\(49=7^2\)

\(121=11^2\)

\(225=15^2\)

\(196=14^2\)

\(1024=32^2\)

( 3x + 1 )² : \(\left(-\frac{1}{4}\right)\)= \(-\frac{49}{4}\)

( 3x + 1 )² = \(-\frac{49}{4}\times\left(-\frac{1}{4}\right)\)

( 3x + 1 )² = \(\frac{49}{16}\)

\(\Rightarrow\)( 3x + 1 )² = \(\left(\frac{7}{4}\right)^2\)hoặc \(\left(-\frac{7}{4}\right)^2\)

Ta xét 2 trường hợp

Th 1 :

( 3x + 1 )² = \(\left(\frac{7}{4}\right)^2\)

=> 3x + 1 = \(\frac{7}{4}\)

=> 3x = \(\frac{7}{4}-1\)

=> 3x = \(\frac{3}{4}\)

=> x = \(\frac{3}{4}\): 3

=> x = \(\frac{1}{4}\)

Th 2 :

( 3x + 1 )² = \(\left(-\frac{7}{4}\right)^2\)

=> 3x + 1  = \(-\frac{7}{4}\)

=> 3x = \(-\frac{7}{4}-1\)

=> 3x = \(-\frac{11}{4}\)

=> x = \(-\frac{11}{4}\): 3

=> x = \(-\frac{11}{12}\)

Vậy x \(\in\){ \(\frac{1}{4}\); \(-\frac{11}{12}\)}

Ta có :

60 = 2² . 3 . 5

126 = 2 . 3² . 7

=> ƯCLN ( 60 ; 126 ) = 2 . 3 = 6

Vậy ..........

Giải :

Vì 60 = 2² . 3 . 5

126 = 2 . 3² . 7

=> ƯCLN( 60 , 126 ) = 2 . 3 = 6

~~Học tốt~~

\(A=\){ C ; H ; A ; M ; O ;  }

#Winter

A = { c ; h ; ă ; m ; h ; o ; c } 

\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}=\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{100.100}\)

\(< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{99.100}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)

\(=1-\frac{1}{100}< 1\)(ĐPCM)

\(\frac{4n+5}{2n-1}=\frac{2\left(2n-1\right)+7}{2n-1}=1+\frac{7}{2n-1}.\)

\(4n+5⋮2n-1\) khi \(7⋮3n-1\)

\(\Rightarrow2n-1=\left\{-7;-1;1;7\right\}\Rightarrow n=\left\{-3;0;1;4\right\}\)