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a)Đang suy nghĩ...
b)\(M\left(x\right)=\left(x^2-3x\right)+\left(x-3\right)=0\)
\(\Leftrightarrow x\left(x-3\right)+\left(x-3\right)=0\Leftrightarrow\left(x+1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=-1\\x=3\end{cases}}\)
a) \(12x^{11}-15x^7-6x^5+2018\)
\(=3x^5.\left(4x^6-5x^2-2\right)+2018\)
\(=3x^5.0+2018\)
\(=2018\)
\(\frac{4^2.4^3}{2^{10}}=\frac{\left(2^2\right)^2.\left(2^2\right)^3}{2^{10}}=\frac{2^4.2^6}{2^{10}}=\frac{2^{10}}{2^{10}}=1\)
a, \(\frac{4^2.4^3}{2^{10}}=\frac{\left(2^2\right)^2.\left(2^2\right)^3}{2^{10}}=\frac{2^4.2^6}{2^{10}}=\frac{2^{4+6}}{2^{10}}=\frac{2^{10}}{2^{10}}=1\)
b,\(\frac{\left(0,6\right)^5}{\left(0,2\right)^6}=\frac{\left(0,2.3\right)^5}{\left(0,2\right)^6}=\frac{\left(0,2\right)^5.3^5}{\left(0,2\right)^6}=\frac{3^5}{0,2}\)
c, \(\frac{2^7.9^3}{6^5.8^2}=\frac{2^7.\left(3^2\right)^3}{\left(2.3\right)^5.\left(2^3\right)^2}=\frac{2^7.3^6}{2^5.3^5.2^6}=\frac{2^7.3^6}{3^5.2^{11}}=\frac{3}{2^4}\)
d, \(\frac{6^3+3.6^2+3^3}{-13}=\frac{\left(2.3\right)^3+3\left(2.3\right)^2+3^3}{-13}=\frac{2^3.3^3+3.2^2.3^2+3^3}{-13}\)
\(=\frac{2^3.3^3+3^3.2^2+3^3}{-13}=\frac{3^9\left(2^3+2^2+1\right)}{-13}=\frac{3^3.13}{-13}=3^3=27\)
\(A=12x^{11}-15x^7-6x^5+2018\)
\(=3x^5.\left(4x^6-3x^2-2\right)+2018\)
\(=3x^5.0+2018\)
\(=2018\)
\(A=-\frac{\frac{-6}{5}+\frac{6}{19}-\frac{6}{23}}{\frac{9}{5}-\frac{9}{19}+\frac{9}{23}}\)
\(=\frac{-6.\left(\frac{1}{5}-\frac{1}{19}+\frac{1}{23}\right)}{9.\left(\frac{1}{5}-\frac{1}{19}+\frac{1}{23}\right)}\)
\(=-\frac{6}{9}=-\frac{2}{3}\)
Bài 2:
a: \(=7^4\left(7^2+7-1\right)=7^4\cdot55⋮55\)
b: \(5A=5+5^2+...+5^{51}\)
\(\Leftrightarrow4A=5^{51}-1\)
hay \(A=\dfrac{5^{51}-1}{4}\)
Bài 3:
\(S=\left(1^2+2^3+3^3+...+10^2\right)\cdot2=385\cdot2=770\)
đề có sai ko bạn
\(\frac{27^2.8^5}{6^6.32^3}=\frac{\left(3^3\right)^2.\left(2^3\right)^5}{2^3.3^3.\left(2^5\right)^3}=\frac{3^6.2^{15}}{2^3.3^3.2^{15}}=\frac{27}{8}\)
học tốt