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1.
2|x-6|+7x-2=|x-6|+7x
2|x-6| - |x-6|=7x-(7x-2)
|x-6| = 2
=>x-6 = +2
*x-6=2 *x-6 = -2
x =2+6 x = (-2)+6
x =8 x = 4
2.
|x-5|-7(x+4)=5-7x
|x-5|-7x-28 =5-7x
|x-5|-28 =5-7x+7x
|x-5|-28 = 5
|x-5| = 5+28
|x-5| = 33
=>x-5 = +33
*x-5=33 *x-5=-33
x =38 x = -28
3.
3|x+4|-2(x-1)=7-2x
3|x+4|-2x+2 =7-2x
3|x+4|-2 =7-2x+2x
3|x+4|-2 =7
3|x+4| =7+2
3|x+4| = 9
|x+4| =9:3
|x+4| = 3
=>x+4 = +3
*x+4=3 *x+4=-3
x =-1 x = -7
a)B = ( 2 + 4 + 6 + 8 +........+ 2014 ) - ( 3 + 5 + 7 + 9 +.......+ 2011 )
= 1015056 - 1012035
= 3021
Mk nhanh nhất đó
a, 11 + 112 + 113 + ... + 117 + 118
= (11 + 112) + (113 + 114) + ... + (117 + 118)
= 11(1 + 11) + 113(1 + 11) + ... + 117(1 + 11)
= 11.12 + 113.12 + .... + 117.12
= 12(11 + 113 + ... + 117) chia hết cho 12
b, 7 + 72 + 73 + 74
= (7 + 73) + (72 + 74)
= 7(1 + 72) + 72(1 + 72)
= 7.50 + 72.50
= 50(7 + 72) chia hết cho 50
c, 3 + 32 + 33 + 34 + 35 + 36
= (3 + 32 + 33) + (34 + 35 + 36)
= 3(1 + 3 + 32) + 34(1 + 3 + 32)
= 3.13 + 34.13
= 13(3 + 34) chia hết cho 13
a: =152,3+7,7+2021,19-2021,19
=160
b: =7/15*3/14*20/13
\(=\dfrac{7}{14}\cdot\dfrac{3}{15}\cdot\dfrac{20}{13}=\dfrac{1}{2}\cdot\dfrac{1}{5}\cdot\dfrac{20}{13}=\dfrac{2}{13}\)
c: \(=\dfrac{7}{4}\left(\dfrac{13}{12}-\dfrac{10}{12}\right)+\dfrac{5}{6}=\dfrac{7}{16}+\dfrac{5}{6}=\dfrac{61}{48}\)
x=(1+2+3-4-5-6)+...+(97+98+99-100-101-102)
x=-9+...+-9
x=-9.17
x=-153
#)Giải :
\(A=1+2+2^2+...+2^{100}\)
\(2A=2+2^2+2^3+...+2^{101}\)
\(2A-A=\left(2+2^2+2^3+...+2^{101}\right)-\left(1+2+2^2+...+2^{100}\right)\)
\(A=2^{101}-1\)
\(B=1+3^2+3^4+...+3^{100}\)
\(3^2B=3^2+3^4+3^6+...+3^{102}\)
\(3^2B-B=\left(3^2+3^4+3^6+...+3^{102}\right)-\left(1+3^2+3^4+...+3^{100}\right)\)
\(8B=3^{102}-1\)
\(B=\frac{3^{102}-1}{8}\)
\(C=1+5^3+5^6+...+5^{99}\)
\(5^2C=5^3+5^6+5^9+...+5^{102}\)
\(5^2C-C=\left(5^3+5^6+5^9...+5^{102}\right)-\left(1+5^3+5^6+...+5^{99}\right)\)
\(24C=5^{102}-1\)
\(C=\frac{5^{102}-1}{24}\)
a) A = 1 + 22 + ... + 2100
=> 2A = 22 + 23 + ... + 2101
Lấy 2A - A = (2 + 22 + ... + 2101) - (1 + 22 + ... 2100)
A = 2101 - 1
b) B = 1 + 32 + 34 + ... + 3100
=> 32B = 32 + 34 + 36 + ..... + 3102
=> 9B = 32 + 34 + 36 + ..... + 3102
Lấy 9B - B = ( 32 + 34 + 36 + ..... + 3102) - (1 + 32 + 34 + ... + 3100)
8B = 3102 - 1
B = \(\frac{3^{102}-1}{8}\)
c) C = 1 + 53 + 56 + ... + 599
=> 53.C = 53 . 56 . 59 + ... + 5102
=> 125.C = 53 . 56 . 59 + ... + 5102
Lấy 125.C - C = (53 . 56 . 59 + ... + 5102) - (1 + 53 + 56 + ... + 599)
124.C = 5102 - 1
=> C = \(\frac{5^{102}-1}{124}\)
=(1-2)-(3-4)+(5-6)-(7-8)+...+(2021-2022)-2023
=(-1)-(-1)+(-1)-...+(-1)-2023
=0-2023
=-2023
\(\frac{547}{140}nha\)
HT