ta có 166/256 lớn hơn phaand tử lớn hơn ko so sánh đc phần bù

ĐKXĐ: \(x\ge-1\)

pt \(\Leftrightarrow-x+4\sqrt{x+1}-5=x^2-5x+14-5\)

\(\Leftrightarrow-\left(x+1-4\sqrt{x+1}+4\right)=x^2-6x+9\)

\(\Leftrightarrow-\left(\sqrt{x+1}-2\right)^2=\left(x-3\right)^2\)

Ta thấy \(VT\le0\forall x\ge-1;VP\ge0\forall x\)

nên pt \(\Leftrightarrow\left\{{}\begin{matrix}\left(\sqrt{x+1}-2\right)^2=0\\\left(x-3\right)^2=0\end{matrix}\right.\)

\(\Leftrightarrow x=3\left(TM\right)\)

Vậy ...

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}\Rightarrow\left(\dfrac{x}{2}\right)^3=\dfrac{xyz}{2.3.5}\Leftrightarrow\dfrac{x^3}{8}=\dfrac{240}{30}=8\Leftrightarrow x^3=64\Leftrightarrow x=4\Rightarrow\left\{{}\begin{matrix}y=6\\z=10\end{matrix}\right.\)

con heo đất

a) 2⁵.8⁴ = 2⁵.2¹² = 2^{5 + 12} = 2¹⁷

b) 12³.3³ = 3³.2⁶.3³ = 3⁶.2⁶ = 6⁶

c) 25⁶.125³ = 5¹².5⁹ = 5²¹

d) 3.3².3³...3⁹⁹.3¹⁰⁰ = 3^{1 + 2 + 3 + ... + 100} = 3⁵⁰⁵⁰