53/7x((13/4-8/43)-53/8:4/7
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7)
53.(51+4)+53.(49+96)+53
=53.55+53.145+53
=53.(55+145+1)
=53.201
=10653
8)
42.(15+96)+6.(25+4).7
=42.111+(6.7).29
=42.111+42.29
=42.(111+29)
=42.140
=5880
9)
45.(13+78)+9.(87+22).5
=45.91+(9.5).109
=45.91+45.109
=45.(91+109)
=45.200
=9000
10)
16.(27+75)+8.(53+25).2
=16.102+(8.2).78
=16.102+16.78
=16.(102+78)
=16.180
=2880
7)
53.(51+4)+53.(49+96)+53
=53.55+53.145+53
=53.(55+145+1)
=53.201
=10653
8)
42.(15+96)+6.(25+4).7
=42.111+(6.7).29
=42.111+42.29
=42.(111+29)
=42.140
=5880
9)
45.(13+78)+9.(87+22).5
=45.91+(9.5).109
=45.91+45.109
=45.(91+109)
=45.200
=9000
10)
16.(27+75)+8.(53+25).2
=16.102+(8.2).78
=16.102+16.78
=16.(102+78)
=16.180
=2880
Phương pháp giải:
Thực hiện phép tính lần lượt từ trái sang phải.
Lời giải chi tiết:
83 − 7 − 6 = 70 83 − 13 = 70
53 − 9 − 4 = 40 53 − 13 = 40
73 − 5 − 8 = 60 73 − 13 = 60
\(\left(\frac{3}{4}+\frac{2}{11}-\frac{8}{13}\right)+\left(\frac{1}{4}-\frac{5}{13}\right)\)
\(=\frac{3}{4}+\frac{2}{11}-\frac{8}{13}+\frac{1}{4}-\frac{5}{13}\)
\(=\left(\frac{3}{4}+\frac{1}{4}\right)+\frac{2}{11}-\left(\frac{8}{13}+\frac{5}{13}\right)\)
\(=1+\frac{2}{11}-1\)
\(=\frac{2}{11}\)
\(\left(\frac{9}{53}+\frac{21}{31}\right)+\left(\frac{44}{53}-\frac{16}{7}\right)+\frac{10}{31}\)
\(=\frac{9}{53}+\frac{21}{31}+\frac{44}{53}-\frac{16}{7}+\frac{10}{31}\)
\(=\left(\frac{9}{53}+\frac{44}{53}\right)+\left(\frac{21}{31}+\frac{10}{31}\right)-\frac{16}{7}\)
\(=1+1-\frac{16}{7}\)
\(=2-\frac{16}{7}\)
\(=\frac{14}{7}-\frac{16}{7}\)
\(=\frac{-2}{7}\)
a) \(\left(-4\right)\left(-3\right)\left(-125\right).25.\left(-8\right)=\left[25\left(-4\right)\right]\left[\left(-125\right)\left(-8\right)\right]\left(-3\right)\)
\(=\left(-100\right).1000.\left(-3\right)=300.1000=300000\)
b) \(\left(-4\right).9\left(-125\right).25.\left(-8\right)=\left[25\left(-4\right)\right]\left[\left(-125\right)\left(-8\right)\right].9\)
\(=\left(-100\right).1000.9=\left(-100\right).9000=-900000\)
c) \(7.\left(-25\right)\left(-3\right).2.\left(-4\right)=\left[\left(-25\right)\left(-4\right)\left(-3\right)\right].7.2=-4200\)
d) \(\left[93-\left(20-7\right)\right]\div16=\left(93-13\right)\div16=80\div16=5\)
e) \(53-\left(-51\right)+\left(-53\right)+49=\left(53-53\right)+\left(51+49\right)=100\)
f) \(168-49+\left(-68\right)+4=168-49-68+4=55\)
\(\dfrac{3}{8}+\dfrac{3}{4}=\dfrac{3+3\cdot2}{8}=\dfrac{9}{8}\)
\(\dfrac{3}{4}\cdot12=\dfrac{3\cdot12}{4}=\dfrac{36}{4}=9\)
\(\dfrac{1}{6}-\dfrac{1}{12}=\dfrac{2}{12}-\dfrac{1}{12}=\dfrac{1}{12}\)
\(\dfrac{13}{7}-\dfrac{9}{5}=\dfrac{13\cdot5-9\cdot7}{7\cdot5}=\dfrac{2}{35}\)
\(\dfrac{3}{8}:2=\dfrac{3}{8\cdot2}=\dfrac{3}{16}\)
\(A=\frac{7}{3\times13}+\frac{7}{13\times23}+...+\frac{7}{53\times63}\)
\(A=\frac{7}{10}.\left[\left(\frac{1}{3}-\frac{1}{13}\right)+\left(\frac{1}{13}-\frac{1}{23}\right)+....+\left(\frac{1}{53}-\frac{1}{63}\right)\right]\)
\(A=\frac{7}{10}.\left(\frac{1}{3}-\frac{1}{13}+\frac{1}{13}-\frac{1}{23}+....+\frac{1}{53}-\frac{1}{63}\right)\)
\(A=\frac{7}{10}.\left(\frac{1}{3}-\frac{1}{63}\right)\)
\(A=\frac{7}{10}.\frac{20}{63}\)
\(A=\frac{2}{9}\)