1: \(=x^2+1\)

3: \(=\left(x-y-z\right)^2\)

a, \(x^4-4x^3-6x^2-4x+1=0\)(*)

<=> \(x^4+4x^2+1-4x^3-4x+2x^2-12x^2=0\)

<=> \(\left(x^2-2x+1\right)^2=12x^2\)

<=>\(\left(x-1\right)^4=12x^2\) <=> \(\left[{}\begin{matrix}\left(x-1\right)^2=\sqrt{12}x\\\left(x-1\right)^2=-\sqrt{12}x\end{matrix}\right.\)<=> \(\left[{}\begin{matrix}x^2-2x+1-\sqrt{12}x=0\left(1\right)\\x^2-2x+1+\sqrt{12}x=0\left(2\right)\end{matrix}\right.\)

Giải (1) có: \(x^2-2x+1-\sqrt{12}x=0\)

<=> \(x^2-2x\left(1+\sqrt{3}\right)+\left(1+\sqrt{3}\right)^2-\left(1+\sqrt{3}\right)^2+1=0\)

<=> \(\left(x-1-\sqrt{3}\right)^2-3-2\sqrt{3}=0\)

<=> \(\left(x-1-\sqrt{3}\right)^2=3+2\sqrt{3}\) <=> \(\left[{}\begin{matrix}x-1-\sqrt{3}=\sqrt{3+2\sqrt{3}}\\x-1-\sqrt{3}=-\sqrt{3+2\sqrt{3}}\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}x=\sqrt{3+2\sqrt{3}}+\sqrt{3}+1\left(ktm\right)\\x=-\sqrt{3+2\sqrt{3}}+\sqrt{3}+1\left(tm\right)\end{matrix}\right.\)

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

Giải (2) có: \(x^2-2x+1+\sqrt{12}x=0\)

<=> \(x^2-2x\left(1-\sqrt{3}\right)+\left(1-\sqrt{3}\right)^2-\left(1-\sqrt{3}\right)^2+1=0\)

<=> \(\left(x+\sqrt{3}-1\right)^2=3-2\sqrt{3}\) .Có VP<0 => PT (2) vô nghiệm

Vậy pt (*) có nghiệm x=\(-\sqrt{3+2\sqrt{3}}+\sqrt{3}+1\)

d: Ta có: \(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)=24\)

\(\Leftrightarrow\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24=0\)

\(\Leftrightarrow x\left(x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)

a) Cách 1: Khai triển HĐT rút gọn được 3 x 2  + 6x + 7 = 0

Vì (3( x 2  + 2x + 1) + 4 < 0 với mọi x nên giải được  x ∈ ∅

Cách 2. Chuyển vế đưa về ( x   +   3 ) 3 =  ( x   - 1 ) 3  Û x + 3 = x - 1

Từ đó tìm được x ∈ ∅

b) Đặt  x 2  = t với t ≥ 0 ta được  t 2  + t - 2 = 0

Giải ra ta được t = 1 (TM) hoặc t = -2 (KTM)

Từ đó tìm được x = ± 1

c) Biến đổi được 

d) Biến đổi về dạng x(x - 2) (x - 4) = 0. Tìm được x ∈ {0; 2; 4}

a,\(x^3-7x+6\)

\(=x^3-2x^2+2x^2-4x-3x+6\)

\(=\left(x^3-2x^2\right)+\left(2x^2-4x\right)-\left(3x-6\right)\)

\(=x^2.\left(x-2\right)+2x.\left(x-2\right)-3.\left(x-2\right)\)

\(=\left(x-2\right).\left(x^2+2x-3\right)\)

\(=\left(x-2\right).\left(x^2-x+3x-3\right)\)

\(=\left(x-2\right).\left[\left(x^2-x\right)+\left(3x-3\right)\right]\)

\(=\left(x-2\right).\left[x.\left(x-1\right)+3.\left(x-1\right)\right]\)

\(=\left(x-2\right).\left(x-1\right).\left(x+3\right)\)

b,\(x^3-9x^2+6x+16\)

\(=x^3-8x^2-x^2+8x-2x+16\)

\(=\left(x^3-8x^2\right)-\left(x^2-8x\right)-\left(2x-16\right)\)

\(=x^2.\left(x-8\right)-x.\left(x-8\right)-2.\left(x-8\right)\)

\(=\left(x-8\right).\left(x^2-x-2\right)\)

\(=\left(x-8\right).\left(x^2+x-2x-2\right)\)

\(=\left(x-8\right).\left[\left(x^2+x\right)-\left(2x+2\right)\right]\)

\(=\left(x-8\right).\left[x.\left(x+1\right)-2.\left(x+1\right)\right]\)

\(=\left(x-8\right).\left(x+1\right).\left(x-2\right)\)

c,\(x^3-6x^2-x+30\)

\(=x^3-5x^2-x^2+5x-6x+30\)

\(=\left(x^3-5x^2\right)-\left(x^2-5x\right)-\left(6x-30\right)\)

\(=x^2.\left(x-5\right)-x.\left(x-5\right)-6.\left(x-5\right)\)

\(=\left(x-5\right).\left(x^2-x-6\right)\)

\(=\left(x-5\right).\left(x^2+2x-3x-6\right)\)

\(=\left(x-5\right).\left[\left(x^2+2x\right)-\left(3x+6\right)\right]\)

\(=\left(x-5\right).\left[x.\left(x+2\right)-3.\left(x+2\right)\right]\)

\(=\left(x-5\right).\left(x+2\right).\left(x-3\right)\)

Chúc bạn học tốt!!!

d,\(2x^3-x^2+5x+3\)

\(=2x^3+x^2-2x^2-x+6x+3\)

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

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

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

e, \(27x^3-27x^2+18x-4\)

\(=27x^3-9x^2-18x^2+6x+12x-4\)

\(=\left(27x^2-9x^2\right)-\left(18x^2-6x\right)+\left(12x-4\right)\)

\(=9x^2.\left(3x-1\right)-6x.\left(3x-1\right)+4.\left(3x-1\right)\)

\(=\left(3x-1\right).\left(9x^2-6x+4\right)\)

Chúc bạn học tốt!!!

7, \(\left(x+2\right).\left(x+3\right).\left(x+4\right).\left(x+5\right)-24\)

\(=\left[\left(x+2\right).\left(x+5\right)\right].\left[\left(x+3\right).\left(x+4\right)\right]-24\)

\(=\left(x^2+5x+2x+10\right).\left(x^2+4x+3x+12\right)-24\)

\(=\left(x^2+7x+10\right).\left(x^2+7x+12\right)-24\)(1)

Đặt \(t=x^2+7x+10\Rightarrow t+2=x^2+7x+12\)

\(\Rightarrow\left(1\right)=t.\left(t+2\right)-24\)

\(=t^2+2t-24=t^2-4t+6t-24\)

\(=\left(t^2-4t\right)+\left(6t-24\right)=t.\left(t-4\right)+6.\left(t-4\right)\)

\(=\left(t-4\right).\left(t+6\right)\) (2)

Vì \(t=x^2+7x+10\) nên:

(2) \(=\left(x^2+7x+10-4\right).\left(x^2+7x+10+6\right)\)

\(=\left(x^2+7x+6\right).\left(x^2+7x+16\right)\)

\(=\left(x^2+x+6x+6\right).\left(x^2+7x+16\right)\)

\(=\left[\left(x^2+x\right)+\left(6x+6\right)\right].\left(x^2+7x+16\right)\)

\(=\left[x.\left(x+1\right)+6.\left(x+1\right)\right].\left(x^2+7x+16\right)\)

\(=\left(x+1\right).\left(x+6\right).\left(x^2+7x+16\right)\)

Chúc bạn học tốt!!!

1)  \(x^2-7x+6=x^3+1-7x-7=\left(x^3+1\right)-7\left(x+1\right)=\left(x+1\right)\left(x^2-x-6\right)\)

2)  \(x^3-9x^2+6x+16\)

\(\left(x^3+1\right)-\left[\left(9x^2-6x+1\right)-16\right]\)

\(=\left(x^3+1\right)-\left[\left(3x-1\right)^2-16\right]=\left(x^3+1\right)-\left(3x-1+4\right)\left(3x-1-4\right)\)\(=\left(x^3+1\right)-3\left(3x-5\right)\left(x+1\right)\)\(=\left(x+1\right)\left[x^2-x+1-9x+15\right]=\left(x+1\right)\left(x^2-10x+16\right)\)

\(=\left(x+1\right)\left[x\left(x-2\right)-8\left(x-2\right)\right]\)\(\left(x+1\right)\left(x-2\right)\left(x-8\right)\)

3)   \(x^3-6x^2-x+30\)

\(=x^3-5x^2-x^2+5x-6x+30\)

\(=x^2\left(x-5\right)-x\left(x-5\right)-6\left(x-5\right)\)

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

4)  \(2x^3-x^2+5x+3=\left(2x^3+x^2\right)-\left(2x^2+x\right)+\left(6x+3\right)\)

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

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

5) \(27x^3-27x^2+18x-4=\left(27x^3-1\right)-\left(27x^2-18x+3\right)\)

\(=\left(3x-1\right)\left(9x^2+3x+1\right)-3\left(9x^2-6x+1\right)\)

\(=\left(3x-1\right)\left(9x^2+3x+1\right)-3\left(3x-1\right)^2\)

\(=\left(3x-1\right)\left(9x^2+3x+1-9x+3\right)=\left(3x-1\right)\left(9x^2-6x+4\right)\)

gửi phần này trước còn lại làm sau !!! tk mk nka !!!

nhiều thế

6) \(\left(x+y\right)^2-\left(x+y\right)-12\)\(=\left(x+y\right)^2-2\cdot\frac{1}{2}\left(x+y\right)+\frac{1}{4}-\frac{49}{4}\)

\(=\left(x+y-\frac{1}{2}\right)^2-\left(\frac{7}{2}\right)^2\)\(=\left(x+y-\frac{1}{2}-\frac{7}{2}\right)\left(x+y-\frac{1}{2}+\frac{7}{2}\right)\)

\(=\left(x-4\right)\left(x+3\right)\)

7)   \(\left(x+2\right)\left(x+5\right)\left(x+3\right)\left(x+4\right)-24\)          (NHÂN x + 2 vs x +  5  và  x + 3 vs x + 4 )

\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)

ĐẶT   \(x^2+7x+11=y\)   ta được :  

\(\left(y+1\right)\left(y-1\right)-24=y^2-1-24\)

\(=y^2-25=\left(y-5\right)\left(y+5\right)\)

8)  \(4x^4-32x^2+1=4x^4+4x^2+1-36x^2\)

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

9) sai đề rùi bạn ơi ! đề đúng nè 

\(3\left(x^4+x^2+1\right)-\left(x^2+x+1\right)^2\)

Ta thấy :  

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

Thay vào biểu thức bài cho ta được : 

\(3\left(x^2-x+1\right)\left(x^2+x+1\right)-\left(x^2+x+1\right)^2\)

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

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

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

bài ở trên câu 3 : kết luận là  \(\left(x-3\right)\left(x^2-x-6\right)\)bạn sửa lại giúp mk nka !!! Th@nk !!! Tk Mk vs  

Bài 1)1)\(x^2+5x+6=x^2+3x+2x+6\)=0

=x(x+3)+2(x+3)=(x+2)(x+3)=0

Dễ rồi

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

=x(x-3)+2(x-3)=0

=(x+2)(x-3)=0

Dễ rồi

3)Phương trình tương đương:\(\left(x^2+1\right)\left(x+2\right)^2=0\)

Vì \(x^2+1>0\)

=>\(\left(x+2\right)^2=0\)

Dễ rồi

4)Phương trình tương đương\(x^2\left(x+1\right)+\left(x+1\right)\)=0

=> \(\left(x^2+1\right)\left(x+1\right)=0Vì\) \(x^2+1>0\)

=>x+1=0

=>..................

5)\(x^2-7x+6=x^2-6x-x+6\) =0

=x(x-6)-(x-6)=0

=(x-1)(x-6)=0

=>.....

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

=2x(x+1)-5(x+1)=0

=(2x-5)(x+1)=0

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

Dễ rồi

Nghỉ đã hôm sau làm mệt