Toán 7 Giá trị tuyệt đối và bài toán tìm x

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a) [tex]\frac{1}{1 . 4}[/tex] + [tex]\frac{1}{4 . 7}[/tex] + ... + [tex]\frac{1}{97 . 100}[/tex] = [tex]\left | \frac{x}{3} \right |[/tex]

[tex]\Leftrightarrow[/tex] [tex]\frac{1}{3}\left ( 1 - \frac{1}{4} + \frac{1}{4} - \frac{1}{7} + ... + \frac{1}{97} - \frac{1}{100} \right )[/tex] = [tex]\left | \frac{x}{3} \right |[/tex]

[tex]\Leftrightarrow[/tex] [tex]\frac{1}{3}\left ( 1 - \frac{1}{100} \right )[/tex] = [tex]\left | \frac{x}{3} \right |[/tex]

[tex]\Leftrightarrow[/tex] |x| = 1 - [tex]\frac{1}{100}[/tex]

[tex]\Leftrightarrow[/tex] |x| = [tex]\frac{99}{100}[/tex]

TH1: x = [tex]\frac{99}{100}[/tex]

TH2: x = [tex]-\frac{99}{100}[/tex]

Vậy x [tex]\epsilon[/tex] [tex]\left \{ \frac{99}{100}; -\frac{99}{100} \right \}[/tex]

b) [tex]\frac{4}{1 . 5}[/tex] + [tex]\frac{4}{5 . 9}[/tex] + ... + [tex]\frac{4}{97 . 101}[/tex] = [tex]\left | \frac{5x - 4}{101} \right |[/tex]

[tex]\Leftrightarrow[/tex] [tex]\left ( 1 - \frac{1}{5} + \frac{1}{5} - \frac{1}{9} + ... + \frac{1}{97} - \frac{1}{101} \right )[/tex] = [tex]\left | \frac{5x - 4}{101} \right |[/tex]

[tex]\Leftrightarrow[/tex] 1 - [tex]\frac{1}{101}[/tex] = [tex]\left | \frac{5x - 4}{101} \right |[/tex]
[tex]\Leftrightarrow[/tex] |5x - 4| = 100
TH1: |5x - 4| = 100 [tex]\Leftrightarrow[/tex] 5x = 104 [tex]\Leftrightarrow[/tex] x = [tex]\frac{104}{5}[/tex]

TH2: |5x - 4| = -100 [tex]\Leftrightarrow[/tex] 5x = -96 [tex]\Leftrightarrow[/tex] x = [tex]\frac{-96}{5}[/tex]
Vậy x [tex]\epsilon \left \{ \frac{104}{5}; \frac{-96}{5} \right \}[/tex]

c) [tex]\left ( 1 - \frac{1}{2} \right )\left ( 1 - \frac{1}{3} \right )\left ( 1 - \frac{1}{4} \right )....\left ( 1 - \frac{1}{100} \right )[/tex] = [tex]\left | x - 1\frac{99}{100} \right |[/tex]

[tex]\Leftrightarrow[/tex] [tex]\left ( 1 - \frac{1}{2} \right )\left ( 1 - \frac{1}{3} \right )....\left ( 1 - \frac{1}{100} \right )[/tex] = [tex]\left | x - \frac{199}{100} \right |[/tex]

[tex]\Leftrightarrow \frac{1}{2}[/tex] . 2 . [tex]\frac{1}{3}[/tex] . 3 . [tex]\frac{1}{4}[/tex] . 4 . [tex]\frac{1}{5}[/tex] .... 99 . [tex]\frac{1}{100}[/tex] = [tex]\left | x - \frac{199}{100} \right |[/tex]

[tex]\Leftrightarrow \left | x - \frac{199}{100} \right |[/tex] = [tex]\frac{1}{100}[/tex]

TH1: [tex]x - \frac{199}{100}[/tex] = [tex]\frac{1}{100}[/tex] [tex]\Leftrightarrow[/tex] x = 2

TH2: [tex]x - \frac{199}{100}[/tex] = [tex]-\frac{1}{100}[/tex] [tex]\Leftrightarrow[/tex] x = [tex]\frac{99}{50}[/tex]

Vậy x [tex]\epsilon \left \{ 2; \frac{99}{50} \right \}[/tex]
 
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