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Trả lời:
a, \(A=5\times\left(\frac{1}{5}+\frac{1}{7}\right)-\left(\frac{2}{5}+\frac{2}{17}+\frac{6}{10}+\frac{9}{51}\right)\)
\(A=5\times\left(\frac{1}{5}+\frac{1}{7}\right)-\left(\frac{2}{5}+\frac{2}{17}+\frac{3}{5}+\frac{3}{17}\right)\)
\(A=5\times\left(\frac{1}{5}+\frac{1}{7}\right)-\left(\frac{5}{5}+\frac{5}{17}\right)\)
\(A=5\times\left(\frac{1}{5}+\frac{1}{7}\right)-5\times\left(\frac{1}{5}+\frac{1}{17}\right)\)
\(A=5\times\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{5}-\frac{1}{7}\right)\)
\(A=5\times\left(\frac{1}{7}-\frac{1}{17}\right)\)
\(A=5\times\frac{10}{119}\)
\(A=\frac{50}{119}\)
b, \(B=\frac{2003\times14+1988+2001\times2002}{2002+2002\times503+504\times2002}\)
\(B=\frac{\left(2002+1\right)\times14+1988+2001\times2002}{2002\times\left(1+503+504\right)}\)
\(B=\frac{2002\times14+14+1988+2001\times2002}{2002\times1008}\)
\(B=\frac{2002\times14+2002+2001\times2002}{2002\times1008}\)
\(B=\frac{2002\times\left(14+1+2001\right)}{2002\times1008}\)
\(B=\frac{2002\times2016}{2002\times1008}\)
\(B=2\)
c, Sửa dề
\(\left(4,58\div3,27+5,23\div3,27\right)\times4,08-4,08\)
\(=\left[\left(4,58+5,23\right)\div3,27\right]\times4,08-4,08\)
\(=\left(9,81\div3,27\right)\times4,08-4,08\)
\(=3\times4,08-4,08\)
\(=4,08\times\left(3-1\right)\)
\(=4,08\times2\)
\(=8,16\)
d, \(\frac{6}{11}+\frac{7}{17}+\frac{8}{25}+\frac{10}{17}+\frac{16}{11}+\frac{17}{25}\)
\(=\left(\frac{6}{11}+\frac{16}{11}\right)+\left(\frac{7}{17}+\frac{10}{17}\right)+\left(\frac{8}{25}+\frac{17}{25}\right)\)
\(=2+1+1\)
\(=4\)
Cho mình xin cách giải chứ đáp số mình biết lâu rùi :)))))(((((
1. Dung
2.Dung
3. Dung
[ minh viet k dau mong ban thong cam k cho minh nha ]
Bài làm
80 : 80 = 1
10 : 10 = 1
20 : 20 = 1
30 : 10 = 3
40 : 20 = 2
50 : 50 = 1
60 : 3 = 20
70 : 7 = 10
100 : 25 = 4
90 : 3 = 30
# Học tốt #
80:80=1
10:10=1
20:20=1
30:10=3
40:20=2
50:50=1
60:3=20
70:7:=10
100:25=4
90:3=30
\(\frac{-48}{x}=\frac{y}{-72}=\frac{36}{z+1}=\frac{-3}{4}\)
Ta có: \(\frac{36}{z+1}=\frac{-3}{4}\)
\(\Leftrightarrow z+1=\frac{36.4}{-3}=-48\)
\(\Leftrightarrow z=-49\)
Lại có: \(\frac{y}{-72}=\frac{36}{-48}\)
\(\Leftrightarrow y=\frac{\left(-72\right).36}{\left(-48\right)}=54\)
Ta có: \(\frac{-48}{x}=\frac{54}{-72}\)
\(\Leftrightarrow x=\frac{\left(-48\right)\left(-72\right)}{54}=64\)
Vậy ...
P/s: Hoq chắc :)
a) 3,54 x 73 + 0,23 x25 +3,54 x 27 +3,54 x 27 +0,17 x 25
= 258,42 + 5,75 + 95,58 + 95,58 + 4,25
= 459,58
b) 563 x 23 +23 x 36 +23
= 12949 + 828 + 23
= 13800
a)Bị lỗi đề bài nhá,nếu là tính nhanh thì vậy,mình xin sửa và làm:
\(3,54\times73+0,23\times25+3,54\times27+0,17\times25\)
\(=3,54\times\left(73+27\right)+25\times\left(0,17+0,23\right)\)
\(=3,54\times100+25\times0,5\)
\(=354+12,5\)
\(=366,5\)
b)
\(563\times23+23\times36+23\)
\(=563\times23+23\times37\)
\(=23\times\left(563+37\right)\)
\(=23\times600\)
\(=13800\)
Cái phần lỗi đề bài đấy mai mình giải không sửa đề.
Không biết nó lỗi thật hay sự cố nhưng mà vẫn : CHÚC BẠN HỌC TỐT
Bài 1 : \(\frac{2}{3}< \left[\frac{1}{6}+\frac{2}{15}+\frac{3}{40}+\frac{4}{96}\right]:5\times x< \frac{5}{6}\)
=> \(\frac{2}{3}< \left[\frac{1}{6}+\frac{2}{15}+\frac{3}{40}+\frac{1}{24}\right]:5\cdot x< \frac{5}{6}\)
=> \(\frac{2}{3}< \left[\frac{1}{6}+\frac{1}{24}+\frac{2}{15}+\frac{3}{40}\right]:5\cdot x< \frac{5}{6}\)
=> \(\frac{2}{3}< \frac{5}{12}:5\cdot x< \frac{5}{6}\)
=> \(\frac{2}{3}< \frac{1}{12}\cdot x< \frac{5}{6}\)
=> \(\frac{2}{3}< \frac{x}{12}< \frac{5}{6}\)
=> \(\frac{8}{12}< \frac{x}{12}< \frac{10}{12}\)
=> x = 9
Bài 2 : \(\frac{\left[\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right]}{x}=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{132}\)
=> \(\frac{\left[1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{16}\right]}{x}=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{11\cdot12}\)
=> \(\frac{\left[1-\frac{1}{16}\right]}{x}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{11}-\frac{1}{12}\)
=> \(\frac{15}{\frac{16}{x}}=1-\frac{1}{12}\)
=> \(\frac{15}{\frac{16}{x}}=\frac{11}{12}\)
=> \(\frac{15}{16}:x=\frac{11}{12}\)
=> \(x=\frac{45}{44}\)
Bài 3 : \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\times(x+1):2}=\frac{399}{400}\)
=> \(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\times(x+1)}=\frac{399}{400}\)
=> \(2\left[\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\times(x+1)}\right]=\frac{399}{400}\)
=> \(2\left[\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{x\times(x+1)}\right]=\frac{399}{400}\)
=> \(\left[\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}\right]=\frac{399}{800}\)
=> \(\frac{1}{2}-\frac{1}{x+1}=\frac{399}{800}\)
=> \(\frac{1}{x+1}=\frac{1}{800}\)
=> x = 799
Bài 2 :
\(\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right):x=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{132}\) (*)
Ta có : \(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}=\frac{8}{16}+\frac{4}{16}+\frac{2}{16}+\frac{1}{16}=\frac{8+4+2+1}{16}=\frac{15}{16}\) (1)
Lại có : \(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{132}\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{11.12}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{11}-\frac{1}{12}\)
\(=1\left(-\frac{1}{2}+\frac{1}{2}\right)+\left(-\frac{1}{3}+\frac{1}{3}\right)+...+\left(-\frac{1}{11}+\frac{1}{11}\right)-\frac{1}{12}\)
\(=1-\frac{1}{12}=\frac{11}{12}\) (2)
Thay (1) và (2) vào biểu thức (*) ta được :
\(\frac{15}{16}:x=\frac{11}{12}\)
\(\Leftrightarrow x=\frac{15}{16}:\frac{11}{12}\)
\(\Leftrightarrow x=\frac{45}{44}\)
Vậy : \(x=\frac{45}{44}\)