b: =x^3+6x^2+9x^2+54x+20x+120

=(x+6)(x^2+9x+20)

=(x+6)(x+4)(x+5)

a: Đa thức này không phân tích được nha bạn

\(\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x-6\right)=15x^2\)

\(\Leftrightarrow\left(x^2-7x+6\right)\left(x^2-5x+6\right)-15x^2=0\) (*)

-Đặt \(t=x^2-5x+6\)

(*) \(\Leftrightarrow t\left(t-2x\right)-15x^2=0\)

\(\Leftrightarrow t^2-2xt-15x^2=0\)

\(\Leftrightarrow t^2-5xt+3xt-15x^2=0\)

\(\Leftrightarrow t\left(t-5x\right)+3x\left(t-5x\right)=0\)

\(\Leftrightarrow\left(t-5x\right)\left(t+3x\right)=0\)

\(\Leftrightarrow t-5x=0\) hay \(t+3x=0\)

\(\Leftrightarrow x^2-5x+6-5x=0\) hay \(x^2-5x+6+3x=0\)

\(\Leftrightarrow x^2-10x+6=0\) hay \(x^2-2x+6=0\)

\(\Leftrightarrow x^2-2.5x+25-19=0\) hay \(\left(x-1\right)^2+5=0\) (pt vô nghiệm)

\(\Leftrightarrow\left(x-5\right)^2-19=0\)

\(\Leftrightarrow\left(x-5-\sqrt{19}\right)\left(x-5+\sqrt{19}\right)=0\)

\(\Leftrightarrow x=5+\sqrt{19}\) hay \(x=5-\sqrt{19}\)

-Vậy \(S=\left\{5+\sqrt{19};5-\sqrt{19}\right\}\)

`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}`

\(\Rightarrow2\left(x-3\right)\left(x^2+1\right)-5x^2+15x=0\)

\(\Rightarrow2\left(x-3\right)\left(x^2+1\right)-5x\left(x-3\right)=0\)

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

\(\Rightarrow\orbr{\begin{cases}x-3=0\\2x^2-5x+2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=3\\x=...\end{cases}}}\)

Dùng máy tính bấm nốt nghiệm phương trình 2 nhé

\(\sqrt[3]{x+1}=x^3-15x^2+75x-125-6=0\)

\(\Leftrightarrow\sqrt[3]{x+1}+6=\left(x-5\right)^3\)

Đặt \(\sqrt[3]{x+1}=a-5\) ta được hệ:

\(\left\{{}\begin{matrix}\left(a-5\right)^3=x+1\\a-5+6=\left(x-5\right)^3\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\left(a-5\right)^3=x+1\\\left(x-5\right)^3=a+1\end{matrix}\right.\)

Trừ vế cho vế ta được:

\(\left(x-5\right)^3-\left(a-5\right)^3=a-x\)

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

\(\Leftrightarrow\left(x-a\right)\left[\left(x-5+\frac{a-5}{2}\right)^2+\frac{3\left(a-5\right)^2}{4}+1\right]=0\)

\(\Leftrightarrow x-a=0\) (phần ngoạc phía sau luôn dương)

\(\Leftrightarrow x=a\Leftrightarrow x=\sqrt[3]{x+1}+5\Leftrightarrow x-5=\sqrt[3]{x+1}\)

\(\Leftrightarrow x^3-15x^2+75x-125=x+1\)

\(\Leftrightarrow x^3-15x^2+74x-126=0\)

\(\Rightarrow x=7\)

@Nguyễn Việt Lâm

ĐK: \(x^3+3x^2-3x+1\ge0\)

\(pt\Leftrightarrow\sqrt[3]{9x^2-15x+9}-\left(2-x\right)+\sqrt{x^3+3x^2-3x+1}=0\)

\(\Leftrightarrow\frac{9x^2-15x+9-\left(2-x\right)^3}{A^2+AB+B^2}+\sqrt{x^3+3x^2-3x+1}=0\)

\(\left(A=\sqrt[3]{9x^2-15x+9};\text{ }B=2-x\right)\)\(\text{(}A^2+AB+B^2=\left(A+\frac{B}{2}\right)^2+\frac{3B^2}{4}>0\text{)}\)

\(\Leftrightarrow\frac{x^3+3x^2-3x+1}{A^2+AB+B^2}+\sqrt{x^3+3x^2-3x+1}=0\)

\(\Leftrightarrow\sqrt{x^3+3x^2-3x+1}\left(\frac{\sqrt{x^3+3x^2-3x+1}}{A^2+AB+B^2}+1\right)=0\)

\(\Leftrightarrow x^3+3x^2-3x+1=0\text{ (do }\frac{\sqrt{x^3+3x^2-3x+1}}{A^2+AB+B^2}+1>0\text{)}\)

\(\Leftrightarrow\left(x+1+\sqrt[3]{2}+\sqrt[3]{4}\right)\left[x^2+\left(2-\sqrt[3]{2}-\sqrt[3]{4}\right)x+\sqrt[3]{2}-1\right]=0\)

\(\Leftrightarrow x+1+\sqrt[3]{2}+\sqrt[3]{4}=0\text{ (}pt\text{ }x^2+\left(2-\sqrt[3]{2}-\sqrt[3]{4}\right)x+\sqrt[3]{2}-1=0\text{ vô nghiệm do }\Delta

\(2\left(x+1\right)=5x+7\\ \Leftrightarrow2x+2=5x+7\\\Leftrightarrow 2x-5x=-2+7\\\Leftrightarrow -3x=5\\ \Leftrightarrow x=-\frac{5}{3}\)

Vậy phương trình trên có nghiệm là \(-\frac{5}{3}\)

\(3x-1=x+3\\ \Leftrightarrow3x-x=1+3\\ \Leftrightarrow2x=4\\\Leftrightarrow x=2\)

Vậy phương trình trên có nghiệm là \(2\)

\(15-7x=9-3x\\\Leftrightarrow -7x+3x=-15+9\\\Leftrightarrow -4x=-6\\ \Leftrightarrow x=\frac{3}{2}\)

Vậy phương trình trên có nghiệm là \(\frac{3}{2}\)

\(2x+1=15x-5\\ \Leftrightarrow2x-15x=-1-5\\ \Leftrightarrow-13x=-6\\ \Leftrightarrow x=\frac{6}{13}\)

Vậy phương trình trên có nghiệm là \(\frac{6}{13}\)

\(3x-2=2x+5\\ \Leftrightarrow3x-2x=2+5\\ \Leftrightarrow x=7\)

Vậy phương trình trên có nghiệm là \(7\)

a) \(\frac{15x-10}{x^2+3}=0\)

<=> 15x - 10 = 0

<=> 5(3x - 2) = 0

<=> 3x - 2 = 0

<=> 3x = 2

<=> x = 2/3

b) ĐKXĐ: \(x\ne1;x\ne-3\)

<=>\(\frac{3x-1}{x-1}-\frac{2x+5}{x+3}-\frac{8}{x^2+2x-3}=0\)

<=> \(\frac{3x-1}{x-1}-\frac{2x+5}{x+3}-\frac{8}{\left(x-1\right)\left(x+3\right)}=0\)

<=> (3x - 1)(x + 3) - (2x + 5)(x - 1) - 8 = (x - 1)(x + 3)

<=> 3x² + 9x - x - 3 - 2x² + 2x - 5x + 5 - 8 = 0

<=> x² + 5x - 6 = 0

<=> (x - 1)(x + 6) = 0

<=> x - 1 = 0 hoặc x + 6 = 0

<=> x = 1 (ktm) hoặc x = -6 (tm)

=> x = -6