b) Bình phương hai vế của PT , ta được :
( x - 1)2 = ( x - 5)2
⇔ ( x - 1)2 - ( x - 5)2 = 0
⇔ ( x - 1 - x + 5)( x - 1 + x - 5) = 0
⇔ 4( 2x - 6) = 0
⇔ 8( x - 3) = 0
⇔ x = 3
KL....
c) Bình phương hai vế của PT , ta được :
( x - 1)2 = ( 3x - 5)2
⇔ ( x - 1)2 - ( 3x - 5)2 = 0
⇔ ( x - 1 - 3x + 5)( x - 1 + 3x - 5) = 0
⇔ ( 4 - 2x )( 4x - 6) = 0
⇔ 2( 2 - x)( 2x - 3) = 0
⇔ x = 2 hoặc : x = \(\dfrac{3}{2}\)
KL....
a. 2|x−3| + (5x−1) = 0 (1)
T/h 1: x - 3 >0 <=> x > 3
(1) <=> 2(x-3) + 5x - 1 = 0
<=> 2x - 6 +5x - 1 = 0
<=> 7x - 7 = 0
<=> x = 1 (loại)
T/h 2: x - 3 < 0 <=> x < 3
(1) <=> 2(3-x) + 5x - 1 = 0
<=> 6 - 2x + 5x - 1 = 0
<=> 3x + 5 = 0
<=> x = \(\dfrac{-5}{3}\) (TĐK)
\(\Rightarrow S=\left\{\dfrac{-5}{3}\right\}\)
b. | x - 1 | = | x - 5 |
\(\Rightarrow\left[{}\begin{matrix}x-1=x-5\\x-1=5-x\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}0x=-4\\2x=6\end{matrix}\right.\)
\(\Rightarrow x=3\)
\(\Rightarrow S=\left\{3\right\}\)
c. | x - 1| = | 3x - 5 |
\(\Rightarrow\left[{}\begin{matrix}x-1=3x-5\\x-1=5-3x\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}-2x=-4\\4x=6\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{3}{2}\end{matrix}\right.\)
\(\Rightarrow S=\left\{\dfrac{3}{2};2\right\}\)
d. 2|x| - | x + 1 | = 2 (2)
T/h 1 : x < -1
(2) <=> - 2x + x + 1 = 2
<=> - x = 1
<=> x = -1 ( TĐK)
T/h 2: \(-1\le x< 0\)
(2) <=> - 2x - x - 1 = 2
<=> - 3x = 3
<=> x = -1 (TĐK)
T/h 3:\(x\ge0\)
(2) <=> 2x - x - 1 = 2
<=> x = 3 (TĐK)
\(\Rightarrow S=\left\{-1;3\right\}\)