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\(=\dfrac{2^{36}.3^{36}\left(2.5^2.5^{40}-2.5.5^{34}\right)}{2^{30}.3^{30}.5^{30}.2^4.5^4\left(2^2.5^2.3^5.5^5-2^2.3^5\right)}=\dfrac{2^{36}.3^{36}\left(2.5^{42}-2.5^{35}\right)}{2^{34}.3^{30}.5^{34}\left(2^2.3^5.5^7-2^2.3^5\right)}\)
\(=\dfrac{2^2.3^6.2.5^{35}\left(5^7-1\right)}{2^2.3^5\left(5^7-1\right)}=3.2.5^{35}=6.5^{35}\)
a) \(\frac{1}{3}.\frac{-6}{13}.\frac{-9}{10}.\frac{-13}{36}\)
\(=\left(\frac{1}{3}.\frac{-9}{10}\right)\left(\frac{-6}{13}.\frac{-13}{36}\right)\)
\(=\frac{-3}{10}.\frac{1}{6}\)
\(=\frac{-1}{20}\)
b) \(\frac{-1}{3}.\frac{-15}{17}.\frac{34}{45}\)
\(=\frac{-1}{3}.\frac{-2}{3}\)
\(=\frac{2}{9}\)
c) \(\left(1-\frac{1}{5}\right)\left(\frac{-3}{10}+\frac{1}{5}\right)\)
\(=\frac{4}{5}.\frac{-1}{10}\)
\(=\frac{-2}{25}\)
d) \(A=\frac{1}{3}.\frac{4}{5}+\frac{1}{3}.\frac{6}{5}+\frac{2}{3}\)
\(=\frac{1}{3}\left(\frac{4}{5}+\frac{6}{5}\right)+\frac{2}{3}\)
\(=\frac{1}{3}.2+\frac{2}{3}\)
\(=\frac{2}{3}+\frac{2}{3}\)
\(=\frac{4}{3}\)
e) \(11\frac{1}{4}-\left(2\frac{5}{7}+5\frac{1}{4}\right)\)
\(=\left(11\frac{1}{4}-5\frac{1}{4}\right)-2\frac{5}{7}\)
\(=6-2\frac{5}{7}\)
\(=5\frac{7}{7}-2\frac{5}{7}\)
\(=3\frac{2}{7}\)
\(A=\dfrac{2^4.3^3+2^3.3^4}{2^5.3^4-2^6.3^3}=\dfrac{2^3.3^3.\left(2+3\right)}{2^5.3^3.\left(3-2\right)}=\dfrac{2^3.3^3.5}{2^5.3^3.1}\)
\(=\dfrac{5}{2^2}=\dfrac{5}{4}\)
\(a.-8:\left(4\dfrac{1}{5}x+\dfrac{3}{10}\right)=4\dfrac{4}{9}\)
\(4\dfrac{1}{5}x+\dfrac{3}{10}=\left(-8\right):4\dfrac{4}{9}\)
\(4\dfrac{1}{5}x+\dfrac{3}{10}=\dfrac{-9}{5}\)
\(4\dfrac{1}{5}x=\dfrac{-9}{5}-\dfrac{3}{10}\)
\(4\dfrac{1}{5}x=\dfrac{-21}{10}\)
\(x=\dfrac{-21}{10}:\dfrac{21}{5}\)
\(x=\dfrac{-1}{2}\)
Vay \(x=\dfrac{-1}{2}\).
\(b.4\dfrac{2}{3}-\left(\dfrac{3}{5}:x\right)=-20\%\)
\(\dfrac{14}{3}-\left(\dfrac{3}{5}:x\right)=\dfrac{-1}{5}\)
\(\dfrac{3}{5}:x=\dfrac{14}{3}-\dfrac{-1}{5}\)
\(\dfrac{3}{5}:x=\dfrac{73}{15}\)
\(x=\dfrac{3}{5}:\dfrac{73}{15}\)
\(x=\dfrac{9}{73}\)
Vay \(x=\dfrac{9}{73}\).
Câu c; d; e tương tự nhé.
a,\(\frac{31}{1000}\)
b,\(\frac{8}{108}\)
c,0
a,\(\frac{49}{97}\)
b,\(\frac{-1}{4751}\)
a) (-17) + 5 + 8 + 17
= [(-17) + 17] + (5 + 8)
= 0 + 13
= 13
b) 30 + 12 + (-20) + (-12)
= [30 + (-20)] + [(-12) + 12]
= 10 + 0
= 10
c) (-4) + (-440) + (-6) + 440
= [(-4) + (-6)] + [440 + (-440)]
= -10 + 0
= -10
d) (-5) + (-10) + 16 + (-1)
= [(-5) + (-10) + (-1)] + 16
= (-16) + 16
= 0
Các bạn có thể bỏ các dấu ngoặc vuông [] đi cũng được vì nó thực sự không quan trọng lắm. Dấu ngoặc vuông [] chỉ giúp các bạn rõ ràng hơn trong các phép tính.
a) (-17) + 5 +8 +17
= [( -17)+17] + ( 5+8)
= 0 +13
=13
b) 30 +12 + (-20) +(-12)
= (30 +-20 ) + ( -12 +12)
= 10+0
=10
c ) (-4) + (-440)+(-6)+440
(-4+-6) = (-440+440)
= -10 + 0
= -10
d) (-5) + (-10 ) +16 +(-1)
= ( 16 + -1 +-5) +(-10)
= 10 + (-10)
= 0
\(=\dfrac{2^{12}\cdot3^5-2^{12}\cdot3^4}{2^{18}\cdot3^6+2^{12}\cdot3^5}-\dfrac{5^{10}\cdot7^3-5^{10}\cdot7^4}{5^9\cdot7^3+5^9\cdot2^3\cdot7^3}\)
\(=\dfrac{2^{12}\cdot3^4\cdot2}{2^{12}\cdot3^5\left(2^6\cdot3+1\right)}-\dfrac{5^{10}\cdot7^3\cdot\left(-6\right)}{5^9\cdot7^3\cdot9}\)
\(=\dfrac{2}{193}+\dfrac{5\cdot2}{3}=\dfrac{1936}{579}\)
Bài 1:
1) \(\frac{11}{3}\): 3\(\frac{1}{3}\)- 3
= \(\frac{11}{3}\): \(\frac{10}{3}\)- 3
= \(\frac{11}{3}\). \(\frac{3}{10}\)- 3
= \(\frac{11}{10}\)- 3
= \(\frac{-19}{10}\)
2) \(\frac{5}{6}\): \(\frac{3}{52}\) - \(\frac{5}{6}\). 47\(\frac{1}{3}\)
= \(\frac{5}{6}\) . \(\frac{52}{3}\)- \(\frac{5}{6}\). 47\(\frac{1}{3}\)
= \(\frac{5}{6}\).(\(\frac{52}{3}\)- 47\(\frac{1}{3}\))
= \(\frac{5}{6}\).( -30)
= -25
không vt lại đề
\(=\frac{2^{12}.3^5-2^{12}.3^4}{2^{12}.3^6+2^{12}.3^5}-\frac{5^{10}.7^3-5^{10}.7^4}{5^6.7^3+5^9.2^3.7^3}\)
\(=\frac{2^{12}\cdot3^4.\left(3-1\right)}{2^{12}.3^5.\left(3+1\right)}-\frac{5^{10}.7^3\left(1-7\right)}{5^6.7^3\left(1+2^3\right)}\)
\(=\frac{1}{6}-\frac{-10}{3}=\frac{1}{6}+\frac{20}{6}=\frac{21}{6}\)