Bài 5: Lũy thừa của một số hữu tỉ

\(2^{-1}.2^{n} + 4.2^{n}= 9.2^{5}\)

\(2^{n}(2^{-1} + 4)=9.2^{5}\)

\(2^{n}.\frac{9}{2}=9.2^{5}\)

\(2^{n}=9.2^{5}.\frac{2}{9}\)

\(2^{n}=2^{6}\)

=> n = 6

ta có 2^-1.2ⁿ + 4.2ⁿ=9.2⁵

=> 2ⁿ(2^-1+4)=288

=>2ⁿ.4,5=288

=>2ⁿ=64

=>2ⁿ=2⁶

=>n=6

\(B=-5\left(3x+2\right)^4-\left(x+2y\right)^2+111\)

Ta có :

\(\left(3x+2\right)^4\ge0\Rightarrow-5\left(3x+2\right)^4\le0\left(1\right)\)

\(\left(x+2y\right)^2\ge0\Rightarrow-\left(x+2y\right)^2\le0\left(2\right)\)

Từ (1)(2) \(\Rightarrow-5\left(3x+2\right)^4-\left(x+2y\right)^2\le0\)

\(\Rightarrow-5\left(3x+2\right)^4-\left(x+2y\right)^2+111\le111\)

Dấu = xảy ra khi \(\left\{{}\begin{matrix}5\left(3x+2\right)^4=0\\\left(x+2y\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x+2=0\\x+2y=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{2}{3}\\-\dfrac{2}{3}+2y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{2}{3}\\y=\dfrac{1}{3}\end{matrix}\right.\)

Vậy B đạt GTLN bằng 11 khi \(x=-\dfrac{2}{3};y=\dfrac{1}{3}\)

\(A=-5\left(3x+2\right)^4-\left(x+2y\right)^2+111\le111\)

Dấu "=" xảy ra khi:

\(\left\{{}\begin{matrix}-5\left(3x+2\right)^4=0\\-\left(x+2y\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{2}{3}\\y=\dfrac{1}{3}\end{matrix}\right.\)

\(\left(x-1\right)^{2000}=\left(x-1\right)^{1998}\)

\(\Rightarrow\left(x-1\right)^{2000}-\left(x-1\right)^{1998}=0\)

\(\Rightarrow\left(x-1\right)^{1998}.\left(x-1\right)^2-\left(x-1\right)^{1998}=0\)

\(\Rightarrow\left(x-1\right)^{1998}.\left[\left(x-1\right)^2-1\right]=0\)

\(\Rightarrow\left(x-1\right)^{1998}\left[\left(x-1\right)^2-1^2\right]=0\)

\(\Rightarrow\left(x-1\right)^{1998}\left(x-1-1\right)\left(x-1+1\right)=0\)

\(\Rightarrow x\left(x-1\right)^{1998}.\left(x-2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\\left(x-1\right)^{1998}=0\\x-2=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x-1=0\\x=2\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=1\\x=2\end{matrix}\right.\)

\(\left(x-1\right)^{2000}=\left(x-1\right)^{1998}\)

\(\Rightarrow\left(x-1\right)^{2000}-\left(x-1\right)^{1998}=0\)

\(\Rightarrow\left(x-1\right)^{1998}\left[\left(x-1\right)^2-1\right]=0\)

\(\Rightarrow\left[{}\begin{matrix}\left(x-1\right)^{2998}=0\\\left(x-1\right)^2=1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=1\\\left[{}\begin{matrix}x=2\\x=0\end{matrix}\right.\end{matrix}\right.\)

\(x^2< x\)

\(\Rightarrow x^2-x< 0\)

\(\Rightarrow x\left(x-1\right)< 0\)

Với mọi \(x\in R\) ta có:\(x-1< x\)

Nên \(\left\{{}\begin{matrix}x>0\\x-1< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>0\\x< 1\end{matrix}\right.\)

Xảy ra khi: \(0< x< 1\)

\(a,4.2^5:\left(2^3\cdot\dfrac{1}{16}\right)\\ =4.2^5:\left(2^3\cdot\dfrac{1}{2^4}\right)\\ =2^2.2^5:\dfrac{1}{2}\\ =2^7:\dfrac{1}{2}=2^7.2=2^8\)

\(b,3^2.2^5\left(\dfrac{2}{3}\right)^2\\ =\left(3\cdot\dfrac{2}{3}\right)^2.2^5\\ =2^2.2^5=2^7\)

\(c,\left(\dfrac{1}{3}\right)^2\cdot\dfrac{1}{3}.9^2\\ =\left(\dfrac{1}{3}\right)^3.3^4\\ =\left(\dfrac{1}{3}\right)^3.3^3.3\\ =\left(\dfrac{1}{3}.3\right)^3.3\\ =1^3.3=1.3\\ =3=3^1\)

a) \(4.2^5:\left(2^3\dfrac{1}{16}\right)\) \(=2^2.2^5:2^3:\dfrac{1}{16}\)

\(=2^7:2^3.16\)

\(= 2^4 . 2^4 = 2^8\)

b) \(3^2.2^5.\left(\dfrac{2}{3}\right)^2\) \(=3^2.2^5\)\(.\dfrac{2^2}{3^2}\)

\(=2^5.2^2=2^7\)

c) \(\left(\dfrac{1}{3}\right)^2.\) \(\dfrac{1}{3}.9^2\) \(=\left(\dfrac{1}{3}\right)^3.\left(3^2\right)^2\)

\(=\dfrac{1^3}{3^3}.3^4\)

\(=1^3.3\) \(= 3^1\)

HỌC TỐT NGHEN ~~~

Lời giải:

Chia cả hai vế cho \(5^x\):

\(\left(\frac{2}{5}\right)^x+\left(\frac{3}{5}\right)^x=1\)

Nếu \(x>1\)

Do \(\frac{2}{5}<1;\frac{3}{5}<1 \Rightarrow \left(\frac{2}{5}\right)^x<\frac{2}{5}; \left(\frac{3}{5}\right)^x< \frac{3}{5}\)

\(\Rightarrow \left(\frac{2}{5}\right)^x+\left(\frac{3}{5}\right)^x< \frac{2}{5}+\frac{3}{5}=1\) (vô lý)

Nếu \(x<1 \)

Do \(\frac{2}{5}; \frac{3}{5}<1 \Rightarrow \left(\frac{2}{5}\right)^x>\frac{2}{5}; \left(\frac{3}{5}\right)^x>\frac{3}{5}\)

\(\Rightarrow \left(\frac{2}{5}\right)^x+\left(\frac{3}{5}\right)^x>\frac{2}{5}+\frac{3}{5}=1\) (vô lý)

Từ 2 TH trên suy ra \(x=1\)