Đặt x^2 + 5x = t
pt <=> t + 2 = \(2\sqrt[3]{t-2}\)
=> ( t+ 2 )^3 = \(8\left(t-2\right)\)
=> t^3 + 6t^2 + 12t + 8 - 8t + 16 = 0
=> t^3 + 6t^2 + 4t + 24 = 0
=> ( t + 6 ) ( t^2 + 4 ) = 0
=> t = -6 ( t^2 + 4 > = 0 )
(+) x^2 + 5x = -6
=> x^2 + 5x + 6 = 0
tự giải nha
gpt:
\(3\left(x^2-3x+1\right)+\sqrt{3\left(x^4+x^2+1\right)}=0\)
\(\sqrt[3]{x^3+5x^2}-1=\sqrt{\frac{5x^2-2}{6}}\)
Gpt: \(2x+2\sqrt{x^2+5x}+\sqrt{x+5}+\sqrt{x}=25\)
Gpt: \(5x^2+3x+6=\left(7x+1\right)\sqrt{x^2+3}\)
GPT :
\(x^3-4x^2+5x-1-\sqrt{2x-3}=0\)
GPT: \(x^3+5x^2+2x=3\left(x+1\right)\sqrt{3x+2}.\)
gpt \(2x^2-5x-1=\sqrt{x+2}+\sqrt{4-x}\)
giúp mk vs : gpt :
A= \(\sqrt{x^2-2x+5}+2\sqrt{4x+5}=x^3-2x^2+5x+4\)
Gpt: \(x^2-5x-3\sqrt{3x}+12=0\)
\(\sqrt{x-1}+\sqrt{2-x}=3x-1\)
gpt:
\(x^4-2x^3+x=\sqrt{(x^2-x).2}\)
\(x(5x^3+2)-2(\sqrt{2x+1}-1)=0\)
\(\sqrt{x+\frac{3}{x}}=\frac{x^2-7}{2x+2}\)
