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Bài 1
a.\(\frac{-3}{4}\)-y:\(\frac{1}{5}\)=\(\frac{9}{28}\)
y:\(\frac{1}{5}\)=\(\frac{-15}{14}\)
y= \(\frac{-3}{14}\)
b.5x + 5x+2=650
5x . 1 + 5x + 52=650
5x(1+25)=650
5x.26=650
5x=25
x=2
Ta có : A = \(\frac{\left(2^2.3\right)^6+8^4.3^5}{2^{12}.3^5-4^6.9^2}\)
= \(\frac{\left(2^2\right)^6.3^6+\left(2^3\right)^4.3^5}{2^{12}.3^5-\left(2^2\right)^6.\left(3^2\right)^2}\)
= \(\frac{2^{12}.3^6+2^{12}.3^5}{2^{12}.3^5-2^{12}.3^4}\)
= \(\frac{2^{12}.\left(3^6+3^5\right)}{2^{12}.\left(3^5-3^4\right)}\)
= \(\frac{3^6+3^5}{3^5-3^4}\)
= \(6\)
\(A=\frac{\left(2^2.3\right)^6+8^4.3^5}{2^{12}.3^5-4^6.9^2}\)
\(A=\frac{2^{12}.3^6+\left(2^3\right)^4.3^5}{2^{12}.3^5-\left(2^2\right)^6.\left(3^2\right)^2}\)
\(A=\frac{2^{12}.3^6+2^{12}.3^5}{2^{12}.3^5-2^{12}.3^4}\)
\(A=\frac{2^{12}.\left(3^6+3^5\right)}{2^{12}.\left(3^5-3^4\right)}\)
\(\Rightarrow A=\frac{3^5.\left(3+1\right)}{3^4.\left(3-1\right)}=\frac{3^5.4}{3^4.2}=\frac{3.3^4.2.2}{3^4.2}=\frac{3.2}{1}=6\)
\(A=\frac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6}-\frac{5^{10}.7^4-25^5.49^2}{\left(125.7\right)3+5^9.\left(14\right)^3}\)
\(=\frac{2^{12}.3^5-\left(2^2\right)^6.\left(3^2\right)^2}{\left(2^2\right)^6.3^6+\left(2^3\right)^4.3^5}-\frac{5^{10}.7^3-\left(5^2\right)^5.\left(7^2\right)^2}{125^3.7^3+5^9.\left(2.7\right)^3}\)
\(=\frac{2^{12}.3^5-2^{12}.3^4}{2^{12}.3^6+2^{12}.3^5}-\frac{5^{10}.7^3-5^{10}.7^4}{5^9.7^3+5^9.7^3.2^3}\)
\(=\frac{2^{12}.3^4\left(3-1\right)}{2^{12}.3^5\left(3+1\right)}-\frac{5^{10}.7^3\left(1-7\right)}{5^9.7^3\left(1+8\right)}\)
\(=\frac{2}{3.4}-\frac{5.\left(-6\right)}{9}=\frac{2}{12}-\frac{-30}{9}\)
\(=\frac{1}{6}+\frac{10}{3}=\frac{1}{6}+\frac{20}{6}=\frac{21}{6}=\frac{7}{2}\)
Bạn ơi cho mk hỏi chỗ đoạn kia bạn lấy 1-7 ở đâu và 1 + 8 ở đâu