Question

Tìm x:

a)(x+1)²=64

b)\(\frac{9}{2.4}+\frac{9}{4.6}+...+\frac{9}{96.98}+\frac{9}{98.100}\)

Answer 1

a) \(\left(x+1\right)^2=64\Leftrightarrow\left|x+1\right|=8\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x+1=8\\x+1=-8\end{array}\right.\) \(\Leftrightarrow\left[\begin{array}{nghiempt}x=7\\x=-9\end{array}\right.\)

b) \(\frac{9}{2.4}+\frac{9}{4.6}+...+\frac{9}{96.98}+\frac{9}{98.100}=\frac{9}{4}\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{48.49}+\frac{1}{49.50}\right)\)

\(=\frac{9}{4}\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{48}-\frac{1}{49}+\frac{1}{49}-\frac{1}{50}\right)\)

\(=\frac{9}{4}\left(1-\frac{1}{50}\right)=\frac{441}{200}\)

Answer 2

 

a)(x+1)²=64

x + 1      = 8

x            = 8 - 1

x            = 7

Answer 3

a)(x+1)²=64

=>(x+1)²=8²

=>x+1=8

=>x=7

b)\(\frac{9}{2.4}+\frac{9}{4.6}+...+\frac{9}{98.100}\)

\(=\frac{9}{2}\left(\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{98.100}\right)\)

\(=\frac{9}{2}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{98}-\frac{1}{100}\right)\)

\(=\frac{9}{2}\left(\frac{1}{2}-\frac{1}{100}\right)\)

\(=\frac{9}{2}\cdot\frac{49}{100}\)

\(=\frac{441}{200}\)

 

Answer 4

\(\frac{9}{2.4}+\frac{9}{4.6}+...+\frac{9}{96.98}+\frac{9}{98.100}\)

=\(\frac{9}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{96}-\frac{1}{98}+\frac{1}{98}-\frac{1}{100}\right)\)

=\(\frac{9}{2}.\left(\frac{1}{2}-\frac{1}{100}\right)\)

=\(\frac{9}{2}.\left(\frac{50}{100}-\frac{1}{100}\right)\)

=\(\frac{9}{2}.\left(\frac{50}{100}+\frac{-1}{100}\right)\)

=\(\frac{9}{2}.\frac{49}{100}\)

=\(\frac{441}{200}\)