\(\frac{92}{15}\) \(×\)\(\)\(\) \(\frac{21}{123}\)= ???
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a) 96 + 78 + 4 = 96 + 4 + 78 = 100 + 78 = 178
67 + 21 + 79 = 67 + 100 = 167
408 + 85 + 92 = 405 + 92 + 85 = 500 + 85 = 585
b) 789 + 285 + 15 = 789 + 300 = 1089
448 + 594 + 52 = 448 + 52 + 594 = 500 + 594 = 1094
677 + 969 + 123 = 677 + 123 + 969 = 1769
a) 96 + 78 + 4 = 96 + 4 + 78 = 100 + 78 = 178
67 + 21 + 79 = 67 + 100 = 167
408 + 85 + 92 = 405 + 92 + 85 = 500 + 85 = 585
b) 789 + 285 + 15 = 789 + 300 = 1089
448 + 594 + 52 = 448 + 52 + 594 = 500 + 594 = 1094
677 + 969 + 123 = 677 + 123 + 969 = 1769
\(\frac{100}{123}:\left(\frac{3}{4}+\frac{7}{2}\right)+\frac{23}{123}:\left(\frac{9}{5}-\frac{7}{15}\right)\)
\(=\frac{100}{123}:\frac{17}{4}+\frac{23}{123}:\frac{4}{3}\)
\(=\frac{100}{123}\times\frac{4}{17}+\frac{23}{123}\times\frac{3}{4}\)
\(=\frac{400}{2091}+\frac{23}{164}=\frac{2773}{8364}\)
\(B=\frac{5}{18\cdot21}+\frac{5}{21\cdot24}+\frac{5}{24\cdot27}+...+\frac{5}{123\cdot126}\\ B=\frac{5}{3}\cdot\left(\frac{3}{18\cdot21}+\frac{3}{21\cdot24}+\frac{3}{24\cdot27}+...+\frac{3}{123\cdot126}\right)\\ B=\frac{5}{3}\cdot\left(\frac{1}{18}-\frac{1}{21}+\frac{1}{21}-\frac{1}{24}+\frac{1}{24}-\frac{1}{27}+...+\frac{1}{123}-\frac{1}{126}\right)\\ B=\frac{5}{3}\cdot\left(\frac{1}{18}-\frac{1}{126}\right)\\ B=\frac{5}{3}\cdot\left(\frac{7}{126}-\frac{1}{126}\right)\\ B=\frac{5}{3}\cdot\frac{1}{21}\\ B=\frac{5}{63}\)
\(y=\frac{4\frac{6}{11}x11\frac{8}{9}+4\frac{12}{13}:3\frac{2}{5}}{123\frac{34}{45}:21\frac{1}{8}}\)
Lời giải:
\(y=\frac{\frac{50}{11}.\frac{107}{9}+\frac{64}{13}.\frac{5}{17}}{\frac{5569}{45}.\frac{8}{169}}=\frac{1214030}{21879}:\frac{44552}{7605}=\frac{39455975}{4165612}\)
\(a,\left(\frac{2}{5}\right)^6.\left(\frac{25}{4}\right)^2=\left(\frac{2}{2.3}\right)^6.\left(\frac{5}{2}\right)^4\)
\(=\frac{1}{3^6}.\frac{5^4}{2^4}=\frac{5^4}{3^6.2^4}\)
\(b,\frac{100}{123}:\left(\frac{3}{4}+\frac{7}{12}\right)+\frac{23}{123}:\left(\frac{9}{5}-\frac{7}{15}\right)\)
\(=\frac{100}{123}:\left(\frac{9+7}{12}\right)+\frac{23}{123}:\left(\frac{27-7}{15}\right)\)
\(=\frac{100}{123}:\frac{16}{12}+\frac{23}{123}:\frac{20}{15}\)
\(=\frac{100.12}{123.16}+\frac{23.15}{123.20}\)
\(=\frac{5.5.4.3.4}{41.3.4.4}+\frac{23.3.5}{41.3.4.5}\)
\(=\frac{25}{41}+\frac{23}{164}=\frac{25.4+23}{164}\)
\(=\frac{123}{164}=\frac{3}{4}\)
(\(\frac{121}{122}\). \(\frac{123}{125}\)) .(\(\frac{1995}{1996}\).\(\frac{17}{16}\)- \(\frac{21}{25}\):\(\frac{16}{17}\)).(\(\frac{42}{30}\).\(\frac{75}{23}\)-\(\frac{19}{23}\).\(\frac{210}{38}\))
= (121/122 . 123/125) . ( 1995/1996 . 17/16 - 21/25 : 16/ 17) . 0
= 0
cộng 1 vào từ số hạng ( hai vế cùng 3 số hạng=> không đổi)
Tử số còn lại x
\(\Leftrightarrow\frac{x}{19}+\frac{x}{20}+\frac{x}{21}=\frac{x}{17}+\frac{x}{16}+\frac{x}{15}\)
\(\Leftrightarrow\left(\frac{1}{19}+\frac{1}{20}+\frac{1}{21}-\frac{1}{17}-\frac{1}{16}-\frac{1}{15}\right)x=0\)
cái (...) khác không=> x =0 là nghiệm duy nhất
Ta có
\(\frac{x-19}{19}+\frac{x-20}{20}+\frac{x-21}{21}=\frac{x-17}{17}+\frac{x-16}{16}+\frac{x-15}{15}\)
\(\Leftrightarrow\frac{x}{19}-1+\frac{x}{20}-1+\frac{x}{21}-1=\frac{x}{17}-1+\frac{x}{16}-1+\frac{x}{15}-1\)
\(\Leftrightarrow\frac{x}{19}+\frac{x}{20}+\frac{x}{21}-3=\frac{x}{17}+\frac{x}{16}+\frac{x}{15}-3\)
\(\Leftrightarrow\frac{x}{19}+\frac{x}{20}+\frac{x}{21}=\frac{x}{17}+\frac{x}{16}+\frac{x}{15}\)
\(\Rightarrow\frac{x}{19}+\frac{x}{20}+\frac{x}{21}-\frac{x}{17}-\frac{x}{16}-\frac{x}{15}=0\)
\(\Leftrightarrow x\left(\frac{1}{19}+\frac{1}{20}+\frac{1}{21}-\frac{1}{17}-\frac{1}{16}-\frac{1}{15}\right)=0\)
Vì \(\left(\frac{1}{19}+\frac{1}{20}+\frac{1}{21}-\frac{1}{17}-\frac{1}{16}-\frac{1}{15}\right)\ne0\)
Nên phương trình chỉ co nghiệm duy nhất là x=0
Vậy x=0
1932/1845