39+61+8=
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(-245) + (-32) + 45 + 132 + (-8)
=(45 - 245) + (132 - 32) - 8
= -200 + 100 - 8
= -108
112 + (-61) + (-95) + (-12) + (-39)
=(112 - 12) - (61 + 39) - 95
= 100 - 100 - 95
= -95
Phương pháp giải:
- Đặt tính : Viết phép tính sao cho các chữ số cùng hàng thẳng cột với nhau.
- Tính : Cộng hoặc trừ các số theo thứ tự từ phải sang trái.
Lời giải chi tiết:
\(\dfrac{x+3}{97}+\dfrac{x+5}{95}+\dfrac{x+9}{91}=\dfrac{x+91}{9}+\dfrac{x+92}{8}+\dfrac{x+61}{39}\)
=> \(\dfrac{x+3}{97}+1+\dfrac{x+5}{95}+1+\dfrac{x+9}{91}+1=\dfrac{x+91}{9}+1+\dfrac{x+92}{8}+1+\dfrac{x+61}{39}+1\)
=> \(\dfrac{x+100}{97}+\dfrac{x+100}{95}+\dfrac{x+100}{91}=\dfrac{x+100}{9}+\dfrac{x+100}{8}+\dfrac{x+100}{39}\)
=> \(\dfrac{x+100}{97}+\dfrac{x+100}{95}+\dfrac{x+100}{91}-\dfrac{x+100}{9}-\dfrac{x+100}{8}-\dfrac{x+100}{39}=0\)
=> \(\left(x+100\right).\left(\dfrac{1}{97}+\dfrac{1}{95}+\dfrac{1}{91}-\dfrac{1}{9}-\dfrac{1}{8}-\dfrac{1}{39}\right)=0\)
=> x = - 100 (do \(\dfrac{1}{97}+\dfrac{1}{95}+\dfrac{1}{91}-\dfrac{1}{9}-\dfrac{1}{8}-\dfrac{1}{39}\ne0\)
Ta có: \(\dfrac{x+3}{97}+\dfrac{x+5}{95}+\dfrac{x+9}{91}=\dfrac{x+91}{9}+\dfrac{x+92}{8}+\dfrac{x+61}{39}\)
\(\Leftrightarrow\dfrac{x+3}{97}+1+\dfrac{x+5}{95}+1+\dfrac{x+9}{91}+1=\dfrac{x+91}{9}+1+\dfrac{x+92}{8}+1+\dfrac{x+61}{39}+1\)
\(\Leftrightarrow\dfrac{x+100}{97}+\dfrac{x+100}{95}+\dfrac{x+100}{91}=\dfrac{x+100}{9}+\dfrac{x+100}{8}+\dfrac{x+100}{39}\)
\(\Leftrightarrow\dfrac{x+100}{97}+\dfrac{x+100}{95}+\dfrac{x+100}{91}-\dfrac{x+100}{9}-\dfrac{x+100}{8}-\dfrac{x+100}{39}=0\)
\(\Leftrightarrow\left(x+100\right)\left(\dfrac{1}{97}+\dfrac{1}{95}+\dfrac{1}{91}-\dfrac{1}{9}-\dfrac{1}{8}-\dfrac{1}{39}\right)=0\)
mà \(\dfrac{1}{97}+\dfrac{1}{95}+\dfrac{1}{91}-\dfrac{1}{9}-\dfrac{1}{8}-\dfrac{1}{39}\ne0\)
nên x+100=0
hay x=-100
Vậy: S={-100}
So Sánh:
a) 9200 và 27133
9200 = 3400
27133 = 3399
3399 < 3400. Vậy 27133 < 9200
b) 2180 và 861
861 = 2183
2180 < 2183. Vậy 2180 < 861
Còn mấy câu kia mk chưa nghĩ ra
Hk tốt
Ta có\(\frac{x+3}{97}+\frac{x+5}{95}+\frac{x+9}{91}=\frac{x+91}{9}+\frac{x+92}{8}+\frac{x+61}{39}\)
<=> \(\left(\frac{x+3}{97}+1\right)+\left(\frac{x+5}{95}+1\right)+\left(\frac{x+9}{91}+1\right)=\left(\frac{x+91}{9}+1\right)+\left(\frac{x+92}{8}+1\right)+\left(\frac{x+61}{39}+1\right)\)
<=>\(\frac{x+100}{97}+\frac{x+100}{95}+\frac{x+100}{91}=\frac{x+100}{9}+\frac{x+100}{8}+\frac{x+100}{39}\)
<=>\(\frac{x+100}{97}+\frac{x+100}{95}+\frac{x+100}{91}-\frac{x+100}{9}-\frac{x+100}{8}-\frac{x+100}{39}=0\)
<=> \(\left(x+100\right)\left(\frac{1}{97}+\frac{1}{95}+\frac{1}{91}-\frac{1}{9}-\frac{1}{8}-\frac{1}{39}\right)=0\)
Do \(\frac{1}{97}+\frac{1}{95}+\frac{1}{91}-\frac{1}{9}-\frac{1}{8}-\frac{1}{39}\ne0\)
Nên x+100=0 => x=-100
2 . 53 . 27 + 6 . 9 . 87 - 3 . 18 . 40
= 2. 53 . 9 . 3 + 2. 3 .9 .87 - 3 .2 .9 .40
= 54.53+54.87-54.40
=54(53+87-40)
=54. 100=5400
12 . 25 + 29 . 25 + 59 . 25
= 25 . ( 12+29+59)
=25 . 90
=2250
43 . 37 + 93 . 43 + 57 . 61 + 69 . 57
= 43.(37+93)+57(61+69)
=43.130+57.130
=130.(43+57)
=130.100
=13 000
12 . 53 + 53 . 172 - 53
=53(12+172-1)
=53.184
=9752
35 . 54 + 35 . 58 + 65 . 75 + 65 . 45
=35(54+58)+65(75+45)
=35.112+65.120
=3920+7800
=11720
39 . 8 + 70 . 3 + 31 . 8
=8(39+31)+70.3
=8.70+70.3
=70.(8+3)
=70.11
=770
39.(-67)-110.61+(-39).43
= (-39).67-110.61+(-39).43
=-39.(67+43)-110.61
=39.110-110.61
=110.(39-61)
=110.(-22)
=-2420
\(39\cdot\left(-67\right)-110\cdot61+\left(-39\right)\cdot43=\left(-39\right)\cdot67+\left(-39\right)\cdot43-110\cdot61\)
\(=\left(-39\right)\left(67+43\right)-110\cdot61=\left(-39\right)\left(110\right)-110\cdot61=110\cdot\left[\left(-39\right)-61\right]\)
\(=110\cdot\left(-100\right)=-11000\)
Chúc bạn học tốt ^^!!!
C = 39.42 + 24.61 - 39.18
= 39.(42 - 18) + 24.61
= 39.24 + 24.61
= 24.(39 + 61)
= 24.100
= 2400
bang 108 do
108 bn à