Ano ang halaga ng?

Ano ang halaga ng?
Anonim

Sagot:

Pagpipilian # 4 -> "Wala sa mga ito" #

Paliwanag:

Sundin ito 3 madaling hakbang, ito ay hindi na mahirap bilang tila..

# x ^ 3 - 3b ^ (2/3) x + 9a #

Saan #x = (2a + sqrt (4a ^ 2 - b ^ 2)) ^ (1/3) + (2a - sqrt (4a ^ 2 - b ^ 2)) ^ (1/3) #

Hakbang 1 #-># Ibahin ang halaga ng x sa pangunahing equation..

#color (pula) x ^ 3 - 3b ^ (2/3) kulay (pula) (x) + 9a #

#color (red) (2a + sqrt (4a ^ 2 - b ^ 2)) ^ (1/3) + (2a - sqrt (4a ^ 2 - b ^ 2)) ^ (1/3) ^ 3 - 3b ^ (2/3) kulay (pula) (2a + sqrt (4a ^ 2 - b ^ 2)) ^ (1/3) + (2a - sqrt (4a ^ 2 - b ^ 2)) ^ (1/3) + 9a #

Hakbang 2 #-># Pag-aalis ng mga kapangyarihan..

(2a + sqrt (4a ^ 2 - b ^ 2)) + (2a - sqrt (4a ^ 2 - b ^ 2)) ^ (1/3 xx cancel3) - 3b ^ (2/3) 2a + sqrt (4a ^ 2 - b ^ 2)) + (2a - sqrt (4a ^ 2 - b ^ 2)) ^ (1/3) + 9a #

(2a + sqrt (4a ^ 2 - b ^ 2)) + (2a - sqrt (4a ^ 2 - b ^ 2)) - 3b ^ (2/3) (2a + sqrt (4a ^ 2 - b ^ 2)) + (2a - sqrt (4a ^ 2 - b ^ 2)) ^ (1/3) + 9a #

(2a + sqrt (4a ^ 2 - b ^ 2)) + (2a - sqrt (4a ^ 2 - b ^ 2) (2a - sqrt (4a ^ 2 - b ^ 2)) ^ ((1/3) xx (2/3)) 9a #

(2a + sqrt (4a ^ 2 - b ^ 2)) + (2a - sqrt (4a ^ 2 - b ^ 2) (2a - sqrt (4a ^ 2 - b ^ 2)) ^ (2/9) + 9a #

Hakbang 3 #-># Pagkolekta tulad ng mga tuntunin..

(2a + 2a) + (sqrt (4a ^ 2 - b ^ 2) - sqrt (4a ^ 2 - b ^ 2) - sqrt (4a ^ 2 - b ^ 2)) ^ (2/9) + 9a #

(4a ^ 2 - b ^ 2) - sqrt (4a ^ 2 - b ^ 2)) - 3b 4a + cancel (sqrt (4a ^ 2 - b ^ 2) - sqrt (4a ^ 2 - b ^ 2)) ^ (2/9) + 9a #

# rArr # # 4a + 0 - 3b 4a + 0 ^ (2/9) + 9a #

# rArr # # 4a - 3b 4a ^ (2/9) + 9a #

# rArr # # 4a + 9a - (3b xx 4a) ^ (2/9) #

# rArr # # 13a - 12ab ^ (2/9) -> Sagot #

Kaya nga Pagpipilian 4 ang angkop na Sagutin..