Question

[˙✧˖° 🫧🕊️тнáι нoà✦нв²ᵏ¹⁴🐬🐟˚✧˚]

B= \(\frac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6+8^4.3^5}-\frac{5^{10}.7^3-25^5.49^2}{\left(125.7\right)^3+5^9.14^3}\) = ?

Answer 1

[🧸♠Vũ điệu Samba ♤]

B=\(\frac{(2^12\times3^5-4^6\times9^2)}{\left\lbrack(2^2\times3)^6+8^4\times3^5\right\rbrack}-\frac{(5^10\times7^3-25^5\times49^2)}{\left\lbrack(125\times7)^3+5^9\times14^3\right\rbrack}\)

B=\(\frac{(2^{12}\times3^5-2^{12}\times3^4)}{(2^{12}\times3^6+2^{12}\times3^5)}-\frac{(5^10\times7^3-5^10\times7^4)}{(5^9\times7^3+5^9\times2^3\times7^3)}\)

B=\(\frac{2^{12}\times3^4\times(3-1)}{\left\lbrack2^12\times3^5\times(3+1)\right\rbrack}-\frac{5^{10}\times7^3\times(1-7)}{\left\lbrack\right.5^9\times7^3\times(1+8)]}\)

B=\(\frac{2}{(3\times4)}-5\times\frac{(-6)}{9}\)

B=\(\frac16+\frac{10}{3}\)

B=\(\frac16+\frac{20}{6}\)

B=\(\frac{21}{6}=\frac72\)

Vậy B=\(\frac72\)

Answer 2

[🧎‍♂️🫥🥲]

\(B=\frac{2^{12}\times3^5-4^6\times9^2}{\left(2^2\times3\right)^6+8^4\times3^5}-\frac{5^{10}\times7^3-25^2\times49^2}{\left(125\times7\right)^3+5^9\times14^3}\)

\(B=\frac{2^{12}\times3^5-2^{12}\times3^4}{2^{12}\times3^6+2^{12}\times3^5}-\frac{5^{10}\times7^3-5^{10}\times7^4}{5^9\times7^3+5^9\times2^3\times7^3}\)

\(B=\frac{2^{12}\times3^4\left(3-1\right)}{2^{12}\times3^5\left(3+1\right)}-\frac{5^{10}\times7^3\left(1-7\right)}{5^9\times7^3\left(1+2^3\right)}\)

\(B=\frac{2}{3\times4}-\frac{5\left(-6\right)}{9}\)

\(B=\frac16-\frac{-10}{3}\)

\(B=\frac72\)

Vậy B = 7/2

Answer 3

[˙✧˖° 🫧🕊️тнáι нoà✦нв²ᵏ¹⁴🐬🐟˚✧˚]

đr nhé cmn

Answer 4

[Nguyễn Thị Ngọc Trinh]

B= \(\frac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6+8^4.3^5}-\frac{5^{10}.7^3-25^5.49^2}{\left(125.7\right)^3+5^9.14^3}\)

=\(\frac{2^{12}.3^5-\left(2^{2^{}}\right)^6.\left(3^2\right)^2}{2^{12}.3^6+\left(2^3\right)^4.3^5}-\frac{5^{10}.7^3-\left(5^2\right)^5.\left(7^2\right)^2}{\left(5^3.7\right)^3+5^9.\left(2.7\right)^3}\)

=\(\frac{2^{12}.3^5-2^{12}.3^4}{2^{12}.3^6+2^{12}.3^5}-\frac{5^{10}.7^3-5^{10}.7^4}{5^9.7^3+5^9.2^3.7^3}\)

=\(\frac{2^{12}.3^4.\left(3-1\right)}{2^{12}.3^5.\left(3+1\right)}-\frac{5^{10}.7^3.\left(1-7\right)}{5^9.7^3.\left(1+2^3\right)}\)

=\(\frac{2^{12}.3^4.2}{2^{12}.3^5.4}-\frac{5^{10}.7^3.\left(-6\right)}{5^9.7^3.9}\)

=\(\frac{2^{13}.3^4}{2^{14}.3^5}-\frac{-6.5^{10}.7^3}{9.5^9.7^3}\)

=\(\frac{1}{2.3}-\frac{-2.5}{3}\)

=\(\frac16-\left(-\frac{10}{3}\right)\)

=\(\frac16+\frac{20}{6}\) =\(\frac{21}{6}\) =\(\frac72\)

Answer 5

[KUDOSHINICHI Vn]

Ta tính giá trị từng phân số của biểu thức \(B\):

• Phân số thứ nhất:
  \(\frac{2^{12}\cdot 3^{5}-4^{6}\cdot 9^{2}}{(2^{2}\cdot 3)^{6}+8^{4}\cdot 3^{5}}=\frac{2^{12}\cdot 3^{5}-(2^{2})^{6}\cdot (3^{2})^{2}}{2^{12}\cdot 3^{6}+(2^{3})^{4}\cdot 3^{5}}\)
  \(=\frac{2^{12}\cdot 3^{5}-2^{12}\cdot 3^{4}}{2^{12}\cdot 3^{6}+2^{12}\cdot 3^{5}}\)
  \(=\frac{2^{12}\cdot 3^{4}\cdot (3-1)}{2^{12}\cdot 3^{5}\cdot (3+1)}\)
  \(=\frac{2^{12}\cdot 3^{4}\cdot 2}{2^{12}\cdot 3^{5}\cdot 4}=\frac{2^{13}\cdot 3^{4}}{2^{14}\cdot 3^{5}}=\frac{1}{2\cdot 3}=\frac{1}{6}\)
• Phân số thứ hai:
  \(\frac{5^{10}\cdot 7^{3}-25^{5}\cdot 49^{2}}{(125\cdot 7)^{3}+5^{9}\cdot 14^{3}}=\frac{5^{10}\cdot 7^{3}-(5^{2})^{5}\cdot (7^{2})^{2}}{(5^{3}\cdot 7)^{3}+5^{9}\cdot (2\cdot 7)^{3}}\)
  \(=\frac{5^{10}\cdot 7^{3}-5^{10}\cdot 7^{4}}{5^{9}\cdot 7^{3}+5^{9}\cdot 2^{3}\cdot 7^{3}}\)
  \(=\frac{5^{10}\cdot 7^{3}\cdot (1-7)}{5^{9}\cdot 7^{3}\cdot (1+2^{3})}\)
  \(=\frac{5^{10}\cdot 7^{3}\cdot (-6)}{5^{9}\cdot 7^{3}\cdot 9}=\frac{5\cdot (-6)}{9}=\frac{-30}{9}=\frac{-10}{3}\)

Thay hai giá trị vừa tính vào biểu thức \(B\), ta được:
\(B=\frac{1}{6}-\left(\frac{-10}{3}\right)\)
\(B=\frac{1}{6}+\frac{10}{3}\)
\(B=\frac{1}{6}+\frac{20}{6}=\frac{21}{6}\)
\(B=\frac{7}{2}\)Vậy \(B = \frac{7}{2}\).

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