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1)
a. \(\left(3x^2-50\right)^2=5^4\)
\(\Leftrightarrow3x^4-50=625\)
\(\Leftrightarrow3x^4=675\)
\(\Leftrightarrow x^4=225\)
\(\Leftrightarrow x=\sqrt{15}\)
2)
a. \(\frac{\left(3^4-3^3\right)^4}{27^3}=\frac{3^{16}-3^{12}}{\left(3^3\right)^3}=\frac{3^{12}.3^4-3^{12}}{3^9}=\frac{3^{12}\left(3^4-1\right)}{3^9}\)
\(=\frac{3^{12}.80}{3^9}=3^3.80=27.80=2160\)
b. \(\frac{25^3}{\left(5^5-5^3\right)^2}=\frac{\left(5^2\right)^3}{5^{10}-5^6}=\frac{5^6}{5^6.5^4-5^6}=\frac{5^6}{5^6\left(5^4-1\right)}\)
\(=\frac{5^6}{5^6.624}=\frac{1}{624}\)
\(A=\frac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6+8^4.3^5}+\frac{16^3.3^{10}+120.6^9}{4^6.3^{12}+6^{12}}\)
\(=\frac{2^{12}.3^5-\left(2^2\right)^6.\left(3^2\right)^2}{2^{12}.3^6+\left(2^3\right)^4.3^5}-\frac{\left(2^4\right)^3.3^{10}+2^3.3.5.\left(2.3\right)^9}{\left(2^2\right)^6.3^{12}+\left(2.3\right)^{12}}\)
\(=\frac{2^{12}.3^5-2^{12}.3^4}{2^{12}.3^6-2^{12}.3^5}-\frac{2^{12}.3^{10}-2^3.3.5.2^9.3^9}{2^{12}.3^{12}+2^{12}.3^{12}}\)
\(=\frac{2^{12}.\left(3^5-3^4\right)}{2^{12}.\left(3^6-3^5\right)}-\frac{2^{12}.3^{10}-2^{12}.3^{10}.5}{2^{12}.3^{12}+2^{12}.3^{12}}\)
\(=\frac{3^5-3^4}{3^6-3^5}-\frac{2^{12}.3^{10}.\left(1-5\right)}{2^{13}.3^{12}}\)
\(=\frac{162}{486}-\frac{2^{12}.3^{10}.\left(-4\right)}{2^{13}.3^{10}.3^2}=\frac{1}{3}-\frac{2^{14}.3^{10}.\left(-1\right)}{2^{13}.3^{10}.9}\)
\(=\frac{1}{3}-\frac{2.1.\left(-1\right)}{1.1.9}=\frac{1}{3}-\frac{2}{9}=\frac{1}{9}\)
eo ôi t làm rồi mà bị xoá :v thôi t hướng dẫn :v
Tạc TS và MS ra rồi gộp và triệt tiêu :) nếu k lm đc ibx t làm cho :)
Bài 1:
a) \(\dfrac{19}{12}+\left|\dfrac{-5}{2}\right|+\left(\dfrac{3}{2}\right)^2=\dfrac{19}{12}+\dfrac{5}{2}+\dfrac{9}{4}\)
\(=\dfrac{19+5.6+9.3}{12}=\dfrac{76}{12}=\dfrac{19}{3}\)
b) \(\dfrac{2}{11}.\dfrac{16}{9}-\dfrac{2}{11}.\dfrac{7}{9}=\dfrac{2}{11}\left(\dfrac{16}{9}-\dfrac{7}{9}\right)=\dfrac{2}{11}.1=\dfrac{2}{11}\)
Bài 2:
Áp dụng t/c dtsbn:
\(\dfrac{a}{8}=\dfrac{b}{3}=\dfrac{a-b}{8-3}=\dfrac{55}{5}=11\)
\(\Rightarrow\left\{{}\begin{matrix}x=11.8=88\\b=11.3=33\end{matrix}\right.\)
\(a.\dfrac{3^{27}}{9^6.3^{16}}=\dfrac{3^{27}}{3^{12}.3^{16}}=\dfrac{3^{27}}{3^{28}}=\dfrac{1}{3}\)
\(\left(x-\dfrac{5}{2}\right)^2=\dfrac{9}{4}\\ \Rightarrow x-\dfrac{5}{2}=\pm\dfrac{3}{2}\)
\(TH1:x-\dfrac{5}{2}=\dfrac{3}{2}\Rightarrow x=\dfrac{3}{2}+\dfrac{5}{2}=\dfrac{8}{2}=4\)
\(TH2:x-\dfrac{5}{2}=-\dfrac{3}{2}\Rightarrow x=-\dfrac{3}{2}+\dfrac{5}{2}=\dfrac{2}{2}=1\)
a: \(=\dfrac{3^{27}}{3^{12}\cdot3^{16}}=\dfrac{1}{3}\)