Question

[lu nguyễn]

giải phương trình

a,\(x-5\sqrt{x}-14=0\)

b, \(\left(x^2+x+1\right)\left(x^2+x+2\right)=2\)

c, \(2\left(x+\dfrac{1}{2}\right)^2\left(x+\dfrac{1}{x}\right)+10=0\)

d, \(\left(x+1\right)\left(x-5\right)\left(x+3\right)\left(x-7\right)=63\)

Answer 1

[DƯƠNG PHAN KHÁNH DƯƠNG]

Câu a :

\(x-5\sqrt{x}-14=0\)

\(\Leftrightarrow\left(\sqrt{x}+2\right)\left(\sqrt{x}-7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}+2=0\\\sqrt{x}-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\in\varnothing\\x=49\end{matrix}\right.\)

Vậy \(S=\left\{49\right\}\)

Câu b :

\(\left(x^2+x+1\right)\left(x^2+x+2\right)=2\)

Đặt \(x^2+x+1=t\)

\(\Leftrightarrow t\left(t+1\right)=2\)

\(\Leftrightarrow t^2+t-2=0\)

\(\Leftrightarrow\left(t-1\right)\left(t+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}t-1=0\\t+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}t=1\\t=-2\end{matrix}\right.\)

Với \(t=1\) thì :

\(x^2+x+1=1\)

\(\Leftrightarrow x\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)

Với \(t=-2\) thì :

\(x^2+x+1=-2\)

\(\Leftrightarrow x^2+x+3=0\) ( pt vô nghiệm )

Vậy \(S=\left\{-1;0\right\}\)