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\(2^4.5-\left[131-\left(13-4\right).2\right]\)
\(=16.5-\left(131-9.2\right)\)
\(=80-\left(131-18\right)\)
\(=80-113\)
\(=-33\)
\(2^4.5-\left(131-\left(13-4\right).2\right)\)
\(=16.5-\left(131-18\right)\)
\(=80-118\)
\(=-33\)
a) 27.(-17)+(-17).73
= -17 ( 27+73)
= -17 . 100
= -1700
b) 512 . ( 2- 128) -128. (-512)
= 512.(2-128) - (-128).512
= 512(2-128+128)
= 512 . 2
= 1024
\(3^2\cdot5+2^3\cdot10-3^4:3\\ =9\cdot5+8\cdot10-3^4\\ =45+80-27\\ =98\)
\(3^2.5+2^3.10-3^4:3\)
\(=9.5+8.10-3^3\)
\(=45+80-27\)
\(=125-27\)
\(=98\)
\(#WendyDang\)
B=\(\left[\left(\frac{1}{3}+\frac{1}{4}\right)x\frac{12}{19}+\frac{12}{19}\right]:\frac{4}{5}-\frac{1}{4}+2012\)
B=\(\left(\frac{7}{12}x\frac{12}{19}+\frac{12}{19}\right):\frac{4}{5}-\frac{1}{4}+2012\)
B=\(\left(\frac{7}{19}+\frac{12}{19}\right):\frac{4}{5}-\frac{1}{4}+2012\)
B=\(\frac{5}{4}-\frac{1}{4}+2012\)
B=1+2012
B=2013
\(B=[\left(\frac{1}{3}+\frac{1}{4}\right)\times\frac{12}{19}+\frac{12}{19}]:\frac{4}{5}-\frac{1}{4}+2012\)
\(B=[\frac{7}{12}\times\frac{12}{19}+\frac{12}{19}]:\frac{4}{5}-\frac{1}{4}+2012\)
\(B=[\frac{7}{19}+\frac{12}{19}]:\frac{4}{5}-\frac{1}{4}+2012\)
\(B=1:\frac{4}{5}-\frac{1}{4}+2012\)
\(B=\frac{5}{4}-\frac{1}{4}+2012\)
\(B=1+2012\)
\(B=2013\)
\(60-\left[15.2+\left(4+1^{35}\right)^2\right]\)
\(\Rightarrow60-\left[30+\left(4+1\right)^2\right]\)
\(\Rightarrow60-\left[30+25\right]\)
\(\Rightarrow60-55\)
\(\Rightarrow5\)
chia ra làm các nhóm (13-12+11)+(10-9+8)-(7-6+5) -(4+3+2-1)
hay ta thấy 3 nhóm đầu có tổng =số trừ trong các nhóm đó tức là 12+9-6+(4+3+2-1)
Hay= 23
\(A=19\frac{1}{4}+\frac{1}{2}\times2\frac{1}{3}+5,75-\frac{1}{6}+74\)
MK GHI ĐẦY ĐỦ RA RÙI, BẠN TỰ BẤM MÁY TÍNH LÀM NHA ( MÌNH LƯỜI )
\(A=19\frac{1}{4}+\frac{1}{2}\times2\frac{1}{3}+5,75-\frac{1}{6}+74\)
\(A=\frac{77}{4}+\frac{1}{2}\times\frac{7}{3}+\frac{23}{4}-\frac{1}{6}+74\)
\(A=\frac{77}{4}+\frac{7}{6}+\frac{23}{4}-\frac{1}{6}+74\)
\(A=(\frac{77}{4}+\frac{23}{4})+(\frac{7}{6}-\frac{1}{6})+74\)
\(A=25+1+74\)
\(A=26+74\)
\(A=100\)
= 6/25 nha bong
\(\frac{4}{5.8}+\frac{4}{8.11}+...+\frac{4}{47.50}\)
\(=\frac{4}{3}.\left(\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{47.50}\right)\)
\(=\frac{4}{3}.\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{47}-\frac{1}{50}\right)\)
\(=\frac{4}{3}.\left(\frac{1}{5}-\frac{1}{50}\right)\)
\(=\frac{4}{3}.\left(\frac{10}{50}-\frac{1}{50}\right)\)
\(=\frac{4}{3}.\frac{9}{50}\)
\(=\frac{6}{25}\)