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(5x+1)2=36/49
Mà: (5x+1)2=36/49=(6/7)2 hoặc (5x+1)2=36/49=(-6/7)2
=>5x+1=6/7 hoặc 5x+1= -6/7
=> 5x= 6/7-1 hoặc 5x= -6/7 -1
=> 5x= -1/7 hoặc 5x= -13/7
=> x= -1/7: 5 hoặc x= -13/7: 5
=> x=-1/35 hoặc x= -13/35
(5x+1)2=36/49
Mà: (5x+1)2=36/49= (6/7)2
=>5x+1=6/7
=>5x=6/7-1
=>5x=-1/7
=>x= -1/7 : 5
=>x= -1/ 35
\(\left(5x+1\right)^2=\frac{36}{49}\)
\(\left(5x+1\right)^2=\left(\frac{6}{9}\right)^2\)
\(5x+1=\frac{6}{9}\)
\(5x=\frac{6}{9}-1\)
\(x=\frac{-1}{3}:5=\frac{-1}{3}.\frac{1}{5}=\frac{-1}{15}\)
\(\left(5x+1\right)^2=\frac{36}{49}\)
(+) 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 )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
\(\left(5x+1\right)^2=\frac{36}{49}\)
\(\Rightarrow\orbr{\begin{cases}5x+1=\frac{6}{7}\\5x+1=\frac{-6}{7}\end{cases}}\)\(\)
\(\Rightarrow\orbr{\begin{cases}5x=\frac{-1}{7}\\5x=\frac{-13}{7}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{-1}{35}\\x=\frac{-13}{35}\end{cases}}\)
a)Ta có:
\(\left(x-3,5\right)^2+\left(y-\dfrac{1}{10}\right)^4\le0\)
\(\Rightarrow x-3,5=y-\dfrac{1}{10}=0\Leftrightarrow\left\{{}\begin{matrix}x=3,5\\y=\dfrac{1}{10}=0,1\end{matrix}\right.\)
b) Ta có:
\(\left(5x+1\right)^2=\dfrac{36}{49}\)
\(\Leftrightarrow\left[{}\begin{matrix}5x+1=\dfrac{6}{7}\\5x+1=\dfrac{-6}{7}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{35}\\x=\dfrac{-13}{35}\end{matrix}\right.\)
b: ta có: \(\left(5x+1\right)^2=\dfrac{36}{49}\)
\(\Leftrightarrow\left[{}\begin{matrix}5x+1=\dfrac{6}{7}\\5x+1=-\dfrac{6}{7}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=\dfrac{-1}{7}\\5x=\dfrac{-13}{7}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{35}\\x=\dfrac{-13}{35}\end{matrix}\right.\)
\(\left(5x+1\right)^2=\frac{36}{49}\)
\(\Rightarrow\left(5x+1\right)^2=\left(\frac{6}{7}\right)^2\)
\(\Rightarrow5x+1=\frac{6}{7}\)
\(\Rightarrow5x=\frac{6}{7}-1=-\frac{1}{7}\)
\(\Rightarrow x=-\frac{1}{7}:5\) \(\Rightarrow x=-\frac{1}{7}\cdot\frac{1}{5}=-\frac{1}{35}\)
\(\left(5x+1\right)^2=\frac{36}{49}\)
\(\left(5x+1\right)^2=\left(\frac{6}{7}\right)^2\)
\(\Rightarrow5x+1=\frac{6}{7}\)
\(5x=-\frac{1}{7}\)
\(x=-\frac{1}{35}\)
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