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\(\left(\frac{3}{4}-\frac{13}{11}+\frac{7}{5}\right)-\left(3-\frac{1}{2}-\frac{35}{11}\right)+\left(\frac{11}{4}-\frac{2}{5}\right)\)
= \(\frac{3}{4}-\frac{13}{11}+\frac{7}{5}-3+\frac{1}{2}+\frac{35}{11}+\frac{11}{4}-\frac{2}{5}\)
= \(\left(\frac{3}{4}+\frac{11}{4}+\frac{1}{2}\right)\left(-\frac{13}{11}+\frac{35}{11}\right)+\left(\frac{7}{5}-\frac{2}{5}\right)-3\)
= \(8+2+1-3\)
= \(8\)
#)Giải :
\(\left(\frac{3}{4}-\frac{13}{11}+\frac{7}{5}\right)-\left(3-\frac{1}{2}-\frac{35}{11}\right)+\left(\frac{11}{4}-\frac{2}{5}\right)\)
\(=\frac{3}{4}-\frac{13}{11}+\frac{7}{5}-3+\frac{1}{2}+\frac{35}{11}+\frac{11}{4}-\frac{2}{5}\)
\(=\left(\frac{3}{4}+\frac{11}{4}\right)+\left(-\frac{13}{11}+\frac{35}{11}\right)+\left(\frac{7}{5}-\frac{2}{5}\right)-3+\frac{1}{2}\)
\(=\frac{7}{2}+2+1-3+\frac{1}{2}\)
\(=\frac{7}{2}+\frac{1}{2}\)
\(=4\)
1/4+2/5+6/8+2/15+6/7
=(1/4+6/8)+(2/5+2/15)+6/7
=(2/8+6/8)+(6/15+2/15)+6/7
=1+8/15+6/7
=1+56/105+90/105
=1+146/105
=1+105/105+41/105
=1+1+41/105
=2+41/105
=2 và 41/105
2 và 41/105 là hỗn số nha
1/4+2/5+6/8+2/15+6/7
Ta có:
1/4=1-3/4
6/8=3/4
2/15=2/3*5=1/3-1/5
==> 1-3/4+2/5+3/4+1/3-1/5+6/7
=1+1/3+1/5+6/7
=(105+35+21+90)/105
=251/105.
\(\frac{2n-1}{n+8}-\frac{n-14}{n+8}\)
a, \(=\frac{\left(2n-1\right)-\left(n-14\right)}{n+8}\)
\(=\frac{2n-1-n+14}{n+8}\)
\(=\frac{n+13}{n+8}\)
Có : \(n+13=n+5+8\)
Vì \(n+8⋮n+8\)
\(=>5⋮n+8\)
\(=>n+8\inƯ\left(5\right)=\left\{1;-1;5;-5\right\}\)
TH1 : n + 8 = 1
n = 1 - 8
n = 7 ( thỏa mãn số nguyên tố )
Th2 : n + 8 = -1
n = -1 - 8
n = -9 ( không thỏa mãn )
TH3 : n + 8 = 5
n = 5 - 8
n = -3 ( không thỏa mãn )
Th4 : n + 8 = -5
n = -5 - 8
n = -13 ( thỏa mãn )
b, ( đã tìm ra ở phần a )
\(n\in\left\{7;-9;-3;-13\right\}\)
Tk mk nha :D
ta có
đặt 3/5+3/7-3/11=N=> N=3*(1/5+1/7-1/11)
đặt 4/5+4/7-4/11=P=> P=4*(1/5+1/7-1/11)
=> N/P=3*(1/5+1/7-1/11)/4*(1/5+1/7-1/11)=> M=3/4
\(=\frac{-\frac{1}{9}+1-\frac{2}{10}+1-\frac{3}{11}+1-...-\frac{92}{100}+1}{\frac{1}{9}+\frac{1}{10}+...+\frac{1}{100}}\)
\(=\frac{\frac{8}{9}+\frac{8}{10}+\frac{8}{11}+...+\frac{8}{100}}{\frac{1}{9}+\frac{1}{10}+...+\frac{1}{100}}\)
\(=\frac{8\left(\frac{1}{9}+\frac{1}{10}+\frac{1}{11}+...+\frac{1}{100}\right)}{\frac{1}{9}+\frac{1}{10}+\frac{1}{11}+...+\frac{1}{100}}\)
= 8
\(A=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{200}\)
\(\Rightarrow\frac{1}{A}=2+3+4+...+200\)
\(\frac{1}{A}=\left(200+2\right).\left\{\left[200-2\right]+1\right\}:2=20099\)
Vậy \(A=\frac{1}{20099}\)
ĐTV sai òi
VD A = 1/2 + 1/3
A = 1/6
1/A = 6
nhưng 2 + 3 = 5