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Gửi tạm trước 2 câu !
\(a,\text{ }3^2\cdot\frac{1}{243}\cdot81^2\cdot3^{-3}=3^2\cdot\frac{1}{3^5}\cdot\left(3^4\right)^2\cdot\frac{1}{3^3}=3^2\cdot\frac{1}{3^5}\cdot3^8\cdot\frac{1}{3^3}=3^2=9\)\(b,\text{ }\frac{\left(-3\right)^{10}\cdot15^5}{25^3\cdot\left(-9\right)^7}=\frac{3^{10}\cdot\left(3\cdot5\right)^5}{\left(5^2\right)^3\cdot\left(-3\cdot3\right)^7}=\frac{3^{10}\cdot3^5\cdot5^5}{5^6\cdot3^7\cdot\left(-3\right)^7}=\frac{3^{15}\cdot5^5}{5^6\cdot3^7\cdot\left(-3\right)^7}=\frac{3}{-5}\)
Trả lời :
\(a,\text{ }3^2\cdot\frac{1}{243}\cdot81^2\cdot3^{-3}=3^2\cdot\frac{1}{3^5}\cdot\left(3^4\right)^2\cdot\frac{1}{3^3}=3^2\cdot\frac{1}{3^5}\cdot3^8\cdot\frac{1}{3^3}=3^2=9\)\(b,\text{ }\frac{\left(-3\right)^{10}\cdot15^5}{25^3\cdot\left(-9\right)^7}=\frac{3^{10}\cdot\left(3\cdot5\right)^5}{\left(5^2\right)^3\cdot\left(-3\cdot3\right)^7}=\frac{3^{10}\cdot3^5\cdot5^5}{5^6\cdot3^7\cdot\left(-3\right)^7}=\frac{3^{15}\cdot5^5}{5^6\cdot3^7\cdot\left(-3\right)^7}=\frac{3}{-5}\)
Bài 2:
a: =>x^2=60
=>\(x=\pm2\sqrt{15}\)
b: =>2^2x+3=2^3x
=>3x=2x+3
=>x=3
c: \(\Leftrightarrow\sqrt{\dfrac{1}{2}x-2}\cdot\dfrac{1}{2}=1\)
\(\Leftrightarrow\sqrt{\dfrac{1}{2}x-2}=2\)
=>1/2x-2=4
=>1/2x=6
=>x=12
Thực hiện các phép tính:
a) 9,6.212−(2.125−1512):149,6.212−(2.125−1512):14
b) 518−1,456:725+4,5.45518−1,456:725+4,5.45;
c) (12+0,8−113).(2,3+4725−1,28)(12+0,8−113).(2,3+4725−1,28)
d) (−5).12:[(−14)+12:(−2)]+113(−5).12:[(−14)+12:(−2)]+113.
Hướng dẫn làm bài:
a) 9,6.212−(2.125−1512):149,6.212−(2.125−1512):14
=9,6.52−(250−1712)×4=9,6.52−(250−1712)×4
=4,8.5−(1000−173)=4,8.5−(1000−173)
=24−1000+173=24−1000+173
=−976+173=−976+173
=−97013=−97013
b) 518−1,456:725+4,5.45518−1,456:725+4,5.45;
=518−1,456×257+92.45=518−1,456×257+92.45
=518−0,208×25+185=518−0,208×25+185
=518−5,2+185=518−5,2+185
=25−468+32490=25−468+32490
=−11990=−11990
c) (12+0,8−113).(2,3+4725−1,28)(12+0,8−113).(2,3+4725−1,28)
=(12+45−43).(2310+10725−3225)=(12+45−43).(2310+10725−3225)
=(15+24−4030).(2310+10725−3225)=(15+24−4030).(2310+10725−3225)
=(15+24−4030).(115+214−6450)=(15+24−4030).(115+214−6450)
=−130.26550=−130.26550
=−53300=−53300
d) (−5).12:[(−14)+12:(−2)]+113(−5).12:[(−14)+12:(−2)]+113
=−60:[14+12×(−12)]+1.13=−60:[14+12×(−12)]+1.13
=−60:[−14−14]+113=−60:[−14−14]+113
=−60:(12)+113=−60:(12)+113
=120+113=120+113
=12113
a) \(9,6.2\dfrac{1}{2}-\left(2.125-1\dfrac{5}{12}\right):\dfrac{1}{4}\)
\(=9,6.\dfrac{5}{2}-\left(250-\dfrac{17}{12}\right).4\)
\(=4,8.5-\left(1000-\dfrac{17}{3}\right)\)
\(=24-1000+\dfrac{17}{3}\)
\(=-976+\dfrac{17}{3}=-970\dfrac{1}{3}\)
b) \(\dfrac{5}{18}-1,456:\dfrac{7}{25}+4,5.\dfrac{4}{5}\)
\(=\dfrac{5}{18}-1,456.\dfrac{25}{7}+\dfrac{9}{2}.\dfrac{4}{5}\)
\(=\dfrac{5}{18}-0,208.25+\dfrac{18}{5}\)
\(=\dfrac{5}{18}-5,2+\dfrac{18}{5}\)
\(=-\dfrac{119}{90}\)
c) \(\left(\dfrac{1}{2}+0,8-1\dfrac{1}{3}\right).\left(2,3+4\dfrac{7}{25}-1,28\right)\)
\(=\left(\dfrac{1}{2}+\dfrac{4}{5}-\dfrac{4}{3}\right).\left(\dfrac{23}{10}+\dfrac{107}{25}-\dfrac{32}{25}\right)\)
\(=-\dfrac{1}{30}.\dfrac{265}{50}=-\dfrac{53}{300}\)
d) \(\left(-5\right).12:\left[\left(-\dfrac{1}{4}\right)+\dfrac{1}{2}:\left(-2\right)\right]+1\dfrac{1}{3}\)
\(=-60:\left[\dfrac{1}{4}+\dfrac{1}{2}.\dfrac{-1}{2}\right]+1.\dfrac{1}{3}\)
\(=-60:\left[-\dfrac{1}{4}-\dfrac{1}{4}\right]+1\dfrac{1}{3}\)
\(=-60:\left(\dfrac{1}{2}\right)+1\dfrac{1}{3}\)
\(=121\dfrac{1}{3}\)
Bài 1: Tính B = 1 + 2 + 3 + ... + 98 + 99
Số số hạng của B là : 99 số hạng
Tổng của B là ( 1 + 99 ) x 99 : 2 = 4950
Vậy : B = 1 + 2 + 3 + ... + 98 + 99 = 4950
Bài 2: Tính C = 1 + 3 + 5 + ... + 997 + 999
Số số hạng của C là : ( 999 - 1 ) : 2 + 1 = 500 ( số hạng )
Tổng của C là : ( 1 + 999 ) x 500 : 2 = 250000
Bài 3. Tính D = 10 + 12 + 14 + ... + 994 + 996 + 998
Số số hạng của D là : ( 998 - 10 ) : 2 + 1 = 495 ( số hạng )
Tổng của D là : ( 10 + 998 ) x 495 : 2 = 249480
!)
B=1+2+3+...+98+99
B= 99(99+1):2
B = 4950
( Áp dụng: Nếu B=1+2+3+...+(n-1)+n
thì B=n(n+1):2
B=4950 nha bạn!
2) Tính: C=1+3+5+...+997+999
Ta có: 999= 2(500)-1. n=500
1+2+3+...+(2n-1)= n^2
= 500^2= 250.000
C=25.000
\(=5+\dfrac{1}{5}-\dfrac{2}{9}-2+\dfrac{1}{23}+\dfrac{73}{35}-\dfrac{5}{6}-8-\dfrac{2}{7}+\dfrac{1}{18}\)
\(=-5+\left(\dfrac{1}{5}-\dfrac{2}{7}+\dfrac{73}{35}\right)+\left(\dfrac{-2}{9}-\dfrac{5}{6}+\dfrac{1}{18}\right)+\dfrac{1}{23}\)
\(=-5+\dfrac{7-10+73}{35}+\dfrac{-4-15+1}{18}+\dfrac{1}{23}\)
\(=-5+2-1+\dfrac{1}{23}=-4+\dfrac{1}{23}=-\dfrac{91}{23}\)
1.Tính
a.\(\dfrac{7}{23}\left[(-\dfrac{8}{6})-\dfrac{45}{18}\right]=\dfrac{7}{23}.-\dfrac{12}{6}=-\dfrac{7}{6}\)
b.\(\dfrac{1}{5}\div\dfrac{1}{10}-\dfrac{1}{3}(\dfrac{6}{5}-\dfrac{9}{4})=2-(-\dfrac{7}{20})=\dfrac{47}{20}\)
c.\(\dfrac{3}{5}.(-\dfrac{8}{3})-\dfrac{3}{5}\div(-6)=-\dfrac{3}{2}\)
d.\(\dfrac{1}{2}.(\dfrac{4}{3}+\dfrac{2}{5})-\dfrac{3}{4}.(\dfrac{8}{9}+\dfrac{16}{3})=-\dfrac{19}{5}\)
e.\(\dfrac{6}{7}\div(\dfrac{3}{26}-\dfrac{3}{13})+\dfrac{6}{7}.(\dfrac{1}{10}-\dfrac{8}{5})=-\dfrac{61}{7}\)
Bài 2
a.\(1^2_5x+\dfrac{3}{7}=\dfrac{4}{5}\)
\(x=\dfrac{13}{49}\)
b.\(\left|x-1,5\right|=2\)
Xảy ra 2 trường hợp
TH1
\(x-1,5=2\)
\(x=3,5\)
TH2
\(x-1,5=-2\)
\(x=-0,5\)
Vậy \(x=3,5\) hoặc \(x=-0,5\) .
Ngại làm quá trời ơi,lần sau bn tách ra nhá làm vậy mỏi tay quá.
a. = \(\dfrac{-1}{24}-\left\{\dfrac{1}{4}-\dfrac{-3}{8}\right\}\)
= \(\dfrac{-1}{24}-\left\{\dfrac{1}{4}+\dfrac{3}{8}\right\}\)
= \(\dfrac{-1}{24}-\dfrac{5}{8}\)
= \(\dfrac{-2}{3}\)
b. = \(12\dfrac{7}{88}-3\dfrac{5}{11}\)
= \(8\dfrac{5}{8}\)
c. = \(\dfrac{-28}{9}+\dfrac{-413}{9}\)
= \(-49\)
d. = \(\dfrac{8}{35}:\dfrac{2}{11}+\dfrac{-8}{35}:\dfrac{2}{11}\)
= \(\dfrac{2}{11}:\left(\dfrac{8}{35}+\dfrac{-8}{35}\right)\)
= 0
c) -55/36
d)-67/120