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a,\(\frac{21}{25}.\frac{11}{9}.\frac{5}{7}=\frac{21.11.5}{25.9.7}=\frac{3.7.11.5}{5^2.3^2.7}=\frac{11}{5.3}=\frac{11}{15}\)
b,\(\frac{5}{23}.\frac{17}{26}+\frac{5}{23}.\frac{9}{26}=\frac{5}{23}.\left(\frac{17}{26}+\frac{9}{26}\right)=\frac{5}{23}.1=\frac{5}{23}\)
c, \(\left(\frac{3}{29}-\frac{1}{5}\right).\frac{29}{3}=\frac{3}{29}.\frac{29}{3}-\frac{1}{5}.\frac{29}{3}=1-\frac{29}{15}=-\frac{14}{15}\)
a , \(\frac{21}{25}\times\frac{11}{9}\times\frac{5}{7}\)
\(=\frac{21\times11\times5}{25\times9\times7}\)
\(=\frac{3\times7\times11\times5}{5\times5\times3\times3\times7}\)
\(=\frac{11}{5\times3}\)
\(=\frac{11}{15}\)
b , \(\frac{5}{23}\times\frac{17}{26}+\frac{5}{23}\times\frac{9}{26}\)
\(=\frac{5}{23}\times\left(\frac{17}{26}+\frac{9}{26}\right)\)
\(=\frac{5}{23}\times\frac{26}{26}\)
\(=\frac{5}{23}\times1\)
\(=\frac{5}{23}\)
c , \(\left(\frac{3}{29}-\frac{1}{5}\right)\times\frac{29}{3}\)
\(=\frac{3}{29}\times\frac{29}{3}-\frac{1}{5}\times\frac{29}{3}\)
\(=1-\frac{29}{15}\)
a) \(\dfrac{3}{5}+0,145-\dfrac{1}{200}\)
\(=\dfrac{3}{5}+\dfrac{145}{1000}-\dfrac{1}{200}\)
\(=\dfrac{3}{5}+\dfrac{29}{200}-\dfrac{1}{200}\)
\(=\dfrac{120}{200}+\dfrac{29}{200}-\dfrac{1}{200}\)
\(=\dfrac{148}{200}\)
\(=\dfrac{37}{50}\)
b) \(\left(31\dfrac{6}{13}+5\dfrac{9}{41}\right)-36\dfrac{6}{13}\)
\(=31\dfrac{6}{13}+5\dfrac{9}{41}-36\dfrac{6}{13}\)
\(=\left(31\dfrac{6}{13}-36\dfrac{6}{13}\right)+5\dfrac{5}{41}\)
\(=\left(-5\right)+5\dfrac{5}{41}\)
\(=0\dfrac{5}{41}\)
\(=\dfrac{5}{41}\)
c) \(5.2\dfrac{1}{7}+5.7\dfrac{6}{7}\)
\(=5\left(2\dfrac{1}{7}+7\dfrac{6}{7}\right)\)
\(=5\left(9+\dfrac{1}{7}+\dfrac{6}{7}\right)\)
\(=5\left(9+1\right)\)
\(=5.10\)
\(=50\)
a) \(\dfrac{3}{5}+0,415-\dfrac{1}{200}\)
\(=\dfrac{3}{5}+\dfrac{83}{200}-\dfrac{1}{200}\\ =\dfrac{120}{200}+\dfrac{83}{200}-\dfrac{1}{200}\\ =\dfrac{120+83-1}{200}=\dfrac{202}{200}=\dfrac{101}{100}\)
b)\(\left(31\dfrac{6}{13}+5\dfrac{9}{41}\right)-36\dfrac{6}{13}\)
\(=\left(\dfrac{409}{13}+\dfrac{214}{41}\right)-\dfrac{474}{13}\)
\(=\dfrac{19551}{533}-\dfrac{474}{13}=\dfrac{9}{41}\)
c)\(5.2\dfrac{1}{7}+5.7\dfrac{6}{7}\)
\(=5.\dfrac{15}{7}+5.\dfrac{55}{7}\\ =5\left(\dfrac{15}{7}+\dfrac{55}{7}\right)\\ =5.10=50\)
Bài 3 :
A = 26 + 27 + 28 + 29 + 30 + 31 + 32 + 33
=> A = ( 33 + 26 ) . 8 : 2 = 236
Vậy A = 236
\(\text{#Hok tốt!}\)
a) 2 . 31 . 12 + 4 . 6 . 42 + 8 . 27 . 3
= 24 . 31 + 24 . 42 + 24 . 27
= 24 . ( 31 + 42 + 27 )
= 24 . 100
= 2400
\(A=11\dfrac{3}{13}-\left(2\dfrac{4}{7}+5\dfrac{3}{13}\right)\)
\(A=11\dfrac{3}{13}-5\dfrac{3}{13}-2\dfrac{4}{7}\)
\(A=6-2\dfrac{4}{7}\)
\(A=5\dfrac{7}{7}-2\dfrac{4}{7}\)
\(A=3\dfrac{3}{7}\)
\(B=\left(6\dfrac{4}{9}+3\dfrac{7}{11}\right)-4\dfrac{4}{9}\)
\(B=\left(6\dfrac{4}{9}-4\dfrac{4}{9}\right)+3\dfrac{7}{11}\)
\(B=2+3\dfrac{7}{11}\)
\(B=5\dfrac{7}{11}\)
\(C=\dfrac{-5}{7}.\dfrac{2}{11}+\dfrac{-5}{7}-\dfrac{9}{11}+1\dfrac{5}{7}\)
\(C=\dfrac{-5}{7}.\left(\dfrac{2}{11}+1\right)-\dfrac{9}{11}+1\dfrac{5}{7}\)
\(C=\dfrac{-5}{7}.\dfrac{13}{11}-\dfrac{9}{11}+1\dfrac{5}{7}\)
\(C=\dfrac{-65}{77}-\dfrac{9}{11}+1\dfrac{5}{7}\)
\(C=\dfrac{4}{11}+1\dfrac{5}{7}\)
\(C=\dfrac{160}{11}\)
\(D=0,7.2\dfrac{2}{3}.20.0,375.\dfrac{5}{28}\)
\(D=\dfrac{7}{10}.\dfrac{8}{3}.20.\dfrac{375}{1000}.\dfrac{5}{28}\)
\(D=\dfrac{7}{28}=\dfrac{5}{2}\)
\(E=\left(-6,17+3\dfrac{5}{9}-2\dfrac{36}{97}\right)\left(\dfrac{1}{3}-0,25-\dfrac{1}{12}\right)\)
\(E=\left(-6,17+3\dfrac{5}{9}-2\dfrac{36}{97}\right)\left(\dfrac{1}{3}-\dfrac{1}{4}-\dfrac{1}{12}\right)\)
\(E=\left(-6,17+3\dfrac{5}{9}-2\dfrac{36}{97}\right)\left(\dfrac{1}{12}-\dfrac{1}{12}\right)\)
\(E=\left(-6,17+3\dfrac{5}{9}-2\dfrac{36}{97}\right).0\)
\(\Rightarrow E=0\)
Giusp m với. Tìm số A biết. 121:A dư 10, 61 : A dư 10. giải thích cách làm theo lớp 6 b nhé.Thank
\(a,a^3\cdot a^9=a^{12}\)
\(b,\left(a^5\right)^7=a^{35}\)
\(c,\left(a^6\right)^4\cdot a^{12}=a^{24}\cdot a^{12}=a^{36}\)
\(d,4\cdot5^2-2\cdot3^2=2^2\cdot5-2\cdot3^2=2\cdot\left(2\cdot5+3^2\right)=2\cdot19=38\)
\(e,5^6:5^3+3^3\cdot3^2=5^3+3^5\)
a,a3.a9=a3+9=a12
b,(a5)7=a5.7=a35
Mấy câu tiếp theo bn lám tương tự!
