a, \(\left(x-5\right)\left(x+2\right)+\left(x+1\right)\left(2-x\right)=15\)
\(\Leftrightarrow x^2+2x-5x-10+2x-x^2+2-x=15\Leftrightarrow-2x-23=0\)
\(\Leftrightarrow x=-\frac{23}{2}\)
b, \(\left(2x-3\right)\left(x+5\right)-\left(x-2\right)\left(2x+1\right)=3\)
\(\Leftrightarrow2x^2+10x-3x-15-\left(2x^2+x-4x-2\right)=3\)
\(\Leftrightarrow10x-16=0\Leftrightarrow x=\frac{8}{5}\)
(x -5)(x + 2) + (x + 1)(2 - x) = 15
=> x2 - 3x - 10 + x - x2 + 2 = 15
=> -2x = 23
=> x = - 11,5
b)(2x - 3)(x + 5) - (x - 2)(2x + 1) = 3
=> 2x2 + 7x - 15 - 2x2 + 3x + 2 = 3
=> 10x = 16
=> x = 1,6
Vậy x = 1,6
\(a,\left(x-5\right)\left(x+2\right)+\left(x+1\right)\left(2-x\right)=15\)
\(\Leftrightarrow x^2+2x-5x-5x-10+2x-x^2+2-x-15=0\)
\(\Leftrightarrow-7x-23=0\)
\(\Leftrightarrow-7x=23\)
\(\Leftrightarrow x=\frac{-23}{7}\)
Vậy x = \(\frac{-23}{7}\)
\(b,\left(2x-3\right)\left(x+5\right)-\left(x-2\right)\left(2x+1\right)=3\)
\(\Leftrightarrow2x^2+10x-3x-15-2x^2-x+4x+2-3=0\)
\(\Leftrightarrow10x-16=0\)
\(\Leftrightarrow x=\frac{16}{10}\)
Vậy x = \(\frac{16}{10}\)
Học tốt nhá :))
a) ( x - 5 )( x + 2 ) + ( x + 1 )( 2 - x ) = 15
<=> x2 - 3x - 10 - x2 + x + 2 = 15
<=> -2x - 8 = 15
<=> -2x = 23
<=> x = -23/2
b) ( 2x - 3 )( x + 5 ) - ( x - 2 )( 2x + 1 )
<=> 2x2 + 7x - 15 - ( 2x2 - 3x - 2 ) = 3
<=> 2x2 + 7x - 15 - 2x2 + 3x + 2 = 3
<=> 10x - 13 = 3
<=> 10x = 16
<=> x = 16/10 = 8/5
Bài làm :
a) (x - 5)(x + 2) + (x + 1)(2 - x) = 15
<=> x2 - 3x - 10 + x - x2 + 2 = 15
<=> -2x = 23
<=> x = - 11,5
b) (2x - 3)(x + 5) - (x - 2)(2x + 1) = 3
<=> 2x2 + 7x - 15 - 2x2 + 3x + 2 = 3
<=> 10x = 16
<=> x = 1,6