Question

1. Tìm giá trị n nguyên dương :

\(\dfrac{1}{8} . 16^n=2^n\)

2. Thực hiện phép tính :

\((\dfrac{1}{4.9}+\dfrac{1}{9.14}+\dfrac{1}{14.19}+...+\dfrac{1}{44.49}).\dfrac{1-3-5-7-...-49}{89}\)

3.a, Tìm x biết : \(|2x+3|=x+2\)

b, Tìm giá trị nhỏ nhất của \(A=|x-2006|+|2007-x|\) khi x thay đổi

Answer 1

Bài 1:

\(\frac{1}{8}.16^n=2^n\)

\(\Rightarrow\frac{16^n}{8}=2^n\)

\(\Rightarrow\frac{\left(2^4\right)^n}{2^3}=2^n\)

\(\Rightarrow\frac{2^{4n}}{2^3}=2^n\)

\(\Rightarrow2^{4n-3}=2^n\)

\(\Rightarrow4n-3=n\)

\(\Rightarrow4n-n=3\)

\(\Rightarrow3n=3\)

\(\Rightarrow n=3:3\)

\(\Rightarrow n=1\left(TM\right).\)

Vậy \(n=1.\)

Bài 3:

a) \(\left|2x+3\right|=x+2\)

\(\Rightarrow\left[{}\begin{matrix}2x+3=x+2\\2x+3=-x-2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x-x=2-3\\2x+x=-2-3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}1x=-1\\3x=-5\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\left(-1\right):1\\x=\left(-5\right):3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\x=-\frac{5}{3}\end{matrix}\right.\)

Vậy \(x\in\left\{-1;-\frac{5}{3}\right\}.\)

Chúc bạn học tốt!

Answer 2

Bài 3:

b) \(A=\left|x-2006\right|+\left|2007-x\right|\)

Áp dụng bất đẳng thức \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\) ta có:

\(A=\left|x-2006\right|+\left|2007-x\right|\ge\left|x-2006+2007-x\right|\)

\(\Rightarrow A\ge\left|1\right|\)

\(\Rightarrow A\ge1.\)

Dấu '' = '' xảy ra khi:

\(\left(x-2006\right).\left(2007-x\right)\ge0\)

\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-2006\ge0\\2007-x\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}x-2006\le0\\2007-x\le0\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\ge2006\\x\le2007\end{matrix}\right.\\\left\{{}\begin{matrix}x\le2006\\x\ge2007\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2006\le x\le2007\\x\in\varnothing\end{matrix}\right.\)

Vậy \(MIN_A=1\) khi \(2006\le x\le2007.\)

Chúc bạn học tốt!

Answer 3

Câu 2 :

Đặt : \(P=\left(\frac{1}{4.9}+\frac{1}{9.14}+....+\frac{1}{44.49}\right).\frac{1-3-5-7...-49}{89}\)

\(=\frac{1}{5}.\left(\frac{5}{4.9}+\frac{5}{9.14}+....+\frac{5.}{44.49}\right).\frac{1-3-5-....-49}{89}\)

\(=\frac{1}{5}.\left(\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+....+\frac{1}{44}-\frac{1}{49}\right).\frac{1-3-5-...-49}{89}\)

\(=\frac{1}{5}.\left(\frac{1}{4}-\frac{1}{49}\right).\frac{-623}{89}\)

\(=\frac{1}{5}.\frac{45}{196}.\left(-7\right)\)

\(=\frac{9}{196}.\left(-7\right)=-\frac{9}{28}\)

Vậy : \(P=-\frac{9}{28}\)