`1)x^4 -10x^3 +26x^2 -10x+1=0`
`x=0=>VT=1=>x=0(l)`
Chia 2 vế cho `x^2>0` ta có
`x^2-10x+26-10/x+1/x^2=0`
`=>x^2+1/x^2+26-10(x+1/x)=0`
`=>(x+1/x)^2-10(x+1/x)+24=0`
Đặt `a=x+1/x`
`pt<=>a^2-10a+24=0`
`<=>` $\left[ \begin{array}{l}a=4\\a=6\end{array} \right.$
`a=4<=>x+1/x=4<=>x^2-4x+1=0<=>` $\left[ \begin{array}{l}x=\sqrt3+2\\x=-\sqrt3+2\end{array} \right.$
`a=6<=>x+1/x=6<=>x^2-6x+1=0<=>` $\left[ \begin{array}{l}x=\sqrt8+3\\x=-\sqrt8+3\end{array} \right.$
Vậy `S={\sqrt3+2,-\sqrt3+2,\sqrt8+3,-\sqrt8+3}`

2)Do hệ số chẵn bằng=hệ số lẻ
`=>x=-1`
`pt<=>x^4+x^3+4x^3+4x^2+6x^2+6x+9x+9=0`
`<=>(x+1)(x^3+4x^2+6x+9)=0`
`<=>(x+1)(x^3+3x^2+x^2+6x+9)=0`
`<=>(x+1)[x^2(x+3)+(x+3)^2]=0`
`<=>(x+1)(x+3)(x^2+x+3)=0`
Do `x^2+x+3=(x+1/2)^2+11/4>0`
`=>` $\left[ \begin{array}{l}x=-3\\x=-1\end{array} \right.$
Vậy `S={-1,-3}`

\(a.\frac{4x-3}{x-5}=\frac{29}{3}\\ \Leftrightarrow\frac{3\left(4x-3\right)}{3\left(x-5\right)}=\frac{29\left(x-5\right)}{3\left(x-5\right)}\\ \Leftrightarrow3\left(4x-3\right)=29\left(x-5\right)\\ \Leftrightarrow3\left(4x-3\right)-29\left(x-5\right)=0\\ \Leftrightarrow12x-9-29x+145=0\\ \Leftrightarrow-17x+136=0\\ \Leftrightarrow-17x=-136\\ \Leftrightarrow x=\frac{-136}{-17}=8\)

\(b.\frac{2x-1}{5-3x}=2\\ \Leftrightarrow\frac{2x-1}{5-3x}=\frac{4}{2}\\ \Leftrightarrow\frac{2\left(2x-1\right)}{2\left(5-3x\right)}=\frac{4\left(5-3x\right)}{2\left(5-3x\right)}\\ \Leftrightarrow2\left(2x-1\right)=4\left(5-3x\right)\\ \Leftrightarrow2\left(2x-1\right)-4\left(5-3x\right)=0\\ \Leftrightarrow4x-2-20+12x=0\\ \Leftrightarrow16x-22=0\\ \Leftrightarrow16x=22\\ \Leftrightarrow x=\frac{22}{16}=\frac{11}{8}\)

\(c.\frac{4x-5}{x-1}=\frac{2+x}{x-1}\\ \Leftrightarrow4x-5=2+x\\ \Leftrightarrow4x-5-2-x=0\\ \Leftrightarrow3x-7=0\\ \Leftrightarrow3x=7\\ \Leftrightarrow x=\frac{7}{3}\)

\(d.\frac{7}{x+2}=\frac{3}{x-5}\\ \Leftrightarrow\frac{7\left(x-5\right)}{\left(x+2\right)\left(x-5\right)}=\frac{3\left(x+2\right)}{\left(x+2\right)\left(x-5\right)}\\ \Leftrightarrow7\left(x-5\right)=3\left(x+2\right)\\ \Leftrightarrow7\left(x-5\right)-3\left(x+2\right)=0\\ \Leftrightarrow7x-35-3x-6=0\\ \Leftrightarrow4x-41=0\\ \Leftrightarrow4x=41\\ \Leftrightarrow x=\frac{41}{4}\)

\(e.\frac{2x+5}{2x}-\frac{x}{x+5}=0\\ \Leftrightarrow\frac{\left(2x+5\right)\left(x+5\right)}{2x\left(x+5\right)}-\frac{x.2x}{2x\left(x+5\right)}=0\\ \Leftrightarrow\left(2x+5\right)\left(x+5\right)-2x^2=0\\ \Leftrightarrow2x^2+10x+5x+25-2x^2=0\\ \Leftrightarrow15x+25=0\\ \Leftrightarrow15x=-25\\ \Leftrightarrow x=\frac{-25}{15}=\frac{-5}{3}\)

\(f.\frac{12x+1}{11x-4}+\frac{10x-4}{9}=\frac{20x+17}{18}\\\Leftrightarrow\frac{18\left(12x+1\right)}{18\left(11x-4\right)}+\frac{\left(10x-4\right).2\left(11x-4\right)}{9.2\left(11x-4\right)}=\frac{\left(20x+17\right)\left(11x-4\right)}{18\left(11x-4\right)}\\ \Leftrightarrow18\left(12x+1\right)+\left(10x-4\right).2\left(11x-4\right)=\left(20x+17\right)\left(11x-4\right)\\ \Leftrightarrow220x^2+48x+50=220x^2+107x-68\\ \Leftrightarrow48x+50=107x-68\\ \Leftrightarrow48x-107x=-68-50\\ \Leftrightarrow59x=-118\\ \Leftrightarrow x=-2\)

\(x^4-10x^3-15x^2+20x+4=0\)

\(\Leftrightarrow x^4-x^3-9x^3+9x^2-24x^2+24x-4x+4=0\)

\(\Leftrightarrow x^3\left(x-1\right)-9x^2\left(x-1\right)-24x\left(x-1\right)-4\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^3-9x^2-24x-4\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^3+2x^2-11x^2-22x-2x-4\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left[x^2\left(x+2\right)-11x\left(x+2\right)-2\left(x+2\right)\right]=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(x^2-11x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+2=0\\x^2-11x-2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)(x² - 11x - 2 không có nghiệm hữu tỉ)

Vậy x = 1 hoặc x = -2.

ta có : \(x^4-10x^3-15x^2+20x+4=0\) (*)

\(\Leftrightarrow x^4-x^3-9x^3+9x^2-24x^2+24x-4x+4=0\)

\(\Leftrightarrow x^3\left(x-1\right)-9x^2\left(x-1\right)-24x\left(x-1\right)-4\left(x-1\right)=0\)

\(\Leftrightarrow\left(x^3-9x^2-24x-4\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left(x^3-11x^2-2x+2x^2-22x-4\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[x\left(x^2-11x-2\right)+2\left(x^2-11x-2\right)\right]\left(x-1\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(x^2-11x-2\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\x-1=0\\x^2-11x-2=0\left(xétsau\right)\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=1\end{matrix}\right.\)

ta có : \(x^2-11x-2=0\) (1)

\(\Delta=11^2-4.1.\left(-2\right)=121+8=129>0\)

\(\Rightarrow\) phương trình (1) có 2 nghiệm phân biệt

\(x_1=\dfrac{11+\sqrt{129}}{2}\) ; \(x_2=\dfrac{11-\sqrt{129}}{2}\)

vậy phương trình (*) có 4 nghiệm phân biệt \(x=1;x=-2;x=\dfrac{11+\sqrt{129}}{2};x=\dfrac{11-\sqrt{129}}{2}\)

\(\Leftrightarrow x^3\left(x-2\right)+10x\left(x-2\right)=0\)

\(\Leftrightarrow x\left(x-2\right)\left(x^2+10\right)=0\)

\(\Leftrightarrow x\left(x-2\right)=0\) (do \(x^2+10>0;\forall x\))

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

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)

`x^4-2x^3+10x^2-20x=0`

`<=>x^3(x-2)+10x(x-2)=0`

`<=>(x^3+10x)(x-2)=0`

`<=>x(x^2+10)(x-2)=0`

`<=>`$\left[\begin{matrix} x=0\\ x^2+10=0\\x-2=0\end{matrix}\right.$

`<=>`$\left[\begin{matrix} x=0\\ x^2=-10 \ \rm(loại) \\x=2\end{matrix}\right.$

Vậy `S={0;2}`

Ta có: \(x^4-2x^3+10x^2-20x=0\)

\(\Leftrightarrow x^3\left(x-2\right)+10x\left(x-2\right)=0\)

\(\Leftrightarrow x\left(x-2\right)\left(x^2+10\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)

         \(x^2-5x-4\left(x-5\right)=0\)

\(\Leftrightarrow\)\(x\left(x-5\right)-4\left(x-5\right)=0\)

\(\Leftrightarrow\)\(\left(x-5\right)\left(x-4\right)=0\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x-5=0\\x-4=0\end{cases}}\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=5\\x=4\end{cases}}\)

Vậy....

\(2x\left(x+6\right)=7x+42\)

\(\Leftrightarrow\)\(2x\left(x+6\right)-7x-42=0\)

\(\Leftrightarrow\)\(2x\left(x+6\right)-7\left(x+6\right)=0\)

\(\Leftrightarrow\)\(\left(x+6\right)\left(2x-7\right)=0\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x+6=0\\2x-7=0\end{cases}}\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=-6\\x=\frac{7}{2}\end{cases}}\)

Vậy......

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

\(\Leftrightarrow\)\(x^2\left(x-5\right)+\left(x-5\right)=0\)

\(\Leftrightarrow\)\(\left(x-5\right)\left(x^2+1\right)=0\)

\(\Leftrightarrow\)\(x-5=0\)

\(\Leftrightarrow\)\(x=5\)

\(x^4-2x^3+10x^2-20x=0\)

\(\Leftrightarrow\)\(x^3\left(x-2\right)+10x\left(x-2\right)=0\)

\(\Leftrightarrow\)\(x\left(x-2\right)\left(x^2+10\right)=0\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=0\\x-2=0\end{cases}}\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=0\\x=2\end{cases}}\)

Vậy...

Lời giải:

Ta có:

\((x+3)(x+12)(x-4)(x-16)+20x^2=0\)

\(\Leftrightarrow [(x+3)(x-16)][(x+12)(x-4)]+20x^2=0\)

\(\Leftrightarrow (x^2-13x-48)(x^2+8x-48)+20x^2=0\)

Đặt \(x^2-12x-48=a\). PT trở thành:

\((a-x)(a+20x)+20x^2=0\)

\(\Leftrightarrow a^2+19ax-20x^2+20x^2=0\Leftrightarrow a^2+19ax=0\)

\(\Leftrightarrow a(a+19x)=0\)

\(\Leftrightarrow (x^2-12x-48)(x^2+7x-48)=0\)

\(\Leftrightarrow \left[\begin{matrix} x^2-12x-48=0\\ x^2+7x-48=0\end{matrix}\right.\)

\(\Leftrightarrow \left[\begin{matrix} x=6\pm 2\sqrt{21}\\ x=\frac{-7\pm \sqrt{241}}{2}\end{matrix}\right.\)

Vậy......

1) \(\Leftrightarrow\sqrt{\left(x+5\right)^2}=4\)

\(\Leftrightarrow\left|x+5\right|=4\)

\(\Leftrightarrow\left[{}\begin{matrix}x+5=4\\x+5=-4\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-9\end{matrix}\right.\)

2) \(ĐK:x\ge2\)

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

\(\Leftrightarrow x-2=4\Leftrightarrow x=6\left(tm\right)\)

3) \(\Leftrightarrow\left(x^2-x+4\right)-\sqrt{x^2-x+4}+\dfrac{1}{4}=\dfrac{9}{4}\)

\(\Leftrightarrow\left(\sqrt{x^2-x+4}-\dfrac{1}{2}\right)^2=\dfrac{9}{4}\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2-x+4}-\dfrac{1}{2}=\dfrac{3}{2}\\\sqrt{x^2-x+4}-\dfrac{1}{2}=-\dfrac{3}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2-x+4}=2\\\sqrt{x^2-x+4}=-1\left(VLý\right)\end{matrix}\right.\)

\(\Leftrightarrow x^2-x+4=4\Leftrightarrow x\left(x-1\right)=0\)

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

4) \(ĐK:x\ge0\)

\(\Leftrightarrow3\sqrt{x}-3=\sqrt{x}+2\)

\(\Leftrightarrow\sqrt{x}=\dfrac{5}{2}\Leftrightarrow x=\dfrac{25}{4}\left(tm\right)\)