Ta có:

\(\frac{1}{2}\left(2x-1\right)-5=3x-0,5\)

=> \(x-\frac{1}{2}-5=3x-0,5\)

=> \(x-5,5=3x-0,5\)

=> \(x-3x=-0,5+5,5\)

=> \(-2x=5\)

=> \(x=5:\left(-2\right)=-2,5\)

\(\frac{1}{2}\left(2x-1\right)-5=3x-0,5\)

\(\Leftrightarrow x-\frac{1}{2}-5=3x-0,5\)

\(\Leftrightarrow x-5,5=3x-0,5\)

\(\Leftrightarrow x-3x=5,5-0,5\)

\(\Leftrightarrow-2x=5\)

\(\Leftrightarrow x=\frac{-5}{2}\)

Vậy \(x=\frac{-5}{2}\)

 \(\left|x^2+2-x\right|=x^2-1\)

ĐKXĐ: x² -1 \(\ge\)0  <=> x² \(\ge\)1 <=> \(\orbr{\begin{cases}x\ge1\\x\le-1\end{cases}}\)

TH1: x² + 2 - x = x² - 1

=> x² - x - x² = -1 - 2

=> -x = -3

=> x = 3 (tm)

TH2: x² + 2 - x = 1 - x²

=> x² - x + x² = 1 - 2

=> 2x² - x = -1

=> ko có giá trị x thõa mãn

Vì 2x² \(\ge\)  x <=> 2x² - x \(\ge\)0

Vậy x = 3 tm

Edogawa ConanDài v:

\(ĐK:\orbr{\begin{cases}x\ge1\\x\le-1\end{cases}}\)

Ta có: \(x^2+2-x=\left(x-\frac{1}{2}\right)^2+\frac{7}{4}>0\)

\(\Rightarrow pt\Leftrightarrow x^2+2-x=x^2-1\)

\(\Leftrightarrow2-x=-1\Leftrightarrow x=3\)(t/m)

Ta có : 2⁵⁰⁰ = 2^5.100 = (2⁵)¹⁰⁰ = 32¹⁰⁰

           5²⁰⁰ = 5^2.100 = (5²)¹⁰⁰ = 25¹⁰⁰

Vì 32 > 25 nên 32¹⁰⁰ > 25¹⁰⁰ nên 2⁵⁰⁰ > 5²⁰⁰

Vậy 2⁵⁰⁰ > 5²⁰⁰

Dễ mà bạn, tách ra thoii ạ

\(2^{500}\)

\(=\left(2^5\right)^{100}=32^{100}\)

\(5^{200}\)

\(=\left(5^2\right)^{100}=10^{100}\)

có \(32^{100}>10^{100}\)

\(\Rightarrow2^{500}>5^{200}\)