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A=
7.31
4
+
7.41
6
+
10.41
9
+
10.57
7
⇒
5
A
=
35.31
4
+
35.41
6
+
50.41
9
+
50.57
7
\dfrac{A}{5}=\dfrac{1}{31}-\dfrac{1}{35}+\dfrac{1}{35}-\dfrac{1}{41}+\dfrac{1}{41}-\dfrac{1}{50}+\dfrac{1}{50}-\dfrac{1}{57}
5
A
=
31
1
−
35
1
+
35
1
−
41
1
+
41
1
−
50
1
+
50
1
−
57
1
\dfrac{A}{5}=\dfrac{1}{31}-\dfrac{1}{57}=\dfrac{26}{31.57}\Rightarrow A=\dfrac{130}{31.57}
5
A
=
31
1
−
57
1
=
31.57
26
⇒A=
31.57
130
\(A=\dfrac{4}{7.31}+\dfrac{6}{7.41}+\dfrac{9}{10.41}+\dfrac{7}{10.57}\Rightarrow\dfrac{A}{5}=\dfrac{4}{35.31}+\dfrac{6}{35.41}+\dfrac{9}{50.41}+\dfrac{7}{50.57}\)
\(\dfrac{A}{5}=\dfrac{1}{31}-\dfrac{1}{35}+\dfrac{1}{35}-\dfrac{1}{41}+\dfrac{1}{41}-\dfrac{1}{50}+\dfrac{1}{50}-\dfrac{1}{57}\)
\(\dfrac{A}{5}=\dfrac{1}{31}-\dfrac{1}{57}=\dfrac{26}{31.57}\Rightarrow A=\dfrac{130}{31.57}\)
\(\dfrac{B}{2}=\dfrac{7}{38.31}+\dfrac{5}{38.41}+\dfrac{3}{46.43}+\dfrac{11}{46.57}\Rightarrow\dfrac{B}{2}=\dfrac{1}{31}-\dfrac{1}{57}=\dfrac{26}{31.57}\)
\(\Rightarrow B=\dfrac{52}{31.57}\)
\(\dfrac{A}{B}=\dfrac{130}{31.51}:\dfrac{52}{31.57}=\dfrac{5}{2}\)
Ta có : `A:5=4/35.31+6/35.41++9/50.41+7/50.57`
`A/5 = 1/31 - 1/35 +1/35-1/41+1/41-1/50+1/50-1/57`
`A/5 =1/31-1/57` `(1)`
Lại có : `B:2=7/38.31 +5/38.43+3/46.43+11/46.57`
`B/2 =1/31-1/38+1/38-1/43+1/43-1/46+1/46-1/57`
`B/2 =1/31-1/57` `(2)`
Từ `(1)` và `(2)` `=>A/5 =B/2`
`=>A/B=5/2`
\(\frac{A}{5}=\frac{4}{35.31}+\frac{6}{35.41}+\frac{9}{50.41}+\frac{7}{50.57}=\frac{35-31}{35.31}+\frac{41-35}{35.41}+\frac{50-41}{50.41}+\frac{57-50}{50.57}\)
\(\frac{A}{5}=\frac{1}{31}-\frac{1}{35}+\frac{1}{35}-\frac{1}{41}+\frac{1}{41}-\frac{1}{50}+\frac{1}{50}-\frac{1}{57}=\frac{1}{31}-\frac{1}{57}\)=> A = 5. \(\left(\frac{1}{31}-\frac{1}{57}\right)\)
\(\frac{B}{2}=\frac{7}{38.31}+\frac{5}{38.43}+\frac{3}{43.46}+\frac{11}{46.57}=\frac{38-31}{31.38}+\frac{43-38}{38.43}+\frac{46-43}{43.46}+\frac{57-46}{46.57}\)
\(\frac{B}{2}=\frac{1}{31}-\frac{1}{38}+\frac{1}{38}-\frac{1}{43}+\frac{1}{43}-\frac{1}{46}+\frac{1}{46}-\frac{1}{57}=\frac{1}{31}-\frac{1}{57}\)=> B = 2.\(\left(\frac{1}{31}-\frac{1}{57}\right)\)
A/B = 5/2
A = \(\frac{4}{7}.31+\frac{6}{7}.41+\frac{9}{10}.41+\frac{7}{10}.57\)
= \(\left[\left(\frac{4}{7}+\frac{6}{7}\right).\left(31+41\right)\right]+\left[\left(\frac{9}{10}+\frac{7}{10}\right).\left(41+57\right)\right]\)
= \(\frac{10}{7}.72+\frac{8}{5}.98\)
= \(\left(\frac{10}{7}+\frac{8}{5}\right).\left(72+98\right)\)
= \(\left(\frac{50}{35}+\frac{56}{35}\right).170\)
= \(\frac{106}{35}.170\)
= \(\frac{3604}{7}\)
B = \(\frac{7}{19}.31+\frac{5}{19}.43+\frac{3}{23}.43+\frac{11}{23}.57\)
= \(\left[\left(\frac{7}{19}+\frac{5}{19}\right).\left(31+43\right)\right]+\left[\left(\frac{3}{23}+\frac{11}{23}\right).\left(43+57\right)\right]\)
= \(\frac{12}{19}.74+\frac{14}{23}.100\)
= \(\left(\frac{12}{19}+\frac{14}{23}\right).\left(100+74\right)\)
= \(\left(\frac{276}{437}+\frac{266}{437}\right).174\)
= \(\frac{542}{437}.174\)
= \(\frac{79674}{437}\)
A : B = \(\frac{3604}{7}:\frac{19674}{437}=\frac{3604.437}{7.19674}=\frac{1802.2.437}{7.9837.2.}=\frac{1802.437}{7.9837}\)
\(A=\dfrac{4}{7.31}+\dfrac{6}{7.41}+\dfrac{9}{10.41}+\dfrac{7}{10.57}\)
\(=\dfrac{20}{35.31}+\dfrac{30}{35.41}+\dfrac{45}{50.41}+\dfrac{35}{50.57}\)
\(=5\left(\dfrac{4}{35.31}+\dfrac{6}{35.41}+\dfrac{9}{50.41}+\dfrac{7}{50.57}\right)\)
\(=5\left(\frac{1}{31}-\dfrac{1}{35}+\dfrac{1}{35}-\dfrac{1}{41}+\dfrac{1}{41}-\dfrac{1}{50}+\dfrac{1}{50}-\dfrac{1}{57}\right)\)
\(=5\left(\dfrac{1}{31}-\dfrac{1}{57}\right)\)
\(B=\dfrac{7}{19.31}+\dfrac{5}{19.43}+\dfrac{3}{23.43}+\dfrac{11}{23.57}\)
\(=\dfrac{14}{34.31}+\dfrac{10}{38.43}+\dfrac{6}{46.43}+\dfrac{22}{46.57}\)
\(=2\left(\dfrac{7}{34.31}+\dfrac{5}{38.43}+\dfrac{3}{46.43}+\dfrac{11}{46.57}\right)\)
\(=2\left(\dfrac{1}{31}-\dfrac{1}{34}+\dfrac{1}{34}-\dfrac{1}{43}+\dfrac{1}{43}-\dfrac{1}{46}+\dfrac{1}{46}-\dfrac{1}{57}\right)\)
\(=2\left(\dfrac{1}{31}-\dfrac{1}{57}\right)\)
\(\Rightarrow\dfrac{A}{B}=\frac{5\left(\dfrac{1}{31}-\dfrac{1}{57}\right)}{2\left(\dfrac{1}{31}-\dfrac{1}{57}\right)}=\dfrac{5}{2}\)
A=47.31+67.41+910.41+710.57A=47.31+67.41+910.41+710.57
=2035.31+3035.41+4550.41+3550.57=2035.31+3035.41+4550.41+3550.57
=5(435.31+635.41+950.41+750.57)=5(435.31+635.41+950.41+750.57)
=5(131−135+135−141+141−150+150−157)=5(131−135+135−141+141−150+150−157)
=5(131−157)=5(131−157)
B=719.31+519.43+323.43+1123.57B=719.31+519.43+323.43+1123.57
=1434.31+1038.43+646.43+2246.57=1434.31+1038.43+646.43+2246.57
=2(734.31+538.43+346.43+1146.57)=2(734.31+538.43+346.43+1146.57)
=2(131−134+134−143+143−146+146−157)=2(131−134+134−143+143−146+146−157)
=2(131−157)=2(131−157)
⇒AB=5131−1572131−157=52
a) Ta có: \(A=\dfrac{4}{7\cdot31}+\dfrac{6}{7\cdot41}+\dfrac{9}{10\cdot41}+\dfrac{7}{10\cdot57}\)
\(=\dfrac{20}{31\cdot35}+\dfrac{30}{35\cdot41}+\dfrac{45}{41\cdot50}+\dfrac{35}{50\cdot57}\)
\(=5\left(\dfrac{4}{31\cdot35}+\dfrac{6}{35\cdot41}+\dfrac{9}{41\cdot50}+\dfrac{7}{50\cdot57}\right)\)
\(=5\left(\dfrac{1}{31}-\dfrac{1}{35}+\dfrac{1}{35}-\dfrac{1}{41}+\dfrac{1}{41}-\dfrac{1}{50}+\dfrac{1}{50}-\dfrac{1}{57}\right)\)
\(=5\left(\dfrac{1}{31}-\dfrac{1}{57}\right)\)
Ta có: \(B=\dfrac{7}{19\cdot31}+\dfrac{5}{19\cdot43}+\dfrac{3}{23\cdot43}+\dfrac{11}{23\cdot57}\)
\(=\dfrac{14}{31\cdot38}+\dfrac{10}{38\cdot43}+\dfrac{6}{43\cdot46}+\dfrac{22}{46\cdot57}\)
\(=2\left(\dfrac{7}{31\cdot38}+\dfrac{5}{38\cdot43}+\dfrac{3}{43\cdot46}+\dfrac{11}{46\cdot57}\right)\)
\(=2\left(\dfrac{1}{31}-\dfrac{1}{38}+\dfrac{1}{38}-\dfrac{1}{43}+\dfrac{1}{43}-\dfrac{1}{46}+\dfrac{1}{46}-\dfrac{1}{57}\right)\)
\(=2\left(\dfrac{1}{31}-\dfrac{1}{57}\right)\)
Suy ra: \(\dfrac{A}{B}=\dfrac{5\left(\dfrac{1}{31}-\dfrac{1}{57}\right)}{2\left(\dfrac{1}{31}-\dfrac{1}{57}\right)}=\dfrac{5}{2}\)
