Question

Bài 1 Tìm x :

{ 2x + \(\frac{1}{5}\)}\(^2\)=\(\frac{9}{25}\)

Bài 2: Chứng tỏ 2⁹ + 2⁹⁹ chia hết cho 100

CÁC BẠN ƠI GIÚP MK VỚI.

Answer 1

1) \(\left(2x+\frac{1}{5}\right)^2=\frac{9}{25}\)

\(\Rightarrow\left(2x+\frac{1}{5}\right)^2=\left(\frac{3}{5}\right)^2=\left(\frac{-3}{5}\right)^2\)

\(\Rightarrow\left[\begin{array}{nghiempt}2x+\frac{1}{5}=\frac{3}{5}\\2x+\frac{1}{5}=\frac{-3}{5}\end{array}\right.\) \(\Rightarrow\left[\begin{array}{nghiempt}2x=\frac{2}{5}\\2x=\frac{-4}{5}\end{array}\right.\) \(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{1}{5}\\x=\frac{-2}{5}\end{array}\right.\)

Vậy \(\left[\begin{array}{nghiempt}x=\frac{1}{5}\\y=\frac{-2}{5}\end{array}\right.\)

2) Ta có:

2⁹ + 2⁹⁹

= 2⁹.(1 + 2⁹⁰)

 

= 512.(1 + 2⁸⁰.2¹⁰)

= 512.[1 + (2²⁰)⁴.1024]

= 512.[1 + (...26)⁴.2014)]

= 512.[1 + (...26).1024]

= 512.[1 + (...24)]

= 512.(...25)

= 128.4.(...25)

= 128.(...00)

= (...00) \(⋮100\)

Chứng tỏ \(2^9+2^{99}⋮100\)

Answer 2

Bài 1:

\(\left(2x+\frac{1}{5}\right)^2=\frac{9}{25}\)

\(\Leftrightarrow2x+\frac{1}{5}=\pm\frac{3}{5}\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}2x+\frac{1}{5}=\frac{3}{5}\\2x+\frac{1}{5}=-\frac{3}{5}\end{array}\right.\Leftrightarrow\left[\begin{array}{nghiempt}2x=\frac{2}{5}\\2x=-\frac{4}{5}\end{array}\right.\Leftrightarrow\left[\begin{array}{nghiempt}x=\frac{1}{5}\\x=-\frac{2}{5}\end{array}\right.\)

Vậy ........