Tính nhanh
A= 4 x 2 x 25 x 5 x 175
B=42-104:(50 x 273-50 x 73)
C=3 x 53 x 6+2 x 9 x 87-18 x 40)
D=99 - 97 + 95 - 93 + 91-...+ 7 - 5 + 3 - 1
K
Khách
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9 tháng 8 2017
A = 4 . 2 . 25 . 5 . 175
A = 22 . 2 . 52 . 5 . 52 . 7
A = 22+1 . 52+1+2 . 7
A = 23 . 55 .7
A = 175000
H
9 tháng 8 2018
\(B=4^2-10^4:\left(50\cdot273-50\cdot73\right)\)
\(B=4^2-10^4:\left[50\cdot\left(273-73\right)\right]\)
\(B=4^2-10^4:\left(50\cdot200\right)\)
\(B=4^2-10^4:10^4=4^2-1=15\)
\(C=3\times53\times6+2\times9\times87-18\times40\)
\(C=18\times53+18\times87-18\times40\)
\(C=18\times\left(53+87-40\right)\)
\(C=18\times100=1800\)
8 tháng 7 2023
Bài 1:
\(101\cdot125+101\cdot25-101\cdot50\)
\(=101\cdot\left(125+25-50\right)\)
\(=101\cdot100\)
\(=10100\)
Bài 2:
\(76\cdot115+56\cdot24+59\cdot24\)
\(=76\cdot115+24\cdot\left(56+59\right)\)
\(=76\cdot115+24\cdot115\)
\(=115\cdot\left(76+24\right)\)
\(=115\cdot100\)
\(=11500\)
PN
2
KN
11 tháng 2 2020
\(\frac{x+1}{99}+\frac{x+3}{97}+\frac{x+5}{95}=\frac{x+7}{93}+\frac{x+9}{91}+\frac{x+11}{89}\)
\(\Rightarrow\frac{x+1}{99}+1+\frac{x+3}{97}+1+\frac{x+5}{95}+1\)\(=\frac{x+7}{93}+1+\frac{x+9}{91}+1+\frac{x+11}{89}+1\)
\(\Rightarrow\frac{x+100}{99}+\frac{x+100}{97}+\frac{x+100}{95}\)\(=\frac{x+100}{93}+\frac{x+100}{91}+\frac{x+100}{89}\)
\(\Rightarrow\frac{x+100}{99}+\frac{x+100}{97}+\frac{x+100}{95}\)\(-\frac{x+100}{93}-\frac{x+100}{91}-\frac{x+100}{89}=0\)
\(\Rightarrow\left(x+100\right)\left(\frac{1}{99}+\frac{1}{97}+\frac{1}{95}-\frac{1}{93}-\frac{1}{91}-\frac{1}{89}\right)=0\)
Mà \(\left(\frac{1}{99}< \frac{1}{97}< \frac{1}{95}< \frac{1}{93}< \frac{1}{91}< \frac{1}{89}\right)\)nên \(\left(\frac{1}{99}+\frac{1}{97}+\frac{1}{95}-\frac{1}{93}-\frac{1}{91}-\frac{1}{89}\right)< 0\)
\(\Rightarrow x+100=0\Leftrightarrow x=-100\)
Vậy x = -100
NH
1
KN
4 tháng 3 2020
\(\frac{x+2}{2002}+\frac{x+5}{1999}+\frac{x+201}{1803}=-3\)
\(\Rightarrow\frac{x+2}{2002}+1+\frac{x+5}{1999}+1+\frac{x+201}{1803}+1=0\)
\(\Rightarrow\frac{x+2004}{2002}+\frac{x+2004}{1999}+\frac{x+2004}{1803}=0\)
\(\Rightarrow\left(x+2004\right)\left(\frac{1}{2002}+\frac{1}{1999}+\frac{1}{1803}\right)=0\)
Dễ thấy \(\left(\frac{1}{2002}+\frac{1}{1999}+\frac{1}{1803}\right)>0\)nên x + 2004 = 0
Vậy x = -2004
NH
4
4 tháng 3 2020
\(\frac{x+2}{2002}+\frac{x+5}{1999}+\frac{x+201}{1803}=-3\)
\(\Leftrightarrow\frac{x+2}{2002}+1+\frac{x+5}{1999}+1+\frac{x+201}{1803}+1=-3+1+1+1\)
\(\Leftrightarrow\frac{x+2004}{2002}+\frac{x+2004}{1999}+\frac{x+2004}{1803}=0\)
\(\Leftrightarrow\left(x+2004\right)\left(\frac{1}{2002}+\frac{1}{1999}+\frac{1}{1803}\right)=0\)
\(\Leftrightarrow x+2004=0\left(\frac{1}{2002}+\frac{1}{1999}+\frac{1}{1803}\ne0\right)\)
<=> x=-2004
4 tháng 3 2020
a,\(\frac{x+2}{2002}+\frac{x+5}{1999}+\frac{x+201}{1803}=-3\)
\(< =>\left(\frac{x+2}{2002}+1\right)+\left(\frac{x+5}{1999}+1\right)+\left(\frac{x+201}{1803}+1\right)=0\)
\(< =>\frac{x+2004}{2002}+\frac{x+2004}{1999}+\frac{x+2004}{1803}=0\)
\(< =>\left(x+2004\right).\left(\frac{1}{2002}+\frac{1}{1999}+\frac{1}{1803}\right)=0\)
Do \(\frac{1}{2002}+\frac{1}{1999}+\frac{1}{1803}\ne0\)
\(=>x+2004=0\)
\(=>x=-2004\)
29 tháng 8 2016
a, 2 x 31 x 12 + 4 x 6 x 42 + 8 x 27 x 3
= 2 x 12 x 31 + 4 x 6 x 42 + 8 x 3 x 27
= 24 x 31 + 24 x 42 + 24 x 27
= 24 x ( 31 + 24 + 27 )
= 24 x 82
= 1968
b, 2 x 53 x 12 + 4 x 6 x 87 - 3 x 8 x 40
= 24 x 53 + 24 x 87 - 24 x 40
= 24 x ( 53 + 87 - 40 )
= 24 x 100
= 2400
c, Tương tự
18 tháng 6 2022
9: \(\dfrac{x-49}{50}+\dfrac{x-50}{49}=\dfrac{49}{x-50}+\dfrac{50}{x-49}\)
=>x-99=0
hay x=99
7: \(\Leftrightarrow\left(\dfrac{x+25}{75}+1\right)+\left(\dfrac{x+30}{70}+1\right)=\left(\dfrac{x+35}{65}+1\right)+\left(\dfrac{x+40}{60}+1\right)\)
=>x+100=0
hay x=-100
8:
Sửa đề: \(\dfrac{99-x}{101}+\dfrac{97-x}{103}+\dfrac{95-x}{105}+\dfrac{93-x}{107}=-4\)
\(\Leftrightarrow\left(\dfrac{99-x}{101}+1\right)+\left(\dfrac{97-x}{103}+1\right)+\left(\dfrac{95-x}{105}+1\right)+\left(\dfrac{93-x}{107}+1\right)=0\)
=>200-x=0
hay x=200
