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a) \(\frac{1}{3}-\frac{-1}{6}=\frac{1}{3}+\frac{1}{6}=\frac{1}{2}\)
b) \(2\frac{1}{3}+4\frac{1}{5}=\frac{7}{3}+\frac{21}{5}=\frac{98}{15}\)
c) \(\frac{4}{9}-\frac{13}{3}-\frac{4}{9}-\frac{10}{3}=\left(\frac{4}{9}-\frac{4}{9}\right)-\left(\frac{13}{3}+\frac{10}{3}\right)\)
\(=0-\frac{23}{3}=\frac{-23}{3}\)
d) \(4-\left(2-\frac{5}{2}\right)+0,5=4-2+\frac{5}{2}+\frac{1}{2}=2+3=5\)
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a) \(9^8=\left(3^2\right)^8=3^{16}\)
\(27^3=\left(3^3\right)^3=3^9\)
\(81^2=\left(3^4\right)^2=3^8\)
\(\Leftrightarrow3^{16}:\left(3^9.3^8\right)\)
\(\Leftrightarrow3^{16}:3^{17}\)
\(\Leftrightarrow3^{-1}=\frac{1}{3}\)
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\(2^3.19-2^3.14+1^{2018}\)
\(=2^3\left(19-14\right)+1\)
\(=2^3.5+1\)
\(=41\)
\(10^2-\left[60:\left(5^6:5^4-3.5\right)\right]\)
\(=10^2-\left[60:\left(5^2-3.5\right)\right]\)
\(=10^2-\left[60:10\right]\)
\(=10^2-6\)
\(=94\)
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\(A=3+3^3+3^5+...+3^{75}\)
\(< =>9A=3^3+3^5+3^7+...+3^{77}\)
\(< =>8A=3^{77}-3< =>A=\frac{3^{77}-3}{8}\)
Mình cảm ơn bạn Amasterasu nhưng mk k hiểu cho lắm, bạn giúp mình làm cụ thể hơn ở đoạn từ 9A sao ra được 8A như vậy thế?
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1/ Bg
\(\frac{21^4}{27.\left(-343\right)}\)= \(\frac{\left(3.7\right)^4}{3^3.\left(-7\right)^3}\)
= \(\frac{3^4.7^4}{3^3.\left(-7\right)^3}\)
= \(\frac{3.\left(-7\right)}{1.1}\)
= 3.(-7)
= -21
2/ Bg
Ta có: \(\frac{a}{4}=\frac{b}{5}=\frac{c}{2}\)và a + b - c = 21 (a, b, c thuộc Z)
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\frac{a}{4}=\frac{b}{5}=\frac{c}{2}=\frac{a+b-c}{4+5-2}=\frac{21}{7}\)= 3
=> a = 3.4 = 12
=> b = 3.5 = 15
=> c = 3.2 = 6
Vậy a = 12, b = 15 và c = 6
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a) A = \(\frac{101}{19}.\) \(\frac{61}{218}-\frac{101}{218}.\frac{42}{19}+\frac{117}{218}\)
= \(\frac{101}{218}.\frac{61}{19}-\frac{101}{218}.\frac{42}{19}+\frac{117}{218}\)
=\(\frac{101}{218}.\left(\frac{61}{19}-\frac{42}{19}\right)+\frac{117}{218}\)
=\(\frac{101}{218}.\frac{19}{19}+\frac{117}{218}\)
=\(\frac{101}{218}.1+\frac{117}{218}\)
=\(\frac{101}{218}+\frac{117}{218}\)
=\(\frac{218}{218}\)\(=1\)
b) B = \(\left(\frac{5}{2011^2}+\frac{7}{2012^2}-\frac{9}{2013^2}\right).\left(\frac{4}{5}-\frac{3}{4}-\frac{1}{20}\right)\)
= \(\left(\frac{5}{2011^2}+\frac{7}{2012^2}-\frac{9}{2013^2}\right)\)\(.\left(\frac{1}{20}-\frac{1}{20}\right)\)
= \(\left(\frac{5}{2011^2}+\frac{7}{2012^2}-\frac{9}{2013^2}\right).0\)
= \(0\)
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a) \(22\frac{1}{2}\cdot\frac{7}{9}+50\%-1,25\)
\(=\frac{45}{2}\cdot\frac{7}{9}+\frac{50}{100}-\frac{125}{100}\)
\(=\frac{5}{2}\cdot\frac{7}{1}+\frac{1}{2}-\frac{5}{4}\)
\(=\frac{35}{2}+\frac{1}{2}-\frac{5}{4}=18-\frac{5}{4}=\frac{67}{4}\)
b) \(1,4\cdot\frac{15}{49}-\left(\frac{4}{5}+\frac{2}{3}\right):2\frac{1}{5}\)
\(=\frac{7}{5}\cdot\frac{15}{49}-\frac{22}{15}:\frac{11}{15}\)
\(=\frac{1}{1}\cdot\frac{3}{7}-\frac{22}{15}\cdot\frac{15}{11}\)
\(=\frac{3}{7}-2=\frac{3-14}{7}=\frac{-11}{7}\)
c) \(\left(-\frac{1}{2}\right)^2-\frac{7}{16}:\frac{7}{4}+75\%\)
\(=\frac{1}{4}-\frac{7}{16}\cdot\frac{4}{7}+\frac{75}{100}\)
\(=\frac{1}{4}-\frac{1}{4}+\frac{3}{4}=\frac{3}{4}\)
Bài 2 Bạn tự làm nhé
1.a,\(22\frac{1}{2}.\frac{7}{9}+50\%-1,25\)
\(=\frac{45}{2}.\frac{7}{9}+\frac{1}{2}-\frac{5}{4}\)
\(=\frac{35}{2}+\frac{1}{2}-\frac{5}{4}\)
\(=\frac{67}{4}\)
b,Các phép tính khác làm tương tự
Đổi các số ra hết thành phân số,có ngoặc thì lm ngoặc trc,Xoq đến nhân chia trước dồi mới cộng trừ
c,tương tự
2.
a,\(1\frac{3}{5}+\frac{7}{12}\div x=\frac{-9}{4}\)
\(\frac{8}{5}+\frac{7}{12}\div x=\frac{-9}{4}\)
\(\frac{7}{12}\div x=\frac{-77}{20}\)
Đến đây dễ bạn tự làm
b,\(\left(2\frac{4}{5}.x+50\right)\div\frac{2}{3}=-51\)
\(\left(\frac{14}{5}x+50\right)\div\frac{2}{3}=-51\)
\(\frac{14}{5}x+50=-34\)
\(\frac{14}{5}x=-84\)
Tự làm tiếp
c,\(\left|\frac{3}{4}.x-\frac{1}{2}\right|=\frac{1}{4}\)\(\Rightarrow\left|\frac{3}{4}x-\frac{1}{2}\right|=\varnothing\)
Đáp án cần chọn là: C