Question

[Lê Lan Hương]

Giải phương trình:

a, \(x^2-3x+1=-\frac{\sqrt{3}}{3}\sqrt{x^4+x^2+1}\)

b, \(\left(x+3\right)\sqrt{\left(4-x\right)\left(12+x\right)}=28-x\)

c, \(\sqrt{x^3-x}=2x^2-x-2\)

d, \(2x^2+5+2\sqrt{x^2+x-2}=5\sqrt{x-1}+5\sqrt{x+2}\)

Answer 1

[Nguyễn Việt Lâm]

a/

\(\Leftrightarrow2\left(x^2-x+1\right)-\left(x^2+x+1\right)=-\frac{\sqrt{3}}{3}\sqrt{\left(x^2-x+1\right)\left(x^2+x+1\right)}\)

Đặt \(\left\{{}\begin{matrix}\sqrt{x^2-x+1}=a>0\\\sqrt{x^2+x+1}=b>0\end{matrix}\right.\)

\(\Leftrightarrow6a^2+\sqrt{3}ab-3b^2=0\)

\(\Leftrightarrow\left(3a-\sqrt{3}b\right)\left(2a+\sqrt{3}b\right)=0\)

\(\Leftrightarrow3a-\sqrt{3}b=0\Rightarrow b=\sqrt{3}a\)

\(\Leftrightarrow\sqrt{x^2+x+1}=\sqrt{3}\sqrt{x^2-x+1}\)

\(\Leftrightarrow x^2+x+1=3x^2-3x+3\)

b/ ĐKXĐ: ...

Đặt \(\left\{{}\begin{matrix}x+3=a\\\sqrt{\left(4-x\right)\left(12+x\right)}=b\end{matrix}\right.\)

\(\Rightarrow a^2+b^2=x^2+6x+9+48-8x-x^2=57-2x=2\left(28-x\right)+1\)

\(\Rightarrow28-x=\frac{a^2+b^2-1}{2}\)

Phương trình trở thành:

\(ab=\frac{a^2+b^2-1}{2}\Leftrightarrow\left(a-b\right)^2=1\Leftrightarrow\left[{}\begin{matrix}a+1=b\\a-1=b\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x+4=\sqrt{\left(4-x\right)\left(12+x\right)}\\x+2=\sqrt{\left(4-x\right)\left(12+x\right)}\end{matrix}\right.\) \(\Leftrightarrow...\)

Answer 2

[Nguyễn Việt Lâm]

c/ ĐKXĐ: ...

\(\sqrt{x\left(x^2-1\right)}=2\left(x^2-1\right)-x\)

Đặt \(\left\{{}\begin{matrix}\sqrt{x}=a\ge0\\\sqrt{x^2-1}=b\ge0\end{matrix}\right.\)

\(ab=2a^2-b^2\Leftrightarrow2a^2-ab-b^2=0\)

\(\Leftrightarrow\left(a-b\right)\left(2a+b\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a-b=0\\2a+b=0\left(l\right)\end{matrix}\right.\)

\(\Leftrightarrow a=b\Leftrightarrow\sqrt{x}=\sqrt{x^2-1}\)

\(\Leftrightarrow x^2-x-1=0\)

d/ Là \(2x^2+5\) hay \(2x+5\) bạn?