Question

Giải phương trình:

a)\(\dfrac{x^3}{\sqrt{4-x^2}}+x^2-4=0\)

b)\(x^4+\sqrt{x^2+1993}=1993\)

c)\(\sqrt{x-2}+\sqrt{10-x}=x^2-12x+40\)

Answer 1

a/ ĐKXĐ: ...

Đặt \(\sqrt{4-x^2}=a>0\)

\(\frac{x^3}{a}-a^2=0\Leftrightarrow x^3-a^3=0\)

\(\Leftrightarrow x=a\) (\(x>0\))

\(\Leftrightarrow x=\sqrt{4-x^2}\Leftrightarrow x^2=4-x^2\)

\(\Leftrightarrow x^2=2\Rightarrow x=\sqrt{2}\)

b/ Đặt \(\sqrt{x^2+1993}=a>0\Rightarrow a^2-x^2=1993\)

\(x^4+a=a^2-x^2\)

\(\Leftrightarrow x^4-a^2+x^2+a=0\)

\(\Leftrightarrow\left(x^2+a\right)\left(x^2-a+1\right)=0\)

\(\Leftrightarrow x^2+1=a\Leftrightarrow x^2+1=\sqrt{x^2+1993}\)

\(\Leftrightarrow x^4+2x^2+1=x^2+1993\)

\(\Leftrightarrow x^4+x^2-1992=0\)

Answer 2

c/

ĐKXĐ: \(2\le x\le10\)

Ta có \(VT\le\sqrt{2\left(x-2+10-x\right)}=4\)

\(VP=x^2-12x+36+4=\left(x-6\right)^2+4\ge4\)

\(\Rightarrow VT\le VP\)

Đẳng thức xảy ra khi và chỉ khi:

\(\left\{{}\begin{matrix}x-2=10-x\\x-6=0\end{matrix}\right.\) \(\Rightarrow x=6\)