a, \(\left(3x-5\right)\left(x+1\right)-\left(3x-1\right)\left(x+1\right)=x-4\)
\(\Leftrightarrow\left(x+1\right)\left(3x-5-3x+1\right)=x-4\Leftrightarrow-4\left(x+1\right)=x-4\)
\(\Leftrightarrow-4x-4=x-4\Leftrightarrow-4x-x=0\Leftrightarrow x=0\)
b, \(\left(x-2\right)\left(x+3\right)-\left(x+4\right)\left(x-7\right)=5-x\)
\(\Leftrightarrow x^2+x-6-x^2-3x+28=5-x\Leftrightarrow-2x+22=5-x\Leftrightarrow x=17\)
c, thiếu đề
d, \(3\left(x-7\right)\left(x+7\right)-\left(x-1\right)\left(3x+2\right)=13\)
\(\Leftrightarrow3x^2-147-3x^2+x+2=13\Leftrightarrow x=11+147=158\)
a.\(3x^2-2x-5-\left(3x^2+2x-1\right)=x-4\)
\(\Leftrightarrow-5x=0\Leftrightarrow x=0\)
b.\(x^2+x-6-\left(x^2-3x-28\right)=5-x\)
\(\Leftrightarrow5x=-17\Leftrightarrow x=-\frac{17}{5}\)
c.\(5\left(x^2-10x+21\right)-\left(5x^2-9x-2\right)=0\)
\(\Leftrightarrow-41x+107=0\Leftrightarrow x=\frac{107}{41}\)
d.\(3\left(x^2-49\right)-\left(3x^2-x-2\right)=13\Leftrightarrow x=158\)
sửa dòng 2 phần b ;-;
\(x^2+x-6-x^2+3x+28=5-x\)
\(\Leftrightarrow4x+22=5-x\Leftrightarrow5x=-17\Leftrightarrow x=-\frac{17}{5}\)
Trả lời:
a, ( 3x - 5 ) ( x + 1 ) - ( 3x - 1 ) ( x + 1 ) = x - 4
<=> ( x + 1 ) [ ( 3x - 5 ) - ( 3x - 1 ) ] = x - 4
<=> ( x + 1 ) ( 3x - 5 - 3x + 1 ) = x - 4
<=> ( x + 1 ) ( - 4 ) = x - 4
<=> - 4x - 4 = x - 4
<=> - 4x - x = - 4 + 4
<=> - 5x = 0
<=> x = 0
b, ( x - 2 ) ( x + 3 ) - ( x + 4 ) ( x - 7 ) = 5 - x
<=> x2 + 3x - 2x - 6 - ( x2 - 7x + 4x - 28 ) = 5 - x
<=> x2 + x - 6 - x2 + 7x - 4x + 28 = 5 - x
<=> 4x + 22 = 5 - x
<=> 4x + x = 5 - 22
<=> 5x = - 17
<=> x = - 17/5
d, 3 ( x - 7 ) ( x + 7 ) - ( x - 1 ) (3x + 2 ) = 13
<=> ( 3x - 21 ) ( x + 7 ) - ( x - 1 ) ( 3x + 2 ) = 13
<=> 3x2 + 21x - 21x - 147 - ( 3x2 + 2x - 3x - 2 ) = 13
<=> 3x2 - 147 - 3x2 - 2x + 3x + 2 = 13
<=> x - 145 = 13
<=> x = 158
cảm ơn nhìu nha
câu c) 5.(x-3).(x-7)-(5x+1).(x-2)=8
