a) Nếu bạn chưa học Bézout - Horner thì giải theo chương trình sgk như sau:

  \(x^3-3x^2-x-45\)     

\(=x^3-5x^2+2x^2-10x+9x-45\)

\(=x^2\left(x-5\right)+2x\left(x-5\right)+9\left(x-5\right)\)

\(=\left(x-5\right)\left(x^2+2x +9\right)\)

Nếu học rồi thì dễ thôi: 

\(x^3-3x^2-x-45\)

Nhẩm nghiệm ta được nghiệm x=5

\(\Rightarrow x^3-3x^2-x-45=\left(x-5\right)\left(x^2+2x+9\right)\)

b)+c) (2 câu này mk chỉ giải theo chương trình sgk thôi nhe. Hình như bạn ghi sai đề câu c):

\(6x^3-17x^2+14x-3\)

\(=6x^3-6x^2-11x^2+11x+3x-3\)

\(=6x^2\left(x-1\right)-11x\left(x-1\right)+3\left(x-1\right)\)

\(=\left(x-1\right)\left(6x^2-11x+3\right)\)

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

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

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

Mk sửa lại nhe: \(4x^3-25x^2-53x-24\)

\(=4x^3-32x^2+7x^2-56x+3x-24\)

\(=4x^2\left(x-8\right)+7x\left(x-8\right)+3\left(x-8\right)\)

\(=\left(x-8\right)\left(4x^2+7x+3\right)\)

\(=\left(x-8\right)\left(4x^2+4x+3x+3\right)\)

\(=\left(x-8\right)\left[4x\left(x+1\right)+3\left(x+1\right)\right]\)

\(=\left(x-8\right)\left(x+1\right)\left(4x+3\right)\)

Nếu bạn muốn giải cách Bézout - Horner thì nhắn cho mk nhe.

\(a,=\left(3-x+1\right)\left(9+3x-3+x^2-2x+1\right)\\ =\left(4-x\right)\left(x^2+x+7\right)\\ b,=4x^2-4xy-13xy+13y^2\\ =4x\left(x-y\right)-13y\left(x-y\right)\\ =\left(4x-13y\right)\left(x-y\right)\\ c,=4\left(x^2-xy-2y^2\right)\\ =4\left(x^2+xy-2xy-2y^2\right)\\ =4\left(x+y\right)\left(x-2y\right)\\ d,=x^3+4x^2+5x^2+20x+6x+24\\ =\left(x+4\right)\left(x^2+5x+6\right)\\ =\left(x+4\right)\left(x^2+2x+3x+6\right)\\ =\left(x+4\right)\left(x+2\right)\left(x+3\right)\\ f,=x\left(x+4y\right)-3\left(x+4y\right)=\left(x-3\right)\left(x+4y\right)\\ g,=4x^3+4x^2-29x^2-29x-24x-24\\ =\left(x+1\right)\left(4x^2-29x-24\right)\\ =\left(x+1\right)\left(4x^2-32x+3x-24\right)\\ =\left(x+1\right)\left(x-8\right)\left(4x+3\right)\)

\(a,27-\left(x-1\right)^3=\left(3-x+1\right)\left[9+3\left(x-1\right)+\left(x+1\right)^2\right]=\left(4-x\right)\left(9+3x-3+x^2+2x+1\right)=\left(4-x\right)\left(x^2+5x+7\right)\)

\(b,4x^2-17xy+13y^2=\left(4x^2-4xy\right)-\left(13xy-13y^2\right)=4x\left(x-y\right)-13y\left(x-y\right)=\left(x-y\right)\left(4x-13y\right)\)

\(c,4x^2-4xy-8y^2=4\left(x^2-xy-2y^2\right)\)

\(d,x^3+9x^2+26x+24=\left(x^3+2x^2\right)+\left(7x^2+14x\right)+\left(12x+24\right)=\left(x+2\right)\left(x^2+7x+12\right)=\left(x+2\right)\left[\left(x^2+3x\right)+\left(4x+12\right)\right]=\left(x+2\right)\left(x+3\right)\left(x+4\right)\)

\(f,4xy+x^2-3x-12y=x\left(4y+x\right)-3\left(x+4y\right)=\left(x+4y\right)\left(x-3\right)\)

\(g,4x^3-25x^2-53x-24=\left(4x^3-32x^2\right)+\left(7x^2-56x\right)+\left(3x-24\right)=\left(4x^2+7x+3\right)\left(x-8\right)=\left[\left(4x^2+4x\right)+\left(3x+3\right)\right]=\left(4x+3\right)\left(x+1\right)\left(x-8\right)\)

a)\(a^4+a^2+1=\left(a^2\right)^2+2a^2.1+1^2-a^2=\left(a^2+1\right)^2-a^2=\left(a^2+1+a\right)\left(a^2+1-a\right)\)

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

c)\(x^4+4x^2-5=x^4-x^2+5x^2-5=x^2\left(x^2-1\right)+5\left(x^2-1\right)=\left(x^2+5\right)\left(x+1\right)\left(x-1\right)\)

d)\(\left(x+2\right)\left(x^2-2x-6\right)=x^3-2x^2-6x+2x^2-4x-12=x^3-10x-12\)

\(\Rightarrow x^3-10x-12=\left(x+2\right)\left(x^2-2x-6\right)\)

e)\(6x^3-17x^2+14x-3\)

Ta có: \(\left(ax^2+bx+c\right)\left(dx+e\right)\)

\(=adx^3+aex^2+bdx^2+bex+cdx+ce\)

\(=adx^3+\left(ae+bd\right)x^2+\left(be+cd\right)x+ce\)

Do đó:\(\left\{{}\begin{matrix}ad=6\\ae+bd=-17\\be+cd=14\\ce=-3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=3;b=-4\\c=1;d=2\\e=-3\end{matrix}\right.\)

Suy ra: \(6x^3-17x^2+14x-3=\left(3x^2-4x+1\right)\left(2x-3\right)\)

h)\(x^4-34x^2+225=x^4-15x^2-15x^2+225-4x^2=x^2\left(x^2-15\right)-15\left(x^2-15\right)-\left(2x\right)^2=\left(x^2-15\right)^2-\left(2x\right)^2=\left(x^2+2x-15\right)\left(x^2-2x-15\right)=\left(x^2-3x+5x-15\right)\left(x^2+5x-3x-15\right)=\left[\left(x-3\right)\left(x+5\right)\right]^2\)

1: \(\left(x-2\right)^2-4x+8\)

\(=\left(x-2\right)\left(x-2-4\right)\)

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

3: \(a^3+6a^2+9a-ab^2\)

\(=a\left(a^2+6a+9-b^2\right)\)

\(=a\left(a+3-b\right)\left(a+3+b\right)\)

b) x² - 2x + 1 = 25x²

<=> (x - 1)² - 25x² = 0

<=> (x - 1 - 5x)(x - 1 + 5x)  = 0

<=> (-4x - 1)(6x - 1) = 0

<=> \(\orbr{\begin{cases}-4x-1=0\\6x-1=0\end{cases}}\)

<=> \(\orbr{\begin{cases}x=-\frac{1}{4}\\x=\frac{1}{6}\end{cases}}\)

c) 4x² - 4x = 24

<=> x² - x - 6 = 0

<=> x² - 3x + 2x - 6 = 0

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

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

<=> \(\orbr{\begin{cases}x+2=0\\x-3=0\end{cases}}\)

<=> \(\orbr{\begin{cases}x=-2\\x=3\end{cases}}\)

a: \(\left(x^2+x\right)^2+2\left(x^2+x\right)-8=0\)

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

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

hay \(x\in\left\{-2;1\right\}\)

b: \(\Leftrightarrow\left(x-1\right)\left(x-3\right)\left(x+2\right)\left(x+4\right)+24=0\)

\(\Leftrightarrow\left(x^2+x-2\right)\left(x^2+x-12\right)+24=0\)

\(\Leftrightarrow\left(x^2+x\right)^2-14\left(x^2+x\right)+48=0\)

\(\Leftrightarrow\left(x^2+x-6\right)\left(x^2+x-8\right)=0\)

hay \(x\in\left\{-3;2;\dfrac{-1+\sqrt{33}}{2};\dfrac{-1-\sqrt{33}}{2}\right\}\)

\(a.\sqrt{x-1}+\sqrt{4x-4}-\sqrt{25x-25}+2=0\)( x lớn hơn hoặc =1)
\(\sqrt{x-1}+\sqrt{4\left(x-1\right)}-\sqrt{25\left(x-1\right)}\)+2=0
\(\sqrt{x-1}\left(1+\sqrt{4}-\sqrt{25}\right)=-2\)
\(\sqrt{x-1}\left(1+2-5\right)=-2\)
\(\sqrt{x-1}.\left(-2\right)=-2\)
\(\sqrt{x-1}=-2.2\)
\(\sqrt{x-1}-4\)(ko thỏa mãn)
b)
\(\dfrac{1}{2}\sqrt{x-1}-\dfrac{3}{2}\sqrt{9x-9}+24\sqrt{\dfrac{x-1}{64}}=-17\)
\(\dfrac{1}{2}\sqrt{x-1}-\dfrac{3}{2}\sqrt{9\left(x-1\right)}+24\dfrac{\sqrt{x-1}}{8}=-17\)
\(\dfrac{1}{2}\sqrt{x-1}-\dfrac{3}{2}.3\sqrt{x-1}+3\sqrt{x-1}=-17\)
\(\dfrac{1}{2}\sqrt{x-1}-\dfrac{9}{2}\sqrt{x-1}+3\sqrt{x-1}=-17\)
\(\left(\dfrac{1}{2}-\dfrac{9}{2}+3\right)\sqrt{x-1}=-17\)
\(7\sqrt{x-1}=-17\)
\(\sqrt{x-1}=-\dfrac{17}{7}\)(ko thỏa mãn căn bậc 2 ko có số âm)

a: Ta có: \(\sqrt{x-1}+\sqrt{4x-4}-\sqrt{25x-25}+2=0\)

\(\Leftrightarrow\sqrt{x-1}+2\sqrt{x-1}-5\sqrt{x-1}=-2\)

\(\Leftrightarrow-2\sqrt{x-1}=-2\)

\(\Leftrightarrow x-1=1\)

hay x=2