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\(=18x\left(\frac{19}{21}+\frac{8}{9}\right)\)
\(=18x\frac{113}{63}\)
\(=\frac{226}{7}\)
3 15/16 giờ= 3 giờ 56,25 phút
Nếu mình đúng thì k mình và gửi kết bạn nhé
a, cách 1 : 7/11 : 3/5 + 4/11 : 3/5 b, cách 1 : [ 6,24 + 1,26 ] : 0,75
=[ 7/11 + 4/11 ] : 3/5 = 7,5 : 0,75
= 1 : 3/5 = 10
= 5/3 cách 2 : [ 6,24 + 1,26 ] : 0,75
cách 2 :7/11 : 3/5 + 4/11 : 3/5 = 6,24 : 0,75 + 1,26 : 0,75
=35/33 + 20/33 = 8,32 + 1,68
=55/33 = 10
=5/3
\(\frac{7}{11}:\frac{3}{5}+\frac{4}{11}:\frac{3}{5}\)
= ( \(\frac{7}{11}+\frac{4}{11}\)) : \(\frac{3}{5}\)
= 1 : \(\frac{3}{5}\)
= \(\frac{5}{3}\)
Đáp án là 0 bạn nhá !
Thử quy đồng số 2/5 ra mẫu số là 10 rồi trừ y như trừ số tự nhiên đó !
Vậy thôi hà !
\(\frac{5,4:0,4\times1420+4,5\times780\times3}{3+6+9+12+15+18+21+24+27}\)
\(=\frac{13,5\times1420+13,5\times780}{\left(3+27\right)+\left(6+24\right)+\left(9+21\right)+\left(12+18\right)+15}\)
\(=\frac{13,5\times\left(1420+780\right)}{30+30+30+30+15}\)
\(=\frac{13,5\times2200}{135}\)
\(=\frac{29700}{135}\)
\(=220\)
\(\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+\frac{1}{1+2+3+4+5}\)
\(=\frac{1}{1+2}\times\left(1+\frac{1}{1+2+3}\div\frac{1}{1+2}+\frac{1}{1+2+3+4}\div\frac{1}{1+2}+\frac{1}{1+2+3+4+5}\div\frac{1}{1+2}\right)\)
\(=\frac{1}{1+2}\times\left(1+\frac{1}{2}+\frac{3}{10}+\frac{1}{5}\right)\)
\(=\frac{1}{1+2}\times2\)
\(=\frac{2}{3}\)
\(\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\right).100-\left[\frac{5}{2}:\left(x+\frac{266}{100}\right)\right]:\frac{1}{2}=89\)
\(\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right).100-\left[\frac{5}{2}:\left(x+\frac{266}{100}\right)\right]:\frac{1}{2}=89\)
\(\left(1-\frac{1}{10}\right).100-\left[\frac{5}{2}:\left(x+\frac{266}{100}\right)\right]:\frac{1}{2}=89\)
\(90-\left[\frac{5}{2}:\left(x+\frac{266}{100}\right)\right]:\frac{1}{2}=89\)
\(\left[\frac{5}{2}:\left(x+\frac{266}{100}\right)\right]:\frac{1}{2}=1\)
\(\frac{5}{2}:\left(x+\frac{266}{100}\right)=\frac{1}{2}\Rightarrow x+\frac{266}{100}=5\Rightarrow x=\frac{117}{50}\)
Vậy x = 117/50
Ta có:
\(\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{9.10}\right).100\\ =\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right).100\)
\(=\left(1-\frac{1}{10}\right).100\)
\(=\frac{9}{10}.100\)
= 90
Khi đó đề bài sẽ thành : \(90-\left[\frac{5}{2}:\left(x+\frac{266}{100}\right)\right]:\frac{1}{2}=89\)
\(\Rightarrow\left[\frac{5}{2}:\left(x+\frac{266}{100}\right)\right]:\frac{1}{2}=1\)
\(\Rightarrow\frac{5}{2}:\left(x+\frac{266}{100}\right)=\frac{1}{2}\)
\(\Rightarrow x+\frac{266}{100}=5\)
\(\Rightarrow x=\frac{117}{50}\)
Vậy \(x=\frac{117}{50}\)
\(\frac{317.452-201}{451.317+116}=\frac{317.\left(451+1\right)-201}{451.317+116}=\frac{317.451+317-201}{451.317+116}\)\(=\frac{451.317+\left(317-201\right)}{451.317+116}=\frac{451.317+116}{451.317+116}\)\(=1\)
Vậy: \(\frac{317.452-201}{451.317+116}=1\)
[317 x 452 - 201][451x317+116]=[317x(451+1)-201][451x317+116]
=[317x451+317-201][451x317+116]
=[317x451-116][451x317-116]=1
nhớ nhé!
0,5\(\frac{11}{3}\):0,75 - 30%
=\(\frac{11}{6}\): \(\frac{3}{4}\)- \(\frac{3}{10}\)
=\(\frac{11}{6}\) x \(\frac{4}{3}\)-\(\frac{3}{10}\)
=\(\frac{11x2x2}{2x3x3}\)-\(\frac{27}{90}\)
= \(\frac{22}{9}\)- \(\frac{27}{90}\)
= \(\frac{220}{90}\)-\(\frac{27}{90}\)
=\(\frac{193}{90}\)
nha !