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\(=\frac{\left(3.7\right)^4}{3^3.\left(-7\right)^3}+7\)
\(=\frac{3^4.7^4}{3^3.\left(-7\right)^3}+7\)
=\(\) 3.(-7)+7
= -14
\(\left(5-\frac{2}{3}+\frac{3}{7}\right):\left(24-25+\frac{4}{21}-\frac{8}{21}\right)\)
\(\left(\frac{5.21-14+9}{21}\right):\left(\frac{-21-4}{21}\right)\)=\(\left(\frac{5.21-5}{21}\right).\left(\frac{21}{-25}\right)\)=\(\frac{5\left(21-1\right)}{\left(-5\right).5}=\frac{20}{-5}\)
=-4
a) \(1\frac{3}{19}+\frac{8}{21}-\frac{3}{19}+0.5+\frac{13}{21}\)
\(=\left(1\frac{3}{19}-\frac{3}{19}\right)+\left(\frac{8}{21}+\frac{13}{21}\right)+0.5\)
\(=1+1+0.5=2.5\)
b) \(\left(-\frac{3}{4}+\frac{2}{7}\right):\frac{3}{7}+\left(\frac{5}{7}+\frac{-1}{4}\right):\frac{3}{7}\)
\(=\left(\frac{-3}{4}+\frac{2}{7}+\frac{5}{7}+\frac{-1}{4}\right):\frac{3}{7}\)
\(=0:\frac{3}{7}=0\)
\(A=\frac{1}{5}\left(\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+\frac{1}{14}-\frac{1}{19}+.....+\frac{1}{44}-\frac{1}{49}\right).\frac{1-\left(49+3\right)\left(\left(49-3\right):2+1\right):2}{89}\)
\(A=\frac{1}{5}\left(\frac{1}{4}-\frac{1}{49}\right).\frac{1-26.24}{89}=\frac{45}{4.5.49}.\frac{-623}{89}=-\frac{9}{28}\)
\(A=\left(\frac{1}{4.9}+\frac{1}{9.14}+\frac{1}{14.19}+....+\frac{1}{44.49}\right).\frac{1-3-5-7-...-49}{89}\)
\(\Rightarrow5A=5.\left(\frac{1}{4.9}+\frac{1}{9.14}+\frac{1}{14.19}+...+\frac{1}{44.49}\right).\frac{1-3-5-7-...-49}{89}\)
\(=\left(\frac{5}{4.9}+\frac{5}{9.14}+\frac{5}{14.19}+...+\frac{5}{44.49}\right).\frac{1+\frac{\left(-3-47\right).23}{2}-49}{89}\)
\(=\left(\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+\frac{1}{14}-\frac{1}{19}+...+\frac{1}{44}-\frac{1}{49}\right).\frac{1+\left(-575\right)-49}{89}\)
\(=\left(\frac{1}{4}-\frac{1}{49}\right).\frac{-623}{89}=\frac{45}{196}.\left(-7\right)=-\frac{45}{26}\)
a,\(\frac{-2}{5}+\frac{7}{21}=\frac{-2}{5}+\frac{1}{3}=\frac{-6}{15}+\frac{5}{15}=\frac{-1}{15}\)
b,\(\left(\frac{1}{3}\right)^5.3^5-2020^0=\left(\frac{1}{3}.3\right)^5-1=1^5-1=1-1=0\)
c,\(\left(-\frac{1}{4}\right).6\frac{2}{11}+3\frac{9}{11}.\left(-\frac{1}{4}\right)\)
\(=\left(-\frac{1}{4}\right).\left(6\frac{2}{11}+3\frac{9}{11}\right)=\left(-\frac{1}{4}\right).\left[\left(6+3\right)+\left(\frac{2}{11}+\frac{9}{11}\right)\right]\)
\(=\left(-\frac{1}{4}\right).\left[9+1\right]=\frac{-1}{4}.10=\frac{\left(-1\right).10}{4}=\frac{\left(-1\right).5}{2}=\frac{-5}{2}\)
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 \(H=\frac{-1}{2}\)
Lười làm qá, hì:
Hướng dẫn thôi nha.
B1: Phá bỏ ngoặc của các phép tính.
B2: Ghép những số nguyên vào vs nhau, phân số vào vs nhau
B3: Giao hoán những phân số có cùng mẫu để cộng vào, ở đây chỉ nói cộng vì trừ lp 7 là cộng vs số đối mà
B4: Tính hết ra là xong
B = (8+6-3) - (9/4-5/4-2/4) + (2/7-3/7-9/7)
B = 11 - 1/2 -10/7
B = 21/2 - 10/7
B = 127/14
\(\frac{21^4}{27\cdot(-343)}+7\)
\(=\frac{(3\cdot7)^4}{3^3\cdot(-7)^3}+7\)
\(=\frac{3^4\cdot7^4}{3^3\cdot(-7)^3}+7\)
\(=3\cdot(-7)+7=-14\)