\(4^{x-5}=16\)

\(4^{x-5}=4^2\)

\(x-5=2\)

\(x=2+5\)

\(x=7\)

\(45-2^{x-1}=29\)

\(2^{x-1}=16\)

\(2^{x-1}=2^4\)

\(x-1=4\)

\(x=5\)

\(\left(2+x\right)^2=144\)

\(\left(2+x\right)^2=12^2\)

\(2+x=12\)

\(x=12-2\)

\(x=10\)

\(\left(x-5\right)^2=81\)

\(\left(x-5\right)^2=9^2\)

\(x-5=9\)

\(x=14\)

\(\left(13-x\right)^4=81\)

\(\left(13-x\right)^4=3^4\)

\(13-x=3\)

\(x=13-3\)

\(x=10\)

\(...4^{x-5}=4^2\Rightarrow x-5=2\Rightarrow x=7\)

\(...2^{x-1}=45-29=16\Rightarrow2^{x-1}=2^4\Rightarrow x-1=4\Rightarrow x=5\)

\(...\Rightarrow\left(2+x\right)^2=12^2\Rightarrow\left[{}\begin{matrix}2+x=12\\2+x=-12\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=10\\x=-14\end{matrix}\right.\)

\(...\Rightarrow\left(x-5\right)^2=9^2\Rightarrow\left[{}\begin{matrix}x-5=3\\x-5=-3\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=8\\x=2\end{matrix}\right.\)

\(...\Rightarrow\left(13-x\right)^4=3^4\Rightarrow\left[{}\begin{matrix}13-x=3\\13-x=-3\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=10\\x=16\end{matrix}\right.\)

Mình quên mũ 2 với mũ 4 là chia ra 2 trường hợp, sửa lại nhé:

\(\left(2+x\right)^2=144\)

 \(\left(2+x\right)^2=12^2\)

Th1: 

\(2+x=12\)

\(x=10\)

Th2: 

\(2+x=-12\)

\(x=-14\)

\(\left(x-5\right)^2=81\)

\(\left(x-5\right)^2=9^2\)

Th1: 

\(x-5=9\)

\(x=14\)

Th2: 

\(x-5=-9\)

\(x=-4\)

\(\left(13-x\right)^4=81\)

\(\left(13-x\right)^4=3^4\)

Th1: 

\(13-x=3\)

\(x=10\)

Th2: 

\(13-x=-3\)

\(x=16\)

Bài 2: 

x=13 nên x+1=14

\(f\left(x\right)=x^{14}-x^{13}\left(x+1\right)+x^{12}\left(x+1\right)-...+x^2\left(x+1\right)-x\left(x+1\right)+14\)

\(=x^{14}-x^{14}-x^{13}+x^{13}-...+x^3+x^2-x^2-x+14\)

=14-x=1

x=13 nên x+1=14

f(x)=x14−x13(x+1)+x12(x+1)−...+x2(x+1)−x(x+1)+14f(x)=x14−x13(x+1)+x12(x+1)−...+x2(x+1)−x(x+1)+14

=x14−x14−x13+x13−...+x3+x2−x2−x+14=x14−x14−x13+x13−...+x3+x2−x2−x+14

=14-x=1

tu hoc moi gioi

\(\left(x+\dfrac{2}{9}\right)^2=\dfrac{4}{81}\\ \left(x+\dfrac{2}{9}\right)^2=\left(\dfrac{2}{9}\right)^2\\ x+\dfrac{2}{9}=\dfrac{2}{9}\\ x=\dfrac{2}{9}-\dfrac{2}{9}\\ x=0\)

Mình bổ sung thêm TH2 nữa. 

\(\left(x+\dfrac{2}{9}\right)^2=\left(-\dfrac{2}{9}\right)^2\\ x+\dfrac{2}{9}=-\dfrac{2}{9}\\ x=-\dfrac{2}{9}-\dfrac{2}{9}\\ x=-\dfrac{4}{9}\)

\(...\Rightarrow\left(x+\dfrac{2}{9}\right)^2=\left(\dfrac{2}{9}\right)^2\)

\(\Rightarrow x+\dfrac{2}{9}=\dfrac{2}{9}\Rightarrow x=0\)

\(2^x\cdot2=128\)

\(\Rightarrow2^x=128:2\)

\(\Rightarrow2^x=64\)

\(\Rightarrow2^x=2^6\)

\(\Rightarrow x=6\)

==============

\(4^x:4^3=16\)

\(\Rightarrow4^{x-3}=16\)

\(\Rightarrow4^{x-3}=4^2\)

\(\Rightarrow x-3=2\)

\(\Rightarrow x=3+2=5\)

=============

\(2^x:3=48\)

\(\Rightarrow2^x=48:3\)

\(\Rightarrow2^x=16\)

\(\Rightarrow2^x=2^4\)

\(\Rightarrow x=4\)

==============

\(3^{x-5}=81\)

\(\Rightarrow3^{x-5}=3^4\)

\(\Rightarrow x-5=4\)

\(\Rightarrow x=4+5\)

\(\Rightarrow x=9\)

2ˣ⁻² = 128

2ˣ⁻² = 2⁷

x - 2 = 7

x = 7 + 2

x = 9

Các câu còn lại xem bài bạn Phong

\(125\)

\(\left(x+2\right)^4=81\)

\(\Rightarrow\left(x+2\right)^4=3^4\)

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

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

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

Vậy \(S=\left\{1;-5\right\}\)

\(\left(x+2\right)^4=81\\ \Leftrightarrow\left(x+2\right)^4=3^4\\ \Leftrightarrow\left[{}\begin{matrix}-x-2=3\\x+2=3\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-5\\x=1\end{matrix}\right.\)

Vậy...

(x-2)^4=81

(x-2)^4=3^4

   =>x-2=3

          x=3+2

          x=5

   \(\)Vậy x=5