Bài 6: Lũy thừa của một số hữu tỉ (tiếp theo...)

\(\dfrac{11\cdot3^7\cdot9^7-9^{15}}{\left(2\cdot3^{14}\right)^2}\)

\(=\dfrac{11\cdot3^7\cdot3^{14}-3^{30}}{2^2\cdot3^{28}}\)

\(=\dfrac{3^{21}\left(11-3^9\right)}{2^2\cdot3^{28}}\)

\(=\dfrac{11-3^9}{2^2\cdot3^7}\)

\(\dfrac{4^{17}+4^3}{4^{16}+4^2}=\dfrac{4^3\left(4^{14}+1\right)}{4^2\left(4^{14}+1\right)}=\dfrac{4^3}{4^2}=4\)

\(\dfrac{4^{17}+4^3}{4^{16}+4^2}=\dfrac{4^3\left(4^{14}+1\right)}{4^2\left(4^{14}+1\right)}=4\)

\(B=\dfrac{\dfrac{1}{2020}+\dfrac{1}{2021}-\dfrac{1}{2022}}{\dfrac{3}{2020}+\dfrac{3}{2021}-\dfrac{3}{2022}}-1=\dfrac{\dfrac{1}{2020}+\dfrac{1}{2021}-\dfrac{1}{2022}}{3\left(\dfrac{1}{2020}+\dfrac{1}{2021}-\dfrac{1}{2022}\right)}-1=\dfrac{1}{3}-1=-\dfrac{2}{3}\)

\(B=\dfrac{\dfrac{1}{2021}+\dfrac{1}{2021}-\dfrac{1}{2022}}{\dfrac{3}{2020}+\dfrac{3}{2021}-\dfrac{3}{2022}}-1=\dfrac{\dfrac{1}{2021}+\dfrac{1}{2021}-\dfrac{1}{2022}}{3\left(\dfrac{1}{2020}+\dfrac{1}{2021}-\dfrac{1}{2022}\right)}-1=\dfrac{1}{3}-1=\dfrac{1}{3}-\dfrac{3}{3}=-\dfrac{2}{3}\)

\(\left(2^3\cdot9^4+9^3+45\right):\left(9^2\cdot10-9^2\right)\)

\(=\dfrac{9^3\cdot\left(2^3\cdot9+1\right)+45}{9^3}\)

\(=\dfrac{9^3\cdot73+45}{9^3}=\dfrac{5918}{81}\)

Ta có: \(\left(\dfrac{4}{7}-\dfrac{1}{3}\right)^x=8\)

\(\Leftrightarrow\left(\dfrac{5}{21}\right)^x=8\)

a)Ta có: \(x^2\ge0\Rightarrow x^2+3\ge3\)

Dấu "=" xảy ra \(\Leftrightarrow x=0\)

Vậy \(A_{Min}=3 khi x=0\)

b) \(\left(2x+1\right)^2\ge0\Rightarrow\left(2x+1\right)^2-5\ge-5\)

Dấu "=" xảy ra \(\Leftrightarrow x=-\dfrac{1}{2}\)

Vậy \(B_{Min}=-5khix=-\dfrac{1}{2}\)

c) \(\left(2x-1\right)^{2008}\ge0\)

Dấu "=" xảy ra \(\Leftrightarrow x=\dfrac{1}{2}\)

\(\left(3y-2\right)^{2008}\ge0\)

Dấu "=" xảy ra \(\Leftrightarrow y=\dfrac{2}{3}\)

\(\Rightarrow\left(2x-1\right)^{2008}+\left(3y-2\right)^{2008}\ge0\)

Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=\dfrac{2}{3}\end{matrix}\right.\)

Vậy \(C_{Min}=0khix=\dfrac{1}{2}vày=\dfrac{2}{3}\)

Ta có: \(\dfrac{\left(-3\right)^n}{81}=-27\)

\(\Leftrightarrow\left(-3\right)^n=-243\)

hay n=5

a: \(\left(-\dfrac{1}{16}\right)^{100}=\left(\dfrac{1}{16}\right)^{100}=\left(-\dfrac{1}{2}\right)^{400}\)

\(\left(-\dfrac{1}{2}\right)^{500}=\left(-\dfrac{1}{2}\right)^{500}\)

mà \(400< 500\)

nên \(\left(-\dfrac{1}{16}\right)^{100}< \left(-\dfrac{1}{2}\right)^{500}\)

a) \(64^2\cdot32^4=2^{16}\cdot2^{20}=2^{36}\)

b) \(11^{16}\cdot5^{24}=\left(11^4\right)^4\cdot\left(5^6\right)^4=\left(11^4\cdot5^6\right)^4\)

a)64².32⁴=(2⁶)².(2⁵)⁴=2¹².2²⁰=2³²

b)11¹⁶.5²⁴(ko phân tích đc nữa)

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