1)Решить уравнения a) (2x-6)(x+7)=0 b) x^2-9=0 в) х+3/8x-24=0. Created by Joel44. algebra-ru

(2x-6)(x+7)=0,2х-6=0, х+7=0,2х=6, х=-7.х=6:2, х=3.Ответ: -7; 3.х²-9=0,х²=9,х²=±3,х=3,х=-3.Ответ: -3; 3.х+ х-24=0,8х+8* х-8*24=8*0,8х+3х-192=0,11х-192=0,11х=192,х=192:11,х= = Ответ:

8x-24=0...

wnull (2x-6)(x+7)=0,
2х-6=0, х+7=0,
2х=6, х=-7.
х=6:2,
х=3.
Ответ: -7; 3.

х²-9=0,
х²=9,
х²=±3,
х=3,
х=-3.
Ответ: -3; 3.

х+ $\frac{3}{8}$ х-24=0,
8х+8* $\frac{3}{8}$ х-8*24=8*0,
8х+3х-192=0,
11х-192=0,
11х=192,
х=192:11,
х= $\frac{192}{11}$ = $17 \frac{5}{11}$
Ответ: $17 \frac{5}{11}$

2 votes Thanks 1

Joel44 спс

irreversibility

Verified answer

$(2x-6)(x+7)=0 \\ 
2x-6=0 \\ 
x+7=0 \\ 
x_1= 3 \\ 
x_2=-7 \\ \\
x^2-9=0 \\
x^2=9 \\ 
x_1=-3 \\
x_2=3 \\ \\ 
\frac{x+3}{8x-24}=0 \\ 
\frac{x+3}{8(x-3)}=0; x \neq 3 \\
x+3=0 \\
x=-3
$

2 votes Thanks 2

Joel44 спс

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