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a,\(2,5.16,27.4+7,3=16,27.10+7,3=162,7+7,3=170\)
b,\(8.6.10+4.25.12+3.16.31+2.34.24=48.10+48.25+48.31+48.34\)
\(=48.\left(10+25+31+34\right)=48.100=4800\)
c,\(78.31+78.24+78.17+22.72=78\left(31+24+17\right)+22.72=78.72+22.72\)
\(=72.\left(78+22\right)=72.100=7200\)
d,\(102.67-34.51=102.67-17.102=102.\left(67-17\right)=102.50=5100\)
đéo hiểu thì cút nhé đéo ai khiến m trả lời cả . ngu thì nói ngu đi đừng bày đặt t ghét nhất là cái loại như m đấy ạ
Bài 2
a, \(1-\left(12,5+x-4,25\right):21,75=0\)
\(=>21,75-12,5-x+4,25=0\)
\(=>13,5-x=0=>x=13,5\)
b,\(x.101-x+72.99+72=21050\)
\(=>100x+72.100=21050\)
\(=>100\left(x+72\right)=21050\)
\(=>x+72=210,5\)\(=>x=138,5\)
c,\(151-2\left(x-6\right)=2227:17\)
\(=>151-2\left(x-6\right)=131\)
\(=>2\left(x-6\right)=20\)
\(=>2x=32=>x=16\)
d,\(\left(x+4\right)+\left(x+6\right)+\left(x+8\right)+...+\left(x+26\right)=210\)\(=>12x+\left(4+6+8+...+26\right)=210\)
\(=>12x+180=210=>12x=30\)
\(=>x=\frac{30}{12}=\frac{15}{6}=\frac{5}{2}\)
Bài 2:
\(a,45+170+25+30\)
\(=\left(45+25\right)+\left(170+30\right)\)
\(=60+200=260\)
Bài 3:
\(a,\left(x-6\right).5=150\)
\(x-6=150:5\)
\(x-6=30\)
\(x=30+6\)
\(x=36\)
\(b,2^5.\left(3x-2\right)=2^3.2^6\)
\(2^5.\left(3x-2\right)=2^{3+6}\)
\(2^5.\left(3x-2\right)=2^9\)
\(3x-2=2^9:2^5\)
\(3x-2=2^4=16\)
\(3x=16+2\)
\(3x=22\)
\(x=22:3\)
\(x\approx7,3\)
\(c,100-7.\left(x-5\right)=51\)
\(7.\left(x-5\right)=100-51\)
\(7.\left(x-5\right)=49\)
\(x-5=49:7\)
\(x-5=7\)
\(x=7+5\)
\(x=12\)
Phần d) bạn thiếu dữ liệu ạ.
d) Ta có: \(n^2+5n+9⋮n+3\)
\(\Leftrightarrow n^2+3n+2n+6+3⋮n+3\)
\(\Leftrightarrow n\left(n+3\right)+2\left(n+3\right)+3⋮n+3\)
mà \(n\left(n+3\right)+2\left(n+3\right)⋮n+3\)
nên \(3⋮n+3\)
\(\Leftrightarrow n+3\inƯ\left(3\right)\)
\(\Leftrightarrow n+3\in\left\{1;-1;3;-3\right\}\)
hay \(n\in\left\{-2;-4;0;-6\right\}\)
Vậy: \(n\in\left\{-2;-4;0;-6\right\}\)
d) Ta có: n2+5n+9⋮n+3n2+5n+9⋮n+3
⇔n2+3n+2n+6+3⋮n+3⇔n2+3n+2n+6+3⋮n+3
⇔n(n+3)+2(n+3)+3⋮n+3⇔n(n+3)+2(n+3)+3⋮n+3
mà n(n+3)+2(n+3)⋮n+3n(n+3)+2(n+3)⋮n+3
nên 3⋮n+33⋮n+3
⇔n+3∈Ư(3)⇔n+3∈Ư(3)
⇔n+3∈{1;−1;3;−3}
Đáp án cần chọn là: A