Bài 1:

a)\(\frac{\left(0,8\right)^5}{\left(0,4\right)^6}=\frac{\left(0,2\cdot4\right)^5}{\left(0,2\cdot2\right)^6}=\frac{\left(0,2\right)^5\cdot\left(2^2\right)^5}{\left(0,2\right)^6\cdot2^6}=\frac{\left(0,2\right)^5\cdot2^{10}}{\left(0,2\right)^6\cdot2^6}=\frac{2^4}{0,2}=\frac{16}{\frac{2}{10}}=80\)

b)\(\frac{8^{10}+4^{10}}{8^4+4^{11}}=\frac{\left(2^3\right)^{10}+\left(2^2\right)^{10}}{\left(2^3\right)^4+\left(2^2\right)^{11}}=\frac{2^{30}+2^{20}}{2^{12}+2^{22}}=\frac{2^{20}\left(2^{10}+1\right)}{2^{12}\left(1+2^{10}\right)}=\frac{2^{20}}{2^{12}}=256\)

Bài 2:

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

\(\Rightarrow2^{x-1}=2^4\)

\(\Rightarrow x-1=4\Rightarrow x=5\)

b)\(\left(x-1\right)^2=25\)

\(\Rightarrow\left(x-1\right)^2=5^2=\left(-5\right)^2\)

\(\Rightarrow x-1=5\) hoặc \(x-1=-5\)

\(\Rightarrow x=6\) hoặc \(x=-4\)

Vậy \(x=6\) hoặc \(x=-4\)

c)\(\left(x-1\right)^{x+2}=\left(x-1\right)^{x+6}\)

\(\Rightarrow\left(x-1\right)^{x+2}-\left(x-1\right)^{x+6}=0\)

\(\Leftrightarrow\left(x-1\right)^{x+2}\left[1-\left(x-1\right)^4\right]\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}\left(x-1\right)^{x+2}=0\\1-\left(x-1\right)^4=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x-1=0\\1=\left(x-1\right)^4\end{array}\right.\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=1\\\left(x-1\right)^4=\left(-1\right)^4=1^4\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=1\\x-1=1\\x-1=-1\end{array}\right.\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=1\\x=2\\x=0\end{array}\right.\)

d)\(\left(x+20\right)^{100}+\left|y+4\right|=0\left(1\right)\)

Ta thấy: \(\begin{cases}\left(x+20\right)^{100}\ge0\\\left|y+4\right|\ge0\end{cases}\)

\(\Rightarrow\left(x+20\right)^{100}+\left|y+4\right|\ge0\left(2\right)\)

Từ (1) và (2) suy ra \(\begin{cases}\left(x+20\right)^{100}=0\\\left|y+4\right|=0\end{cases}\)

\(\Rightarrow\begin{cases}x+20=0\\y+4=0\end{cases}\)\(\Rightarrow\begin{cases}x=-20\\y=-4\end{cases}\)

Đặt  z -60 = t 

\(x+y+z=100\Rightarrow x+y+t=40;\)

\(\Leftrightarrow x+y+t\ge3\sqrt[3]{xyt}\Leftrightarrow xyt\le\frac{\left(x+y+t\right)^3}{3^3}=\left(\frac{40}{3}\right)^3\)

\(Max\left(xyt\right)=\left(\frac{40}{3}\right)^3\) khi x =y =t =40/3  => z =60+t =60+40/3=220/3

=>\(xyz\le\frac{40}{3}.\frac{40}{3}.\frac{220}{3}=\frac{352000}{27}\) khi x =y =40/3 ; z =220/3

a, x=65,9

b, x=1/5=0,2

c,x=12300

x = -20

y =-4

(x+20)¹⁰⁰ + |y+4| = 0

Xet thay: (x+20)¹⁰⁰ lon hon hoac bang 0 ( vi co so mu chan) 

               | y+4| lon hon hoac bang 0

Ma: (x+20)¹⁰⁰ + |y+4| = 0

=> (x+20)¹⁰⁰ = 0 => x = -20

=> |y+4| = 0 => y = -4

Vay: x=-20  ;   y=-4

Ban ăn d

<=>(X+20)^100=0

=>X+20=0

<=>|Y+4|=0=>Y+4=0

=>X=-20

=>Y=-4

Vì \(\left(x+20\right)^{100}\ge0;\left|y+4\right|\ge0\)

\(\Rightarrow\left(x+20\right)^{100}+\left|y+4\right|\ge0\)

Dấu "=" xảy ra <=> \(\orbr{\begin{cases}\left(x+20\right)^{100}=0\\\left|y+4\right|=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x+20=0\\y+4=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=-20\\y=-4\end{cases}}}\)

Vậy x = - 20; y = - 4