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a, \(\left|x-3\right|-2\left(-4\right)=\left(-7\right)\left(-12\right)\)
\(\Leftrightarrow\left|x-3\right|+8=84\Leftrightarrow\left|x-3\right|=76\Leftrightarrow x-3=\pm76\)
TH1 : \(x-3=76\Leftrightarrow x=79\)
TH2 : \(x-3=-76\Leftrightarrow x=-73\)
b, \(\left|x+2\right|+3=3^3=27\Leftrightarrow\left|x+2\right|=24\Leftrightarrow x+2=\pm24\)
TH1 : \(x+2=24\Leftrightarrow x=22\)
TH2 : \(x+2=-24\Leftrightarrow x=-26\)
a)7^3.25-7^3.18
=7^3.(25-18)
=7^3.7
=7^3+1
=7^4
b)2^4.15-[173-(14-9)^2]
=16.15-[173-5^2]
=240-[173-25]
=240-148
=92
c)53.147+53^2+47.200
=53.147+53.53+47.200
=53.(147+53)+47.200
=53.200+47.200
=200.(53+47)
=200.100
=20000
d)36:{280:[240-(95+3.25)]}
=36:{280:[240-(95+75)]}
=36:{280:[240-170]}
=36:{280:70}
=36:4
=9
a)=1+(2-3-4+5)+(6-7-8+9)+...+(994-995-996+997)+998
=1+0+0+...+0+998
=999
b)=(1-3)+(5-7)+...+(97-99)+101
=-2+(-2)+...+(-2)+101 (50 số hạng -2)
=-2.50+101
=-100+101
=1
a) = (2-3-4+5) + ..... + (993 - 994 - 995 + 996) + 997 + 998 + 1
= 998 + 997 + 1 = 1996
\(\frac{3}{7.10}+\frac{3}{10.13}+....+\frac{3}{100.103}\)
\(=\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+....+\frac{1}{100}-\frac{1}{103}\)
\(=\frac{1}{7}-\frac{1}{103}\)
\(=\frac{96}{721}\)
\(\frac{2}{7.10}+\frac{2}{10.13}+...+\frac{2}{100.103}\)
\(=\frac{2}{3}\left(\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{100}-\frac{1}{103}\right)\)
\(=\frac{2}{3}\left(\frac{1}{7}-\frac{1}{103}\right)\)
\(=\frac{2}{3}.\frac{96}{721}\)
\(=\frac{64}{721}\)
\(A=\)\(\frac{3}{7.10}+\frac{3}{10.13}+...+\frac{3}{100.103}\)
\(A=\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{100}-\frac{1}{103}\)
\(A=\frac{1}{7}-\frac{1}{103}\)
\(A=\frac{96}{721}\)
\(B=\frac{2}{7.10}+\frac{2}{10.13}+...+\frac{2}{100.103}\)
\(B=2\left(\frac{1}{7.10}+\frac{1}{10.13}+...+\frac{1}{100.103}\right)\)
\(3B=2.3\left(\frac{1}{7.10}+\frac{1}{10.13}+...+\frac{1}{100.103}\right)\)
\(3B=2\left(\frac{3}{7.10}+\frac{3}{10.13}+...+\frac{3}{100.103}\right)\)
\(3B=2\left(\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{100}-\frac{1}{103}\right)\)
\(3B=2\left(\frac{1}{7}-\frac{1}{103}\right)\)
\(3B=2.\frac{96}{721}\)
\(3B=\frac{192}{721}\)
\(\Rightarrow B=\frac{192}{721}:3\)
\(B=\frac{64}{721}\)
\(A=\frac{3}{7.10}+\frac{3}{10.13}+...+\frac{3}{100.103}\)
\(A=\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{100}-\frac{1}{103}\)
\(A=\frac{1}{7}-\frac{1}{103}\)
\(A=\frac{96}{721}\)
Vậy \(A=\frac{96}{721}\)
\(B=\frac{2}{7.10}+\frac{2}{10.13}+...+\frac{2}{100.103}\)
\(B=\frac{2}{3}.\left(\frac{3}{7.10}+\frac{3}{10.13}+...+\frac{3}{100.103}\right)\)
\(B=\frac{2}{3}.\left(\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{100}-\frac{1}{103}\right)\)
\(B=\frac{2}{3}.\left(\frac{1}{7}-\frac{1}{103}\right)\)
\(B=\frac{2}{3}.\frac{96}{721}\)
\(B=\frac{64}{721}\)
Vậy \(B=\frac{64}{721}\)
_Chúc bạn học tốt_
a) 13,6 + 14,5 + 15,4 - 4,6 - 5,5 - 6,4
= (13,6 - 4,6) + (14,5 - 5,5) + (15,4 - 6,4)
= 9 + 9 + 9
= 9 . 3
= 27.
b) 7/8 x 1/2 + 7/8 x 3/2 - 7/8 x 2
= 7/8 x (1/2 + 3/2 - 2)
= 7/8 x 0 = 0
5) - 12 + 3(-x + 7) = -18
3(-x+7) = -18- -12
3(-x+7) = -30
-x+7 = -30 : 3
-x+7 = -10
-x = -10 -7 = -17
Vậy x= 17
6) 24 : (3x – 2) = -3
( 3x - 2) = 24: -3
3x - 2 = -8
3x = -8 +2
3x = -6
x = -6:3 = -2
Vậy x= -2
7) -45 : 5.(-3 – 2x) = 3
5(-3-2x) = -45: 3
5( -3-2x) = -15
-3-2x = -15 :5
-3- 2x = -3
2x = -3--3
2x= 0
Vậy x= 0
\(P=1+2+3+...+50\)
Số số hạng của \(P\)là: \(\frac{50-1}{1}+1=50\)
Tổng của \(P\)là: \(\frac{\left(1+50\right).50}{2}=1275\)
\(Q=1+4+7+...+2014+2017\)
Số số hạng của \(Q\)là: \(\frac{2017-1}{3}+1=673\)
Tổng của \(Q\)là: \(\frac{\left(1+2017\right).673}{2}=679057\)
Số số hạng : \(P=50-1+1=50\)
Tổng : \(P=\left(1+50\right)\times25=1275\)
Số số hạng :\(Q=\frac{2017-1}{3}+1=673\)
Tổng : \(Q=\left(1+2017\right)\times\frac{673}{2}679057\)
\(\dfrac{3}{2}+\dfrac{7}{3}=\dfrac{3.3}{2.3}+\dfrac{7.2}{3.2}\\ =\dfrac{9}{6}+\dfrac{14}{6}=\dfrac{9+14}{6}\\ =\dfrac{23}{6}\)
3/2+7/3 ( quy đồng lên mẫu số chung là 6)
= 9/6 + 14/6
= 9+14/6
= 23/6