Let \(u=x^2\). Then \(u^2-5u+4=0\).

\((u-1)(u-4)=0\), so \(u=1\) or \(u=4\).

Hence \(x^2=1\) or \(x^2=4\).

So \(x=\pm1,\pm2\).

Let \(u=x^2\). Then \(u^2-13u+36=0\).

\((u-4)(u-9)=0\), so \(u=4\) or \(u=9\).

Hence \(x^2=4\) or \(x^2=9\).

So \(x=\pm2,\pm3\).

Let \(u=x^2\). Then \(u^2+u-20=0\).

\((u+5)(u-4)=0\), so \(u=-5\) or \(u=4\).

Since \(u=x^2\ge 0\), only \(u=4\) gives real solutions.

Hence \(x=\pm2\).

Let \(u=x^2\). Then \(2u^2-5u-3=0\).

\((2u+1)(u-3)=0\), so \(u=-\dfrac12\) or \(u=3\).

Only \(u=3\) gives real solutions.

Hence \(x=\pm\sqrt3\).

Let \(u=x^2\). Then \(u^2-10u+9=0\).

\((u-1)(u-9)=0\), so \(u=1\) or \(u=9\).

Hence \(x=\pm1,\pm3\).

Let \(u=x^2\). Then \(3u^2-13u+4=0\).

\((3u-1)(u-4)=0\), so \(u=\dfrac13\) or \(u=4\).

Hence \(x^2=\dfrac13\) or \(x^2=4\).

So \(x=\pm\dfrac{\sqrt3}{3},\pm2\).

Let \(u=x^2\). Then \(u^2-2u-8=0\).

\((u-4)(u+2)=0\), so \(u=4\) or \(u=-2\).

Only \(u=4\) gives real solutions.

Hence \(x=\pm2\).

Let \(u=x^2\). Then \(u^2-8u+12=0\).

\((u-2)(u-6)=0\), so \(u=2\) or \(u=6\).

Hence \(x=\pm\sqrt2,\pm\sqrt6\).

Let \(u=x^2\). Then \(2u^2+u-3=0\).

\((2u+3)(u-1)=0\), so \(u=-\dfrac32\) or \(u=1\).

Only \(u=1\) gives real solutions.

Hence \(x=\pm1\).

Let \(u=x^2\). Then \(u^2-17u+16=0\).

\((u-1)(u-16)=0\), so \(u=1\) or \(u=16\).

Hence \(x=\pm1,\pm4\).

Let \(u=x^2\). Then \(4u^2-20u+9=0\).

\((2u-1)(2u-9)=0\), so \(u=\dfrac12\) or \(u=\dfrac92\).

Hence \(x=\pm\dfrac{\sqrt2}{2},\pm\dfrac{3\sqrt2}{2}\).

Let \(u=x^2\). Then \(u^2+4u+3=0\).

\((u+1)(u+3)=0\), so \(u=-1\) or \(u=-3\).

But \(u=x^2\ge 0\), so there are no real solutions.

Let \(u=x^2\). Then \(5u^2-29u+20=0\).

\((5u-4)(u-5)=0\), so \(u=\dfrac45\) or \(u=5\).

Hence \(x=\pm\dfrac{2\sqrt5}{5},\pm\sqrt5\).

Let \(u=x^3\). Then \(u^2-5u+6=0\).

\((u-2)(u-3)=0\), so \(u=2\) or \(u=3\).

Hence \(x^3=2\) or \(x^3=3\).

So \(x=\sqrt[3]{2}\) or \(x=\sqrt[3]{3}\).

Let \(u=x^3\). Then \(u^2+u-6=0\).

\((u+3)(u-2)=0\), so \(u=-3\) or \(u=2\).

Hence \(x=-\sqrt[3]{3}\) or \(x=\sqrt[3]{2}\).

Let \(u=x^3\). Then \(2u^2-7u+3=0\).

\((2u-1)(u-3)=0\), so \(u=\dfrac12\) or \(u=3\).

Hence \(x=\sqrt[3]{\dfrac12}\) or \(x=\sqrt[3]{3}\).

Also \(\sqrt[3]{\dfrac12}=\dfrac{1}{\sqrt[3]{2}}\).

Let \(u=x^3\). Then \(u^2-9u+20=0\).

\((u-4)(u-5)=0\), so \(u=4\) or \(u=5\).

Hence \(x=\sqrt[3]{4}\) or \(x=\sqrt[3]{5}\).

Let \(u=x^3\). Then \(3u^2+2u-8=0\).

\((3u-4)(u+2)=0\), so \(u=\dfrac43\) or \(u=-2\).

Hence \(x=\sqrt[3]{\dfrac43}\) or \(x=-\sqrt[3]{2}\).

Let \(u=x^3\). Then \(u^2-6u+9=0\).

\((u-3)^2=0\), so \(u=3\).

Hence \(x^3=3\), so \(x=\sqrt[3]{3}\).

Let \(u=x^3\). Then \(4u^2-4u-3=0\).

\((2u-3)(2u+1)=0\), so \(u=\dfrac32\) or \(u=-\dfrac12\).

Hence \(x=\sqrt[3]{\dfrac32}\) or \(x=-\dfrac{1}{\sqrt[3]{2}}\).

Let \(u=x^3\). Then \(u^2+5u+6=0\).

\((u+2)(u+3)=0\), so \(u=-2\) or \(u=-3\).

Hence \(x=-\sqrt[3]{2}\) or \(x=-\sqrt[3]{3}\).

Let \(u=x^3\). Then \(u^2-2u+5=0\).

The discriminant is \((-2)^2-4(1)(5)=4-20=-16\), which is negative.

So there are no real values of \(u\), and hence no real values of \(x\).

Let \(u=\sqrt{x}\), so \(u\ge 0\) and \(x=u^2\).

Then \(u^2-5u+6=0\).

\((u-2)(u-3)=0\), so \(u=2\) or \(u=3\).

Hence \(x=4\) or \(x=9\).

Let \(u=\sqrt{x}\), with \(u\ge0\). Then \(u^2+2u-15=0\).

\((u+5)(u-3)=0\), so \(u=-5\) or \(u=3\).

Since \(u\ge0\), only \(u=3\) is valid.

Hence \(x=9\).

Let \(u=\sqrt{x}\), so \(x=u^2\).

Then \(2u^2-7u+3=0\).

\((2u-1)(u-3)=0\), so \(u=\dfrac12\) or \(u=3\).

Hence \(x=\dfrac14\) or \(x=9\).

Let \(u=\sqrt{x}\). Then \(u^2-8u+12=0\).

\((u-2)(u-6)=0\), so \(u=2\) or \(u=6\).

Hence \(x=4\) or \(x=36\).

Let \(u=\sqrt{x}\), so \(x=u^2\).

Then \(3u^2+2u-8=0\).

\((3u-4)(u+2)=0\), so \(u=\dfrac43\) or \(u=-2\).

Only \(u=\dfrac43\) is valid.

Hence \(x=\dfrac{16}{9}\).

Let \(u=\sqrt{x}\). Then \(4u^2-12u+5=0\).

\((2u-1)(2u-5)=0\), so \(u=\dfrac12\) or \(u=\dfrac52\).

Hence \(x=\dfrac14\) or \(x=\dfrac{25}{4}\).

Let \(u=\sqrt{x}\), so \(x=u^2\).

Then \(u^2+u-2=0\).

\((u+2)(u-1)=0\), so \(u=-2\) or \(u=1\).

Only \(u=1\) is valid.

Hence \(x=1\).

Let \(u=\sqrt{x}\), so \(x=u^2\).

Then \(2u^2-u-3=0\).

\((2u-3)(u+1)=0\), so \(u=\dfrac32\) or \(u=-1\).

Only \(u=\dfrac32\) is valid.

Hence \(x=\dfrac{9}{4}\).

Let \(u=\sqrt{x}\), so \(u\ge0\).

Then \(u^2-2u+5=0\).

The discriminant is \((-2)^2-4(1)(5)=4-20=-16\), which is negative.

So there are no real values of \(u\), and therefore no real solutions for \(x\).

Let \(u=\sqrt{x}\). Then \(u^2-6u+9=0\).

\((u-3)^2=0\), so \(u=3\).

Hence \(x=9\).

Let \(u=x^{1/4}\), so \(u\ge0\) and \(\sqrt{x}=u^2\).

Then \(u^2-5u+6=0\).

\((u-2)(u-3)=0\), so \(u=2\) or \(u=3\).

Hence \(x=2^4=16\) or \(x=3^4=81\).

Let \(u=x^{1/4}\), so \(\sqrt{x}=u^2\).

Then \(u^2+u-2=0\).

\((u+2)(u-1)=0\), so \(u=-2\) or \(u=1\).

Only \(u=1\) is valid.

Hence \(x=1^4=1\).

Let \(u=x^{1/4}\), so \(\sqrt{x}=u^2\).

Then \(2u^2-3u-2=0\).

\((2u+1)(u-2)=0\), so \(u=-\dfrac12\) or \(u=2\).

Only \(u=2\) is valid.

Hence \(x=2^4=16\).

Let \(u=x^{1/4}\), so \(\sqrt{x}=u^2\).

Then \(3u^2-10u+3=0\).

\((3u-1)(u-3)=0\), so \(u=\dfrac13\) or \(u=3\).

Hence \(x=\left(\dfrac13\right)^4=\dfrac{1}{81}\) or \(x=3^4=81\).

Let \(u=x^{1/4}\), so \(\sqrt{x}=u^2\).

Then \(u^2-4u+3=0\).

\((u-1)(u-3)=0\), so \(u=1\) or \(u=3\).

Hence \(x=1^4=1\) or \(x=3^4=81\).

Let \(u=x^{1/4}\), so \(u\ge0\) and \(\sqrt{x}=u^2\).

Then \(5u^2+6u+1=0\).

\((5u+1)(u+1)=0\), so \(u=-\dfrac15\) or \(u=-1\).

Neither value is allowed since \(u\ge0\).

So there are no real solutions.

Let \(u=\dfrac1x\). Then \(2u^2-5u+2=0\).

\((2u-1)(u-2)=0\), so \(u=\dfrac12\) or \(u=2\).

Thus \(\dfrac1x=\dfrac12\) or \(\dfrac1x=2\).

So \(x=2\) or \(x=\dfrac12\).

Let \(u=\dfrac1x\). Then \(3u^2+2u-8=0\).

\((3u-4)(u+2)=0\), so \(u=\dfrac43\) or \(u=-2\).

Thus \(x=\dfrac34\) or \(x=-\dfrac12\).

Let \(u=\dfrac1x\). Then \(u^2-5u+6=0\).

\((u-2)(u-3)=0\), so \(u=2\) or \(u=3\).

Hence \(x=\dfrac12\) or \(x=\dfrac13\).

Let \(u=\dfrac1x\). Then \(4u^2-12u+5=0\).

\((2u-1)(2u-5)=0\), so \(u=\dfrac12\) or \(u=\dfrac52\).

Hence \(x=2\) or \(x=\dfrac25\).

Let \(u=\dfrac1x\). Then \(5u^2+6u+1=0\).

\((5u+1)(u+1)=0\), so \(u=-\dfrac15\) or \(u=-1\).

Hence \(x=-5\) or \(x=-1\).

Let \(u=\dfrac1x\). Then \(2u^2+u-3=0\).

\((2u+3)(u-1)=0\), so \(u=-\dfrac32\) or \(u=1\).

Hence \(x=-\dfrac23\) or \(x=1\).

Multiply both sides by \(x\):

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

\(x^2-3x+1=0\).

Using the quadratic formula,

\(x=\dfrac{3\pm\sqrt{9-4}}{2}=\dfrac{3\pm\sqrt5}{2}\).

Multiply through by \(x\):

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

\(x^2-4x+2=0\).

Using the quadratic formula,

\(x=\dfrac{4\pm\sqrt{16-8}}{2}=2\pm\sqrt2\).

Multiply by \(x\):

\(2x^2+3=5x\)

\(2x^2-5x+3=0\).

\((2x-3)(x-1)=0\).

So \(x=\dfrac32\) or \(x=1\).

Multiply by \(x\):

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

\(x^2-4x-4=0\).

\(x=\dfrac{4\pm\sqrt{16+16}}{2}=2\pm2\sqrt2\).

Multiply by \(x\):

\(3x^2+2=4x\)

\(3x^2-4x+2=0\).

The discriminant is \((-4)^2-4(3)(2)=16-24=-8\), which is negative.

So there are no real solutions.

Multiply by \(x\):

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

\(x^2+2x+1=0\)

\((x+1)^2=0\), so \(x=-1\).