a: =(x-3)(2x+5)

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

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

=>x=2 hoặc x=5

c: =>x-1=0

hay x=1

TK

c)=\(\left(x-1\right)^3=0\)=>x=1

d) \(PT\Leftrightarrow x\left(2x-7\right)-4\left(x-7\right)=0\)

\(\Leftrightarrow\left(2x-7\right)\left(x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-7=0\\x-4=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=4\end{matrix}\right.\)

Vậy: \(S=\left\{\dfrac{7}{2};4\right\}\)

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

\(\Leftrightarrow\left(x-7\right)\left(3x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-7=0\\3x-3=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=7\\x=1\end{matrix}\right.\)

Vậy: \(S=\left\{7;1\right\}\)

f) \(PT\Leftrightarrow x\left(x-1\right)-3\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-3=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)

Vậy: \(S=\left\{1;3\right\}\)

\(d,x\left(2x-7\right)-4x+14=0\)

\(x\left(2x-7\right)-2\left(2x-7\right)=0\)

\(\left(x-2\right)\left(2x-7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{7}{2}\end{matrix}\right.\)

d: =>(2x-7)(x-2)=0

=>x=7/2 hoặc x=2

e: =>(2x-5-x-2)(2x-5+x+2)=0

=>(x-7)(3x-3)=0

=>x=7 hoặc x=1

f: =>x(x-1)-3(x-1)=0

=>(x-1)(x-3)=0

=>x=1 hoặc x=3

\(\begin{cases} x-3y=-2\\2x+y=3 \end{cases} <=> \begin{cases} x-3(3-2x)=-2\\y=3-2x \end{cases} <=> \begin{cases} 7x=7\\y=3-2x \end{cases} \\<=> \begin{cases} x=1\\y=3-2.1 \end{cases} <=>\begin{cases} x=1\\y=1 \end{cases}\)

ta có : x^5+2x^4+3x^3+3x^2+2x+1=0

\(\Leftrightarrow\)x^5+x^4+x^4+x^3+2x^3+2x^2+x^2+x+x+1=0

\(\Leftrightarrow\)(x^5+x^4)+(x^4+x^3)+(2x^3+2x^2)+(x^2+x)+(x+1)=0

\(\Leftrightarrow\)x^4(x+1)+x^3(x+1)+2x^2(x+1)+x(x+1)+(x+1)=0

\(\Leftrightarrow\)(x+1)(x^4+x^3+2x^2+x+1)=0

\(\Leftrightarrow\)(x+1)(x^4+x^3+x^2+x^2+x+1)=0

\(\Leftrightarrow\)(x+1)[x^2(x^2+x+1)+(x^2+x+1)]=0

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

VÌ x^2+x+1=(x+\(\dfrac{1}{2}\))^2+\(\dfrac{3}{4}\)\(\ne0\) và x^2+1\(\ne0\)

\(\Rightarrow\)x+1=0

\(\Rightarrow\)x=-1

CÒN CÂU B TỰ LÀM (02042006)

b: x^4+3x^3-2x^2+x-3=0

=>x^4-x^3+4x^3-4x^2+2x^2-2x+3x-3=0

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

=>x-1=0

=>x=1

bn nhờ bn này giúp nek

bn ấy hc giỏi toán hơn mk

https://olm.vn/thanhvien/cuoidoi09081002

xl bn nha

mẫu mấy câu: 

\(a,x^4y^4+4\)

\(=\left(x^2y^2+2\right)^2-4x^2y^2\)(cái này hok lâu sẽ tự hiêur)

\(=\left(x^2y^2-2xy+2\right)\left(x^2y^2+2xy+2\right)\)

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

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

\(b,x^4+x^3+x^2+1=??\text{ xem lại đề tí đi bn}+x\text{ nữa mói lm đc}\)

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

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

a) \(x^5+2x^4+3x^3+3x^2+2x+1=0\)

\(\Leftrightarrow x^5+x^4+x^4+x^3+2x^3+2x^2+x^2+x+x+1=0\)

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

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

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

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

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

Dễ thấy \(x^2+x+1>0\forall x;x^2+1>0\forall x\)

\(\Rightarrow x+1=0\)

\(\Leftrightarrow x=-1\)

Vậy....

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

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

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

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

...

\(\Leftrightarrow x=1\)

p/s: có bác nào giải đc pt \(x^3+4x^2+2x+3=0\)thì giúp nhé :))

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

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

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

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

bài 1: 

a) ĐKXĐ: x khác 0; x khác -1

 \(\frac{x-1}{x}+\frac{1-2x}{x^2+x}=\frac{1}{x+1}\)

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

<=> (x - 1)(x + 1) + 1 - 2x = x

<=> x^2 - 2x = x

<=> x^2 - 2x - x = 0

<=> x^2 - 3x = 0

<=> x(x - 3) = 0

<=> x = 0 hoặc x - 3 = 0

<=> x = 0 hoặc x = 0 + 3

<=> x = 0 (ktm) hoặc x = 3 (tm)

=> x = 3

b) ĐKXĐ: x khác +-3; x khác -7/2

\(\frac{13}{\left(x-3\right)\left(2x+7\right)}+\frac{1}{2x+7}=\frac{6}{x^2-9}\)

<=> \(\frac{13}{\left(x-3\right)\left(2x+7\right)}+\frac{1}{2x+7}=\frac{6}{\left(x-3\right)\left(x+3\right)}\)

<=> 13(x + 3) + (x - 3)(x + 3) = 6(2x + 7)

<=> 13x + 30 + x^2 = 12x + 42

<=> 13x + 30 + x^2 - 12x - 42 = 0

<=> x - 12 + x^2 = 0

<=> (x - 3)(x + 4) = 0

<=> x - 3 = 0 hoặc x + 4 = 0

<=> x = 0 + 3 hoặc x = 0 - 4

<=> x = 3 (ktm) hoặc x = -4 (tm)

=> x = -4

c) ĐKXĐ: x khác +-1

\(\frac{x}{x-1}-\frac{2x}{\left(x-1\right)\left(x+1\right)}=0\)

<=> x(x + 1) - 2x = 0

<=> x^2 + x - 2x = 0

<=> x^2 - x = 0

<=> x(x - 1) = 0

<=> x = 0 hoặc x - 1 = 0

<=> x = 0 hoặc x = 0 + 1

<=> x = 0 (tm) hoặc x = 1 (ktm)

=> x = 0

d) \(\frac{x^2+2x}{x^2+1}-2x=0\)

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

<=> x(x + 2) - 2x(x^2 + 1) = 0

<=> x^2 - 2x^3 = 0

<=> x^2(1 - 2x) = 0

<=> x^2 = 0 hoặc 1 - 2x = 0

<=> x = 0 hoặc -2x = 0 - 1

<=> x = 0 hoặc -2x = -1

<=> x = 0 hoặc x = 1/2

bài 2: 

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

<=> x^3 + 3x^2 - 2x - x^2 - 3x + 2 - x^2 + 1 = 0

<=> 2x^2 - 2x - 3x + 3 = 0

<=> 2x(x - 1) - 3(x - 1) = 0

<=> (2x - 3)(x - 1) = 0

<=> 2x - 3 = 0 hoặc x - 1 = 0

<=> 2x = 0 + 3 hoặc x = 0 + 1

<=> 2x = 3 hoặc x = 1

<=> x = 3/2 hoặc x = 1

bài 3:

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

<=> x^3 + x^2 + x^2 + x = 0

<=> x^3 + 2x^2 + x = 0

<=> x(x^2 + 2x + 1) = 0

<=> x(x + 1)^2 = 0

<=> x = 0 hoặc x + 1 = 0

<=> x = 0 hoặc x = 0 - 1

<=> x = 0 hoặc x = -1

x³-2x²-5x=x(x²-2x-5)

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