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\(B=\frac{1-\frac{1}{\sqrt{49}}+\frac{1}{49}-\frac{1}{\left(7\sqrt{7}\right)^2}}{\frac{\sqrt{64}}{2}-\frac{4}{7}+\frac{2^2}{7^2}-\frac{4}{343}}\)
\(B=\frac{1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}}{\frac{8}{2}-\frac{4}{7}+\frac{4}{49}-\frac{4}{343}}\)
\(B=\frac{\frac{343}{343}-\frac{49}{343}+\frac{7}{343}-\frac{1}{343}}{4-\frac{4}{7}+\frac{28}{343}-\frac{4}{343}}\)
\(B=\frac{\frac{300}{343}}{\frac{28}{7}-\frac{4}{7}+\frac{24}{343}}\)
\(B=\frac{\frac{300}{343}}{\frac{24}{7}+\frac{24}{343}}\)
\(B=\frac{\frac{300}{343}}{\frac{1323}{343}+\frac{24}{343}}\)
\(B=\frac{300}{343}:\frac{1347}{343}\)
\(B=\frac{100}{449}\)
\(A=\frac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6+8^4.3^5}-\frac{5^{10}.7^3-25^5.49^2}{\left(125.7\right)^3+5^9.14^3}\)
\(A=\frac{2^{12}.3^5-2^{12}.3^6}{2^{12}.3^6+2^{12}.3^5}-\frac{5^{10}.7^3-5^{10}.7^6}{5^9.7^3+5^9.2^3.7^3}\)
\(A=\frac{2^{12}.3^5\left(1-3\right)}{2^{12}.3^5.\left(3+1\right)}-\frac{5^{10}.7^3.\left(1-7^3\right)}{5^9.7^3.\left(1+8\right)}\)
\(A=\frac{-2}{4}-\frac{5.\left(-342\right)}{9}\)
\(A=\frac{-1}{2}+\frac{1710}{9}\)
\(A=\frac{-1}{2}+190\)
\(A=\frac{-1}{2}+\frac{380}{2}\)
\(A=\frac{379}{2}\)
1: \(A=\dfrac{-25}{27}-\dfrac{31}{42}+\dfrac{7}{27}+\dfrac{3}{42}=\dfrac{-2}{3}-\dfrac{2}{3}=\dfrac{-4}{3}\)
2: \(B=\dfrac{10.3-\left(9.5-4.5\right)\cdot2}{1.2-1.5}=\dfrac{10.3-10}{-0.3}=-1\)
c: \(=\dfrac{3}{49}\left(\dfrac{19}{2}-\dfrac{5}{2}\right)-\left(\dfrac{1}{20}-\dfrac{5}{20}\right)^2\cdot\left(\dfrac{-7}{14}-\dfrac{193}{14}\right)\)
\(=\dfrac{3}{49}\cdot7-\dfrac{1}{25}\cdot\dfrac{-200}{14}\)
\(=\dfrac{3}{7}+\dfrac{8}{14}=1\)
Tử số: 2^19 x (3^3)^3 x 5+15 x 4^9 x(3^2)^4
=2^19 x3^9x5 + 15 x(2^2)^9 x 3^8
= 2^19 x 3^9 x 5 +3 x 5 x 2^18 x 3^8
= 2^19 x 3^9 x 5+ 3^9 x 5 x 2^18
= 5 x 3^9 x 2^18 (2+1)
=5 x 3^10 x 2^18
Mẫu số
= (2 x 3)^9 x 2^10 -12^10
= 2^9 x 3^9 x 2^10 - (2^2x3)^10
= 2^9 x 3^9 x 2^10 -2^20 x 3^10
= 2^19 x 3^9 - 2^20 x 3^10
= 2^19 x 3^9 (1-2 x 3)
= 2^19 x 3^9 x(-5)
Chia cả tử và mẫu ta có
(5 x 3^10 x 2^18) / (2^19 x 3^9 x (-5)) = -3/2
\(H=\frac{2^{19}.27^3.5-15.\left(-4\right)^9.9^4}{6^9.2^{10}-\left(-12\right)^{10}}\)
\(\Rightarrow\)\(H=\frac{2^{19}.3^9.5-3.5-1.2^{18}.3^8}{2^9.3^9.2^{10}-6^{10}.2^{10}}=\frac{2^{19}.3^9.5-3^9.5-2^{18}}{2^{19}.3^9-3^{10}.2^{10}.2^{10}}=\frac{2^{19}.3^9.5-3^9.5-2^{18}}{2^{19}.3^9-3^{10}.2^{20}}\)
\(\Rightarrow H=\frac{2^{18}.3^9.5\left(2-1\right)}{2^{19}.3^9.\left(1-3.2\right)}=\frac{5}{2.\left(-5\right)}=\frac{-1}{2}\)
Vậy \(H=\frac{-1}{2}\)
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) \(\left(-5\right).12:\left[\left(-\dfrac{1}{4}\right)+\dfrac{1}{2}:\left(-2\right)\right]+1\dfrac{1}{3}\)
\(=-60:\left[\dfrac{1}{4}+\dfrac{1}{2}.\dfrac{-1}{2}\right]+1.\dfrac{1}{3}\)
\(=-60:\left[-\dfrac{1}{4}-\dfrac{1}{4}\right]+1\dfrac{1}{3}\)
\(=-60:\left(\dfrac{1}{2}\right)+1\dfrac{1}{3}\)
\(=121\dfrac{1}{3}\)
a) \(4\frac{5}{9}:\left(-\frac{5}{7}\right)+\frac{49}{9}:\left(-\frac{5}{7}\right)=\frac{41}{9}:\left(-\frac{5}{7}\right)+\frac{49}{9}:\left(-\frac{5}{7}\right)\)
\(=\frac{41}{9}\cdot\left(-\frac{7}{5}\right)+\frac{49}{9}\cdot\left(-\frac{7}{5}\right)=\left(\frac{41}{9}+\frac{49}{9}\right)\cdot\left(-\frac{7}{5}\right)=10\cdot\left(-\frac{7}{5}\right)=-14\)
b) \(\left(\frac{-3}{5}+\frac{4}{9}\right):\frac{7}{11}+\left(\frac{-2}{5}+\frac{5}{9}\right):\frac{7}{11}\)
\(=\left(\frac{-3}{5}+\frac{4}{9}+\frac{-2}{5}+\frac{5}{9}\right):\frac{7}{11}\)
\(=\left(\frac{-3}{5}+\frac{-2}{5}+\frac{4}{9}+\frac{5}{9}\right):\frac{7}{11}\)
\(=\left(-1+1\right):\frac{7}{11}=0\cdot\frac{11}{7}=0\)
c) \(\left(\frac{3}{4}\right)^4\cdot\left(\frac{8}{9}\right)^2=\left(\frac{3}{4}\right)^2\cdot\left(\frac{3}{4}\right)^2\cdot\left(\frac{8}{9}\right)^2=\left(\frac{3}{4}\cdot\frac{3}{4}\cdot\frac{8}{9}\right)^2\)
\(=\left(\frac{1}{2}\right)^2=\frac{1}{4}\)
d) \(\left(-\frac{3}{5}\right)^6\cdot\left(-\frac{5}{3}\right)^5=\left(-\frac{3}{5}\right)^5\cdot\left(-\frac{3}{5}\right)\cdot\left(-\frac{5}{3}\right)^5=\left[\left(-\frac{3}{5}\right)\cdot\left(-\frac{5}{3}\right)\right]^5\cdot\left(-\frac{3}{5}\right)\)
\(=1^5\cdot\left(-\frac{3}{5}\right)=1\cdot\left(-\frac{3}{5}\right)=-\frac{3}{5}\)
e) \(\frac{8^{14}}{4^4\cdot64^5}=\frac{\left(2^3\right)^{14}}{\left(2^2\right)^4\cdot\left(2^6\right)^5}=\frac{2^{42}}{2^8\cdot2^{30}}=\frac{2^{42}}{2^{38}}=2^4=16\)
f) \(\frac{9^{10}\cdot27^7}{81^7\cdot3^{15}}=\frac{\left(3^2\right)^{10}\cdot\left(3^3\right)^7}{\left(3^4\right)^7\cdot3^{15}}=\frac{3^{20}\cdot3^{21}}{3^{28}\cdot3^{15}}=\frac{3^{41}}{3^{43}}=3^{-2}=\frac{1}{3^2}=\frac{1}{9}\)
b)=1/5.(1/4-1/9+1/9-1/14+1/14-1/19+...+1/44-1/49).2-1-3-5-7-...-49/89
=1/5.(1/4-1/49).2-(1+3+5+7...+49)/89
=1/5.45/196.2-625/89
=9/196.-623/89
=9/196.-7
=9/28
h cho mình nha ! Chúc bạn học tốt
\(a,\frac{27^4\cdot2^3-3^{10}\cdot4^3}{6^4\cdot9^3}=\frac{3^{12}\cdot2^3-3^{10}\cdot2^6}{2^3\cdot3^4\cdot3^6}=\frac{3^{10}\cdot2^3\cdot\left(3^2-2^3\right)}{2^3\cdot3^{10}}=3^2-2^3=1\)
\(b,\left(\frac{1}{4\cdot9}+\frac{1}{9\cdot14}+\frac{1}{14\cdot19}+...+\frac{1}{44\cdot49}\right)\cdot\frac{1-3-5-7-...-49}{89}\)
\(=\frac{1}{5}\left(\frac{5}{4\cdot9}+\frac{5}{9\cdot14}+\frac{5}{14\cdot19}+...+\frac{1}{44\cdot49}\right)\cdot\frac{1-\left(3+5+7+...+49\right)}{89}\)
\(=\frac{1}{5}\left(\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+\frac{1}{14}-\frac{1}{19}+...+\frac{1}{44}-\frac{1}{49}\right)\cdot\frac{1-\left(3+49\right)\cdot24\div2}{89}\)
\(=\frac{1}{5}\left(\frac{1}{4}-\frac{1}{49}\right)\cdot\frac{505}{89}\)
\(=\frac{1}{5}\cdot\frac{45}{196}\cdot\frac{505}{89}\)
\(\dfrac{20^3.\left(-49\right)^2}{14^3.5^4}=\dfrac{2^3.2^3.5^3.7^4}{2^3.7^3.5^4}=\dfrac{2^3.7}{5}=\dfrac{8.7}{5}=\dfrac{56}{5}\)
Cho mình hỏi là bạn lớp mấy rồi được không ạ :3?