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\(\frac{2x+1}{3}=\frac{5}{2}\)
\(2x+1=\frac{5.3}{2}=\frac{15}{2}\)
2x= 15/2 - 1 = 13/2
x = 13/2 : 2
x = 13/4
b) 2x + 2x+1 + 2x+2 + 2x+3 = 480
2x.(1+ 2 +22 + 23) = 480
2x . 15 = 480
2x = 480 : 15 = 32
2x = 25 => x = 5
c) \(\left(\frac{3x}{7}+1\right):\left(-4\right)=-\frac{1}{28}\)
\(\frac{3x}{7}+1=\frac{-1}{28}.\left(-4\right)=\frac{1}{7}\)
\(\frac{3x}{7}=\frac{1}{7}-1=-\frac{6}{7}\)
< = > 3x= -6 => x = -2
\(\frac{1}{6}.\frac{2}{5}.\frac{25}{7}=\frac{1.2.25}{6.5.7}=\frac{1.2.5.5}{2.3.5.7}=\frac{5}{21}\)
minh lam bai nay roi nhung minh ko co nha nen minh ko nho
=>B=\(\frac{12.\left(\frac{1}{13}+\frac{1}{1313}+\frac{1}{131}+\frac{1}{1313}\right)}{15.\left(\frac{1}{13}+\frac{1}{1313}+\frac{1}{131}+\frac{1}{1313}\right)}\)
=>B=\(\frac{12}{15}\)
=>B=\(\frac{4}{5}\)
B = \(\frac{12.\left(\frac{1}{13}+\frac{1}{1313}+\frac{1}{131}-\frac{1}{1313}\right)}{15.\left(\frac{1}{13}+\frac{1}{131}-\frac{1}{1313}+\frac{1}{1313}\right)}\)
=\(\frac{12.\left(\frac{1}{13}+\frac{1}{131}\right)}{15.\left(\frac{1}{13}+\frac{1}{131}\right)}\)
=\(\frac{12}{15}=\frac{4}{5}\)
Vậy B = 4/5.
\(\frac{1}{7}\)B=\(\frac{5}{2.7.1}+\frac{4}{1.7.11}+\frac{3}{11.2.7}+\frac{1}{2.7.15}+\frac{13}{15.4.7}\)
\(\frac{1}{7}\)B=\(\frac{5}{2.7}+\frac{4}{7.11}+\frac{3}{11.14}+\frac{1}{14.15}+\frac{13}{15.28}\)
\(\frac{1}{7}B=\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{15}+\frac{1}{15}-\frac{1}{28}\)
\(\frac{1}{7}B=\frac{1}{2}-\frac{1}{28}\)
\(\frac{1}{7}B=\frac{13}{28}\)
B=\(\frac{13}{28}:\frac{1}{7}\)
B=\(\frac{13}{4}\)
1717B=52.7.1+41.7.11+311.2.7+12.7.15+1315.4.752.7.1+41.7.11+311.2.7+12.7.15+1315.4.7
1717B=52.7+47.11+311.14+114.15+1315.2852.7+47.11+311.14+114.15+1315.28
17B=12−17+17−111+111−114+114−115+115−12817B=12−17+17−111+111−114+114−115+115−128
17B=12−12817B=12−128
17B=132817B=1328
B=1328:171328:17
B=134
1717B=52.7.1+41.7.11+311.2.7+12.7.15+1315.4.752.7.1+41.7.11+311.2.7+12.7.15+1315.4.7
1717B=52.7+47.11+311.14+114.15+1315.2852.7+47.11+311.14+114.15+1315.28
17B=12−17+17−111+111−114+114−115+115−12817B=12−17+17−111+111−114+114−115+115−128
17B=12−12817B=12−128
17B=132817B=1328
B=1328:171328:17
B=134
1717B=52.7.1+41.7.11+311.2.7+12.7.15+1315.4.752.7.1+41.7.11+311.2.7+12.7.15+1315.4.7
1717B=52.7+47.11+311.14+114.15+1315.2852.7+47.11+311.14+114.15+1315.28
17B=12−17+17−111+111−114+114−115+115−12817B=12−17+17−111+111−114+114−115+115−128
17B=12−12817B=12−128
17B=132817B=1328
B=1328:171328:17
B=134
1717B=52.7.1+41.7.11+311.2.7+12.7.15+1315.4.752.7.1+41.7.11+311.2.7+12.7.15+1315.4.7
1717B=52.7+47.11+311.14+114.15+1315.2852.7+47.11+311.14+114.15+1315.28
17B=12−17+17−111+111−114+114−115+115−12817B=12−17+17−111+111−114+114−115+115−128
17B=12−12817B=12−128
17B=132817B=1328
B=1328:171328:17
B=134
1717B=52.7.1+41.7.11+311.2.7+12.7.15+1315.4.752.7.1+41.7.11+311.2.7+12.7.15+1315.4.7
1717B=52.7+47.11+311.14+114.15+1315.2852.7+47.11+311.14+114.15+1315.28
17B=12−17+17−111+111−114+114−115+115−12817B=12−17+17−111+111−114+114−115+115−128
17B=12−12817B=12−128
17B=132817B=1328
B=1328:171328:17
B=134
1717B=52.7.1+41.7.11+311.2.7+12.7.15+1315.4.752.7.1+41.7.11+311.2.7+12.7.15+1315.4.7
1717B=52.7+47.11+311.14+114.15+1315.2852.7+47.11+311.14+114.15+1315.28
17B=12−17+17−111+111−114+114−115+115−12817B=12−17+17−111+111−114+114−115+115−128
17B=12−12817B=12−128
17B=132817B=1328
B=1328:171328:17
B=134
1717B=52.7.1+41.7.11+311.2.7+12.7.15+1315.4.752.7.1+41.7.11+311.2.7+12.7.15+1315.4.7
1717B=52.7+47.11+311.14+114.15+1315.2852.7+47.11+311.14+114.15+1315.28
17B=12−17+17−111+111−114+114−115+115−12817B=12−17+17−111+111−114+114−115+115−128
17B=12−12817B=12−128
17B=132817B=1328
B=1328:171328:17
B=134
1717B=52.7.1+41.7.11+311.2.7+12.7.15+1315.4.752.7.1+41.7.11+311.2.7+12.7.15+1315.4.7
1717B=52.7+47.11+311.14+114.15+1315.2852.7+47.11+311.14+114.15+1315.28
17B=12−17+17−111+111−114+114−115+115−12817B=12−17+17−111+111−114+114−115+115−128
17B=12−12817B=12−128
17B=132817B=1328
B=1328:171328:17
B=134
\(\frac{6}{x}=\frac{24}{x-27}\)
=> 6.( x - 27 ) = 24x
=> 6x - 162 = 24x
=> 162 = 6x - 24x
=> 162 = -18x
=> x = 162 : (-18)
=> x = -9
Ta có :
\(B=\frac{5}{2\cdot1}+\frac{4}{1\cdot11}+\frac{3}{11\cdot2}+\frac{1}{2\cdot15}+\frac{13}{15\cdot4}\)
\(=>\frac{B}{7}=\frac{5}{2\cdot7}+\frac{4}{7\cdot11}+\frac{3}{11\cdot14}+\frac{1}{14\cdot15}+\frac{13}{15\cdot28}\)
\(=>\frac{B}{7}=\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{15}+\frac{1}{15}-\frac{1}{28}\)
\(=>\frac{B}{7}=\frac{1}{2}-\frac{1}{28}=\frac{14}{28}-\frac{1}{28}=\frac{13}{28}\)
\(=>B=\frac{13}{28}\cdot7=\frac{13}{4}\)
Ta có:\(\frac{\left(3^4+3^3\right).\left(3^4+3^3\right)}{9^2}\)=\(\frac{\left(3^4+3^3\right)^2}{9^2}\)=\(\frac{3^4+3^3}{9}\)=\(\frac{81+27}{9}\)=\(\frac{108}{9}\)=12
Rất chi tiết, cái thứ 3 mình bỏ dấu mũ 2 vì cả ưử và mẫu đều có mũ 2 nên ta chia đều cho cả 2 số mũ đó nên mới ra \(\frac{3^4+3^3}{9}\)bạn ạ! Chúc bạn thành công!
qui luật số hạng cuối kiểu gì vậy ???
Bạn ấy ghi sai đề rồi!