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\(\left(\dfrac{-10}{3}\right)^5.\left(\dfrac{6}{5}\right)^4\\ =\left(\dfrac{-10}{3}\right).\left(\dfrac{10}{3}\right)^4.\left(\dfrac{6}{5}\right)^4\\ =\left(\dfrac{-10}{3}\right).\left(\dfrac{10}{3}.\dfrac{6}{5}\right)^4\\ =\left(\dfrac{-10}{3}\right).4^4\\ =\left(\dfrac{-10}{3}\right).256\\ =\dfrac{-2560}{3}\)
\(\frac{16}{2^n}=2\)
=> 2n=16:2
=> 2n=8
=> 2n=23
=> n=3
b;
\(\frac{\left(-3\right)^n}{81}=-27\)
\(\Rightarrow\left(-3\right)^n=-27.81\)
=> (-3)n=-2187
=> (-3)n=(-3)7
=> n=7
c ; \(8^n:2^n=4\)
\(\Rightarrow\left(8:2\right)^n=4\)
\(\Rightarrow\left(4\right)^n=4\)
Mà : 4=41
=> 4n=41
=> n=1
\(\frac{16}{2^n}=2\)
\(16:2^n=2\)
\(2^4:2^n=2\)
\(2^n=2^4:2\)
\(2^n=2^3\)
\(=>n=3\)
\(\frac{\left(-3\right)^n}{81}=-27\)
\(\left(-3\right)^n:81=-27\)
\(\left(-3\right)^n=-27\cdot81\)
\(\left(-3\right)^n=-2187\)
\(\left(-3\right)^n=\left(-3\right)^7\)
\(=>n=7\)
\(\frac{2^7.9^3}{6^5.8^2}=\frac{2^7.\left(3^2\right)^3}{\left(2.3\right)^5.\left(2^3\right)^2}=\frac{2^7.3^6}{2^5.3^5.2^6}=\frac{2^7.3^6}{2^{11}.3^5}=\frac{3}{2^4}=\frac{3}{16}\)
\(=\frac{2^7.3^3.3^3}{2^5.3^5.4^2.2^2}=\frac{3^3.3^3}{3^5.4^2}=3.4^2=3.16=48\)
\(\frac{2^7.9^3}{6^5.8^2}=\frac{2^7.\left(3^2\right)^3}{2^5.3^5.\left(2^3\right)^2}=\frac{2^7.3^6}{2^{11}.3^5}=\frac{3}{2^4}=\frac{3}{16}\)
Ta thấy: \(\begin{cases}\left(x-2\right)^{2012}\ge0\\\left|y^2-9\right|^{201}\ge0\end{cases}\)
\(\Rightarrow\left(x-2\right)^{2012}+\left|y^2-9\right|^{201}\ge0\)
\(\Rightarrow\left(x-2\right)^{2012}+\left|y^2-9\right|^{201}+5\ge5\)
\(\Rightarrow A\ge5\). Dấu "=" xảy ra khi
\(\begin{cases}\left(x-2\right)^{2012}=0\\\left|y^2-9\right|^{201}=0\end{cases}\)\(\Leftrightarrow\begin{cases}x-2=0\\y^2-9=0\end{cases}\)\(\Leftrightarrow\begin{cases}x=2\\y=\pm3\end{cases}\)
Vậy \(\begin{cases}x=2\\y=\pm3\end{cases}\)
Theo đề bài ta có x = a/m, y = b/m (a, b, m ∈ Z, b # 0)
Vì x < y nên ta suy ra a < b
Ta có: x = 2a/2m, y = 2b/2m; z = (a+b)/2m
Vì a < b => a + a < a + b => 2a < a + b
Do 2a < a + b nên x < z (1)
Vì a < b => a + b < b + b => a + b < 2b
Do a + b < 2b nên z < y (2)
Từ (1) và (2) ta suy ra x < z < y