Помогите решить (0,2у-1) (0,2у+1)(8+x)(8-x)[tex]6x + 9 + x {}^{2} [/tex]25х+20х+9. Created by naskabruskovskaa2437. algebra-ru

tex]25х+20х+9...

naskabruskovskaa2437 @naskabruskovskaa2437

July 2022 2 5 Report

Помогите решить
(0,2у-1) (0,2у+1)
(8+x)(8-x)
[tex]6x + 9 + x {}^{2} [/tex]
25х+20х+9

Verified answer

$1) \ (0,2y-1)(0,2y+1)=(0,2y)^{2}-1^{2}=\boxed{0,04y^{2}-1} \\\\\\2) \ (8+x)(8-x)=8^{2} -x^{2} =\boxed{64-x^{2}}\\\\\\3) \ 6x+9+x^{2}=x^{2} +6x+9=x^{2}+2\cdot3\cdot x+3^{2} =\boxed{(x+3)^{2} } \\\\\\4) ???$

1 votes Thanks 1

stepanfedorov782

1.(0,2у-1)(0,2у+1)=0,04у в квадрате+0,2у-0,2у-1=0,04у в квадрате-у

2.(8+х)(8-х)=64-8х+8х-х=64-х

3.6х+9+х в квадрате

х в квадрате+6х+9=0

Х 1,2=-в+-корень в2-4ас/2а=-6+-корень 6 в квадрате-4•1•9/2•1=-6+-корень 36-36/2=-6/2=-3.

4.25х+20х+9

25х+20х=-9

45х=-9

х=45/-9

х=-5

1 votes Thanks 1

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