`(5x+1)=36/49`

`<=> 5x = 36/49-1`

`<=> 5x = -13/49`.

`<=> x = -13/245.`

Vậy `x = -13/245`.

`b, x-2/9 = 2/3`.

`<=> x = 2/3 + 2/9`

`<=> x = 8/9`.

Vậy `x = 8/9`.

c: (8x-1)^(2x+1)=5^(2x+1)

=>8x-1=5

=>8x=6

=>x=3/4

d: Sửa đề: (x-3,5)^2+(y-1/10)^4=0

=>x-3,5=0 và y-0,1=0

=>x=3,5 và y=0,1

\(\left(5x+1\right)^2=\frac{36}{49}\)

\(\left(5x+1\right)^2=\left(\frac{6}{9}\right)^2\)

\(5x+1=\frac{6}{9}\)

\(5x=\frac{6}{9}-1\)

\(x=\frac{-1}{3}:5=\frac{-1}{3}.\frac{1}{5}=\frac{-1}{15}\)

a,x=-2

b,minh chiu

  \(\left(5x+1\right)^2=\frac{36}{49}\)

(+) TH 1: 5x + 1 = 6/7

              5x        = 6/7 - 1

               5x        =  -1/7

                 x        = -1/7 : 5 

                 x          = -1 /35

(+) TH2 : 5x + 1 = - 6/7

             5x         = -6/7 - 1

             5x         = -13/7

               x         =-13/7 : 5 

               x          = -13/35

( 5x + 1 )² = 36/49

<=> ( 5x + 1 )² = ( ±6/7 )²

<=> 5x + 1 = 6/7 hoặc 5x + 1 = -6/7

<=> x = -1/35 hoặc x = -13/35

\(\left(5x+1\right)^2=\frac{36}{49}\)

\(\Rightarrow\orbr{\begin{cases}5x+1=\frac{6}{7}\\5x+1=\frac{-6}{7}\end{cases}}\)\(\)

\(\Rightarrow\orbr{\begin{cases}5x=\frac{-1}{7}\\5x=\frac{-13}{7}\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{-1}{35}\\x=\frac{-13}{35}\end{cases}}\)

\(\left(5x+1\right)=\pm\sqrt{\frac{36}{49}}=\pm\frac{6}{7}\)                         

\(\orbr{\begin{cases}5x+1=\frac{6}{7}\\5x+1=-\frac{6}{7}\end{cases}}\)      

\(\orbr{\begin{cases}5x=\frac{-1}{7}\\5x=\frac{-13}{7}\end{cases}}\)    

\(\orbr{\begin{cases}x=\frac{-1}{35}\\x=\frac{-13}{35}\end{cases}}\)

\(5^x.\left(5^3\right)^2=625\)

\(\Rightarrow5^x.5^6=5^4\)

\(\Rightarrow5^x=5^4:5^6\)

\(\Rightarrow5^x=\frac{1}{25}\)

.......................

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