Toán 6 Ôn Tập So Sánh và Tìm x

Nguyễn Thị Quỳnh Lan said:

Bài 1: Tìm x:
\frac{2}{3x} - \frac{2}{3} (x - \frac{1}{2}) = \frac{5}{12}

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ĐKXĐ: [tex]x\neq 0[/tex]
Có:
[tex]\frac{2}{3x} - \frac{2}{3} (x - \frac{1}{2}) = \frac{5}{12}[/tex]
[tex]\Leftrightarrow \frac{2}{3}(\frac{1}{x}-x)+\frac{1}{3}=\frac{5}{12}[/tex]
[tex]\Leftrightarrow \frac{1}{x}-x=\frac{1}{8}[/tex]
[tex]\Leftrightarrow x^2+\frac{1}{8}x-1=0[/tex]
[tex]\Leftrightarrow (x+\frac{1}{16})^2-\frac{257}{256}=0[/tex]
[tex]\Leftrightarrow (x+\frac{1}{16}-\frac{\sqrt{257}}{16})(x+\frac{1}{16}+\frac{\sqrt{257}}{16})=0[/tex]
[tex]\Leftrightarrow x=...[/tex]

2. [tex]A=1+\frac{2020^{200}-2020^{199}}{2020^{199}+3};B=\frac{2020^{199}+3}{2020^{198}+3}=\frac{2020^{200}+3.2020}{2020^{199}+3.2020}=1+\frac{2020^{200}-2020^{199}}{2020^{199}+3.2020}[/tex]
Ta thấy: [TEX]\frac{2020^{200}-2020^{199}}{2020^{199}+3} > \frac{2020^{200}-2020^{199}}{2020^{199}+3.2020} \Rightarrow A > B[/TEX]