Giải pt
$\sqrt[3]{x-16}-\sqrt[3]{x+3}=-1$
$\sqrt[3]{x-16}-\sqrt[3]{x+3}=-1$
Started By Sudden123, 02-01-2018 - 20:14
#1
Posted 02-01-2018 - 20:14
#2
Posted 02-01-2018 - 20:49
Đặt √x -16 = a; √x +3= b
Ta có hệ a -b =-1; b3 - a3=19
#3
Posted 02-01-2018 - 20:51
Đặt $a=\sqrt[3]{x+3}, b=\sqrt[3]{x-16}$, ta được $a-b=1$ và $a^3-b^3=19$.
Thay $a=b+1$: $3b^2+3b+19$, hay $(b-2)(b+3)=0$.
Nếu $b=2$ thì $x=24$.
Nếu $b=-3$ thì $x=-11$.
Thay $a=b+1$: $3b^2+3b+19$, hay $(b-2)(b+3)=0$.
Nếu $b=2$ thì $x=24$.
Nếu $b=-3$ thì $x=-11$.
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