
a)\(\dfrac{7}{8}.\left(\dfrac{2}{12}+\dfrac{4}{10}\right)\)
=\(\dfrac{7}{8}.\left(\dfrac{1}{6}+\dfrac{2}{5}\right)\)
=\(\dfrac{7}{8}.\)\(\dfrac{17}{30}\)
=\(\dfrac{119}{240}\)
Câu 2
(a+3)(b-4)-(a-3)(b+4)=0
=>ab-4a+3b-12-ab-4a+3b+12=0
=>-8a=-6b
=>a/b=3/4
=>a/3=b/4
\(a.\)
\(\left[6.\left(-\dfrac{1}{3}\right)^2-3\left(-\dfrac{1}{3}\right)+1\right]:\left(-\dfrac{1}{3}-1\right)\)
\(=\left[6.\dfrac{1}{9}+1+1\right]:\left(-\dfrac{4}{3}\right)\)
\(=\left(\dfrac{8}{3}\right):\left(-\dfrac{4}{3}\right)\)
\(=\left(\dfrac{8}{3}\right).\left(-\dfrac{3}{4}\right)\)
\(=-2\)
\(b.\)
\(\dfrac{\left(\dfrac{2}{3}\right)^3.\left(-\dfrac{3}{4}\right)^2.\left(-1\right)^{2003}}{\left(\dfrac{2}{5}\right)^2.\left(-\dfrac{5}{12}\right)^3}\)
\(=\dfrac{\dfrac{8}{27}.\dfrac{9}{16}.\left(-1\right)}{\dfrac{4}{25}.\left(-\dfrac{125}{1728}\right)}\)
\(=\dfrac{-\dfrac{1}{6}}{-\dfrac{5}{432}}\)
\(=\dfrac{72}{5}\)