\(\frac{x-3}{x+5}=\frac{5}{7}\)
=> 7 ( x - 3 ) = 5 ( x + 5 )
7x - 21 = 5x + 25
7x - 5x = 25 + 21
2x = 46
x = 23
\(\frac{x-3}{x+5}=\frac{5}{7}\)
=> \(7\left(x-3\right)=5\left(x+5\right)\)
<=> \(7x-21=5x+25\)
<=> \(2x=46\)
<=> \(x=23\)
Vậy...
\(\frac{7}{x-1}=\frac{x+1}{9}\)
<=> \(\left(x-1\right)\left(x+1\right)=63\)
<=> \(x^2-1=63\)
<=> \(x^2=64\)
<=> \(x=\pm8\)
Vậy...
a, (=) (x-3).7=5.(x+5)
(=)7x-21=5x+25
(=)7x-5x=25+21
(=)2x=46
(=)x=23
b,(=)(x-1)(x+1)=63
mà 63=1.63=7.9
mà (x-1)(x+1) là 2 số tự nhiên cách nhau 2 đơn vị
=>\(\orbr{\begin{cases}x-1=7\\x+1=9\end{cases}\left(=\right)\orbr{\begin{cases}x=8\\x=8\end{cases}}}\)
=>x=8
Tìm x , biết :
x - 3 / x + 5 = 5/7
<=> ( x - 3 ) . 7 = ( x + 5 ) . 5
<=> 7x - 21 = 5x + 25
<=> 7x - 5x = 25 + 21
<=> 2x = 46
<=> x = 46 : 2
<=> x = 23
Vậy x = 23
7 / x - 1 = x + 1 / 9
<=> ( x - 1 ) . ( x + 1 ) = 7 . 9
<=> x . ( x + 1 ) - 1 . ( x + 1 ) = 63
<=> x 2 + x - x - 1 = 63
<=> x2 + 0 - 1 = 63
<=> x2 + 0 = 63 + 1
<=> x2 + 0 = 64
<=> x2 = 64 - 0
<=> x2 = 64
<=> x2 = 82 hoặc (-8)2
=> x = 8 hoặc -8
Vậy x € { 8 ; -8 }