Question

Giải các phương trình sau:

a, \(\sqrt{3+x}+\sqrt{6-x}\) =3

b, \(\sqrt{x+4\sqrt{x-4}}+x+2+\sqrt{x-4}=8\)

c, \(\sqrt{x^2+3x+2}-2\sqrt{x^2+6x+8}+\sqrt{x^2+5x}+6\)

d, \(x^4-2x^2+2x+1=0\)

e, \(\left(x+3\right)^4+\left(x-5\right)^4=1312\)

Answer 1

a, ĐKXĐ: \(-3\le x\le6\)

\(pt\Leftrightarrow3+x+6-x+2\sqrt{\left(3+x\right)\left(6-x\right)}=9\)

\(\Leftrightarrow\sqrt{\left(3+x\right)\left(6-x\right)}=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\left(tm\right)\\x=6\left(tm\right)\end{matrix}\right.\)

b, ĐKXĐ: \(x\ge4\)

\(pt\Leftrightarrow\sqrt{x-4+4\sqrt{x-4}+4}+x+2+\sqrt{x-4}=8\)

\(\Leftrightarrow\sqrt{\left(\sqrt{x-4}+2\right)^2}+x+2+\sqrt{x-4}=8\)

\(\Leftrightarrow\sqrt{x-4}+2+x+2+\sqrt{x-4}=8\)

\(\Leftrightarrow2\sqrt{x-4}=4-x\)

\(\Leftrightarrow\left\{{}\begin{matrix}4-x\ge0\\4\left(x-4\right)=\left(4-x\right)^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\le4\\x^2-12x+32=0\end{matrix}\right.\Leftrightarrow x=4\left(tm\right)\)

Answer 2

e, Đặt \(y=x-1\) ta có

\(pt\Leftrightarrow\left(y+4\right)^4+\left(y-4\right)^4=1312\)

\(\Leftrightarrow2y^4+192y^2-800=0\)

\(\Leftrightarrow\left[{}\begin{matrix}y^2=4\\y^2=-100\left(l\right)\end{matrix}\right.\Leftrightarrow y=\pm2\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\)