Tìm x biết:
(5x + 1)2 = 36/49
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\(\left(5x+1\right)^2=\frac{36}{49}\)
\(\left(5x+1\right)^2=\left(\frac{6}{9}\right)^2\)
\(5x+1=\frac{6}{9}\)
\(5x=\frac{6}{9}-1\)
\(x=\frac{-1}{3}:5=\frac{-1}{3}.\frac{1}{5}=\frac{-1}{15}\)
\(\left(5x+1\right)^2=\frac{36}{49}\)
(+) 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)=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
a) (5x + 1)2 = 36/49
(5x + 1)2 = (6/7)2
suy ra: 5x + 1= 6/7 hoặc 5x +1 = -6/7
5x = 6/7 - 1 5x = -6/7 -1
5x = -1/7 5x = -13/7
x = -1/7 : 5 x = -13/7 : 5
x = -1/35 x = -13/35
a, \(\left(5x+1\right)^2=\frac{36}{49}\)
\(\left(5x+1\right)^2=\left(\frac{6}{7}\right)^2\)
=> 5x + 1 = \(\frac{6}{7}\)
\(\Rightarrow5x=\frac{6}{7}-1\)
\(\Rightarrow5x=\frac{-1}{7}\)
\(\Rightarrow x=\frac{-1}{7}:5=\frac{-1}{35}\)
b,, \(\left(2x-\frac{1}{2}\right)^2=\frac{1}{4}\)
\(\left(2x-\frac{1}{2}\right)^2=\left(\frac{1}{2}\right)^2\)
\(\Rightarrow2x-\frac{1}{2}=\frac{1}{2}\)
\(\Rightarrow2x=\frac{1}{2}+\frac{1}{2}=1\)
\(\Rightarrow x=1:2=\frac{1}{2}\)
\(\left(5x+1\right)^2=\left(\frac{\pm6}{7}\right)^2\)
+) 5x + 1 = 6/7
5x = -1/7
x = -1/35
+) 5x + 1 = -6/7
5x = -13/7
x = -13/35
Vậy,.........
( 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
\(\left(5x+1\right)^2=\frac{36}{49}\)
\(\Rightarrow\orbr{\begin{cases}5x+1=\frac{6}{7}\\5x+1=\frac{-6}{7}\end{cases}}\)\(\)
\(\Rightarrow\orbr{\begin{cases}5x=\frac{-1}{7}\\5x=\frac{-13}{7}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{-1}{35}\\x=\frac{-13}{35}\end{cases}}\)