Toán Lớp 8: c)(x+2)(x-3)-x^2+5x=18 3x-9-2x(x-3)=0 2(x+1)-x=8 x(x+3)+2x+6=0

Question

Toán Lớp 8:  c)(x+2)(x-3)-x^2+5x=18 3x-9-2x(x-3)=0  2(x+1)-x=8 x(x+3)+2x+6=0

annhocbichchaubich · Answer

Giải đáp:     1) (x+2)(x-3)-x^2+5x=18        $x^{2}$ - x -6 - $x^{2}$+5x =0       => 4x -6=0       => x= $ frac{6}{4}$        2) 3x-9-2x(x-3)=0          => 3(x-3)-2x(x-3)=0         =>(x-3)(3-2x)=0         => ( left[  begin{array}{l}x-3=0  3-2x=0 end{array}  right. )          =>  ( left[  begin{array}{l}x=3  x=3/2 end{array}  right. )          3)2(x+1)-x=8         => 2x+2-x = 8         => x+ 2 =8         => x= 6         4) x(x+3)+2x+6=0         => x(x+3 ) + 2(x+3)=0         => (x+3)(x+2)=0         =>  ( left[  begin{array}{l}x+3=0  x+2=0 end{array}  right. )          =>  ( left[  begin{array}{l}x=-3  x=-2 end{array}  right. )        @Thomasedison2

doanthulan · Answer

(x+2)(x-3) - x^2 + 5x = 18   => x^2 - 3x+2x-6-x^2+5x=18   => 4x - 6 = 18   => 4x = 24   =>x=6   Vậy x=6      3x - 9 - 2x (x-3) =0   => 3 (x-3) - 2x (x-3) =0   => (x-3)(3-2x)=0   =>x-3=0 hoặc 3-2x=0   +)x-3=0=>x=3   +)3-2x=0=>2x=3=>x=3/2   Vậy x  in {3 ; 3/2}      2 (x+1) - x = 8   => 2x + 2 - x=  8   => x + 2 = 8   =>x=6   Vậy x=6      x (x+3) + 2x+6=0   => x (x+3)+ 2 (x+3) =0   =>(x+2)(x+3)=0   =>x+2=0 hoặc x+3=0   +)x+2=0=>x=-2   +)x+3=0=>x=-3   Vậy x  in {-2 ; -3}