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d=43/37d=43/37 × 23/15−43/3723/15−43/37 × 8/15+31/378/15+31/37
=43/37=43/37 x 23/15+43/3723/15+43/37 x (−8/15)+31/37(−8/15)+31/37
=43/37(23/15−8/15)+31/37=43/37(23/15−8/15)+31/37
=43/37.1+31/37=43/37.1+31/37
=2=2
E=−5/24E=−5/24 × 9/11−5/249/11−5/24 × 2/11+7/122/11+7/12
=−5/24(9/11+2/11)+7/12=−5/24(9/11+2/11)+7/12
=−5/24.1+7/12=−5/24.1+7/12
=3/8=3/8
a)3/7–−11/14−7/3×6/7a)3/7–−11/14−7/3×6/7
=3/7−−11/14−2=3/7−−11/14−2
=3/7+11/14−2=3/7+11/14−2
=−11/14=−11/14
b)6/5:3/2+−2/3×9/14b)6/5:3/2+−2/3×9/14
=4/5−3/7=4/5−3/7
=13/35
d=43/37d=43/37 × 23/15−43/3723/15−43/37 × 8/15+31/378/15+31/37
=43/37=43/37 x 23/15+43/3723/15+43/37 x (−8/15)+31/37(−8/15)+31/37
=43/37(23/15−8/15)+31/37=43/37(23/15−8/15)+31/37
=43/37.1+31/37=43/37.1+31/37
=2=2
E=−5/24E=−5/24 × 9/11−5/249/11−5/24 × 2/11+7/122/11+7/12
=−5/24(9/11+2/11)+7/12=−5/24(9/11+2/11)+7/12
=−5/24.1+7/12=−5/24.1+7/12
=3/8=3/8
a)3/7–−11/14−7/3×6/7a)3/7–−11/14−7/3×6/7
=3/7−−11/14−2=3/7−−11/14−2
=3/7+11/14−2=3/7+11/14−2
=−11/14=−11/14
b)6/5:3/2+−2/3×9/14b)6/5:3/2+−2/3×9/14
=4/5−3/7=4/5−3/7
=13/35
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\(\text{24) 15 (4-7)- 15 (5-3)}\)
\(=15.\left[\left(4-7\right)-\left(5-3\right)\right]\)
\(=15.\left[4-7-5+3\right]\)
\(=15.\left(-5\right)\)
\(=-75\)
\(\text{25) 73(8-59) -59(8-73)}\)
\(=73.8-73.59-59.8+59.73\)
\(=8\left(73-59\right)+59\left(-73+73\right)\)
\(=8.14+59.0\)
\(=112\)
24) 15 (4-7)- 15 (5-3)
=15. 4 − 7 − 5 − 3
= 15. 4 − 7 − 5 + 3
= 15. −5
= −75
25) 73(8-59) -59(8-73)
= 73.8 − 73.59 − 59.8 + 59.73
= 8 73 − 59 + 59 −73 + 73
= 8.14 + 59.0
= 112
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\(A=\frac{19}{24}-\frac{1}{2}-\frac{1}{3}-\frac{7}{24}\)
\(A=\frac{19}{24}+\frac{-1}{2}+\frac{-1}{3}+\frac{-7}{24}\)
\(A=\left(\frac{19}{24}+\frac{-7}{24}\right)+\frac{-1}{2}+\frac{-1}{3}\)
\(A=\frac{1}{2}+\frac{-1}{2}+\frac{-1}{3}\)
\(A=0+\frac{-1}{3}=\frac{-1}{3}\)
\(B=\frac{7}{24}+\frac{5}{6}+\frac{1}{4}-\frac{3}{7}-\frac{5}{15}\)
\(B=\frac{7}{24}+\frac{5}{6}+\frac{1}{4}+\frac{-3}{7}+\frac{-1}{3}\)
\(B=\left(\frac{49}{168}+\frac{140}{168}+\frac{42}{168}\right)+\left(\frac{-72}{168}+\frac{-56}{168}\right)\)
\(B=\frac{231}{168}+\frac{-128}{168}=\frac{103}{168}\)
Có: \(A=\frac{-1}{3}=\frac{\left(-1\right)\cdot56}{3\cdot56}=\frac{-56}{168}\)
Mặt khác: \(-56< 103\)
\(\Rightarrow\)\(\frac{-56}{168}< \frac{103}{168}\)
\(hay\) \(A< B\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\frac{19}{24}-\frac{1}{2}-\frac{1}{3}-\frac{7}{24}\)
\(A=\frac{19}{24}-\frac{12}{24}-\frac{8}{24}-\frac{7}{24}\)
\(A=\frac{19-12-8-7}{24}\)
\(A=\frac{-8}{24}\)\(=\frac{-1}{3}\)
\(B=\frac{7}{24}+\frac{5}{6}+\frac{1}{4}-\frac{3}{7}-\frac{5}{15}\)
\(B=\frac{7}{24}+\frac{30}{24}+\frac{6}{24}-\frac{3}{7}-\frac{1}{3}\)
\(B=\frac{7}{24}+\frac{30}{24}+\frac{6}{24}-\frac{3}{7}-\frac{8}{24}\)
\(B=\frac{7+30+6-8}{24}-\frac{3}{7}\)
\(B=\frac{21}{24}-\frac{3}{7}\)
\(B=\frac{7}{8}-\frac{3}{7}\)
\(\Rightarrow B=\frac{25}{56}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
2x - 3 = 7 3 . ( x + 7 ) = 15 24 . 3 . ( 5 - x ) = 27 15 + 4 . ( x - 2 ) = 95
2x = 3 + 7 x + 7 = 15 : 3 3 . ( 5 - x ) = 27 : 24 4 . ( x - 2 ) = 95 - 15
2x = 10 x + 7 = 5 3 . ( 5 - x ) = 1,125 4 . ( x - 2 ) = 80
x = 10 : 2 x = 5 - 7 5 - x = 1,125 : 3 x - 2 = 80 : 4
x = 5 x = ( -2 ) 5 - x = 0,375 x - 2 = 20
x = 5 - 0,375 x = 20 + 2
x = 4,625 x = 22
TL :
( 15 + 7 ) . 3 - 24
= 22 . 3 - 24
= 66 - 24
= 42
_HT_