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I -2-x I = -15-37-(-30)
I-2-xI =-22
Vì giá trị tuyệt đối của một số luôn luôn là số nguyên dương (hoặc = 0 khi số đó là 0)nên không có giá trị nào của x thỏa mãn đề bài.
S = 5 + 52 + 53 + 54 + .......... + 599
a) S = ( 5 + 52 + 53 ) + ( 54 + 55 + 56 ) + .... + ( 597 + 598 + 599 )
= 5. ( 1 + 5 + 52 ) + 54 . ( 1 + 5 + 52 ) + .... + 597 . ( 1 + 5 + 52 )
= ( 1 + 5 + 52 ). ( 5 + 54 + .. + 597 )
= 31 . ( 5 + 54 + .... + 597 ) chia hết cho 31 ( đpcm )
c ) 5S = 52 + 53 + .. + 5100
=> 5S - S = 4S = 5100 + 599 + ........ + 53 + 52 - 5 - 52 - 53 - ..... - 599
= 5100 - 5
25x - 5 = 4S
=> 25x - 5 = 5100 - 5
=> 25x = 5100
=> 25x = ( 52 )50
=> 25x = 2550
=> x = 50
Vậy x = 50
Câu b quên cách làm rồi
a) S=5+52+53+54+...+599
=(5+52+53)+(54+55+56)+...+(597+598+599)
=5(1+5+52)+54(1+5+52)+...+597(1+5+52)
=5.31+54.31+...+597.31
=31(5+54+...+597)⋮31(đpcm)
b) S=5+52+53+54+...+599
=5+(52+53)+(54+55)+...+(598+599)
=5+5(5+52)+53(5+52)+...+597(5+52)
=5+5.30+53.30+...+597.30
=5+30.(5+53+...+597)
Mà 5⋮̸30 nên S⋮̸30(đpcm)
c) Ta có: 5S=52+53+54+55+...+5100
5S−S=(52+53+54+55+...+5100)−(5+52+53+54+...+599)
4S=5100−5
⇒25x−5=5100−5
⇒25x=5100
⇒25x=2550
⇒x=50
a)\(\frac{7}{13}.\frac{7}{15}-\frac{5}{12}.\frac{21}{39}+\frac{49}{21}.\frac{8}{15}\)
=\(\frac{7}{13}.\frac{7}{15}-\frac{5}{12}.\frac{7}{13}+\frac{7}{13}.\frac{8}{15}\)
=\(\frac{7}{13}.\left(\frac{7}{15}-\frac{5}{12}-\frac{8}{15}\right)\)
=\(\frac{7}{13}.\frac{7}{12}\)
=\(\frac{49}{156}\)
b)\(\left(\frac{12}{199}+\frac{23}{200}-\frac{34}{201}\right).\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)\)
=\(a.\left(\frac{3}{6}-\frac{2}{6}-\frac{1}{6}\right)\)
=a . 0
=0
Bài 2
a)Có
\(3^{200}=3^{2.100}=\left(3^2\right)^{100}=9^{100}\)
\(2^{300}=2^{3.100}=\left(2^3\right)^{100}=8^{100}\)
Vì 8<9 =>\(8^{100}< 9^{100}\) =>\(3^{200}>2^{300}\)
I, 65.5+65.42-65=6500
II,
1,420-6x=330
<=>-6x=-90
<=>x=15
2, 10+2x=65536
<=>2x=65526
<=>x=32763
3, 37-2x-10=9
<=>-2x=-18
x=9
III, 2333<3222
Bài 1 .
65 . 59 - 65 . 42 - 65
= 65 . ( 59 - 42 ) - 65
= 65 . 17 - 65
= 1105 - 65
= 1040
Bài 2 .
a) 6 . ( 70 - x ) = 330
<=> 70 - x = 330 : 6
<=> 70 - x = 55
<=> x = 70 - 55
<=> x = 15
Vậy x = 15
b) 10 + 2x = 45 : 43
<=> 10 + 2x = 42
<=> 10 + 2x = 16
<=> 2x = 16 - 10
<=> 2x = 6
<=> x = 6 : 2
<=> x = 3
Vậy x = 3.
c) 37 - 2( x + 5 ) = 9
<=> 2( x + 5 ) = 37 - 9
<=> 2( x + 5 ) = 28
<=> x + 5 = 28 : 2
<=> x + 5 = 14
<=> x = 14 - 5
<=> x = 9
Vậy x = 9.
k)(-37)+14+26+37
=(-37)+37+14+26
=0+14+26
=40
l)15+23+(-25)+(-23)
=15+(-25)+23+(-23)
=-10+0
=-10
n)25+37-48-25-37
=25-25+37-37-48
=0+0-48
=-48
o)24.(16-5)-16.(24-5)
=24.16-24.5-16.24-16.5
=0-24.5-16.5
=0-(24-16).5
=0-8.5
=0-40
=-40
p)31.(-18)+31.(-81)-31
=31.(-18)+31.(-81)+31.(-1)
=31.[-18+(-81)+(-1)]
=31.(-100)
=-3100
q)-48+48.(-78)+48.(-21)
=48.(-1)+48.(-78)+48.(-21)
=48.[(-1)+(-78)+(-21)]
=48.(-100)
=-4800
r)(7.3-3):(-6)
=(7.3-1.3):(-6)
=(6.3):(-6)
=-3
s)-3.7-4.(-5)+1
=-21-(-20)+1
=-20-(-20)
=0
t(6.8-10:5)+3.(-7)
=(48-2)+(-21)
=46+(-21)
=25
Viết mỏi cả tay rồi.tick đi nhé
\(A=\frac{\left(23\frac{11}{15}-26\frac{13}{20}\right)}{12^2+5^2}\cdot\frac{1-\frac{1}{30}-\frac{1}{42}-\frac{1}{56}}{3^2.13.2-13.5}-\frac{19}{37}\)
\(A=\frac{\left(23+\frac{11}{15}-26+\frac{13}{20}\right)}{144+25}\cdot\frac{1-\frac{1}{5.6}-\frac{1}{6.7}-\frac{1}{7.8}}{9.13.2-13.5}-\frac{19}{37}\)
\(A=\frac{\left(23+26+\frac{11}{15}-\frac{13}{20}\right)}{169}\cdot\frac{1-\left(\frac{1}{5}-\frac{1}{6}\right)-\left(\frac{1}{6}-\frac{1}{7}\right)-\left(\frac{1}{7}-\frac{1}{8}\right)}{13.\left(9.2-5\right)}-\frac{19}{37}\)
\(A=\frac{49+\frac{44}{60}-\frac{39}{60}}{169}\cdot\frac{1-\frac{1}{5}+\frac{1}{6}-\frac{1}{6}+\frac{1}{7}-\frac{1}{7}+\frac{1}{8}}{13.13}-\frac{19}{37}\)
\(A=\frac{49+\frac{1}{20}}{169}\cdot\frac{1-\frac{1}{5}+\frac{1}{8}}{169}-\frac{19}{37}\)
\(A=\frac{49\frac{1}{20}}{169}\cdot\frac{\frac{4}{5}+\frac{5}{40}}{169}-\frac{19}{37}\)
\(A=\frac{981}{169}\cdot\frac{\frac{32}{40}+\frac{5}{40}}{169}-\frac{19}{37}\)
\(A=\frac{981}{169}\cdot\frac{\frac{37}{40}}{169}-\frac{19}{37}\)
\(A=\frac{981.\frac{37}{40}}{169^2}-\frac{19}{37}\)
\(A=\frac{\frac{36297}{40}}{28561}-\frac{19}{37}\)
\(A=\frac{907,425}{28561}-\frac{19}{37}\)
\(A=\frac{33574,725}{1056757}-\frac{542659}{1056757}\)
\(A=\frac{-509084,275}{1056757}=-0,04604282...\)
Mik chỉ làm đc thế này thôi, ôn thi học kì II tốt nha bạn!
a: \(S=\dfrac{-1}{2}\cdot\dfrac{-2}{3}\cdot...\cdot\dfrac{-99}{100}=-\dfrac{1}{100}\)
c: \(5S_3=5^6+5^7+...+5^{101}\)
\(\Leftrightarrow4\cdot S_3=5^{101}-5^5\)
hay \(S_3=\dfrac{5^{101}-5^5}{4}\)
d: \(S_4=7\cdot\left(\dfrac{1}{10}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{12}+...+\dfrac{1}{69}-\dfrac{1}{70}\right)\)
\(=7\left(\dfrac{1}{10}-\dfrac{1}{70}\right)=7\cdot\dfrac{6}{70}=\dfrac{6}{10}=\dfrac{3}{5}\)
\(37-2.\left(x+1\right)=15\)
\(2.\left(x+1\right)=37-15\)
\(2.\left(x+1\right)=22\)
\(x+1=22:2\)
\(x+1=11\)
\(x=11-1\)
\(x=10\)
-2 ( x + 1 ) = 37 - 15
-2 ( x + 1) = 22
x + 1 = -11
x = -12