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

Đặt a = 3 - x b = 2 - x
=>a ^ 4 + b ^ 4 = (a + b) ^ 4 và a - b=1
(=) a ^ 4 + b ^ 4 = a ^ 4 + 4a ^ 3b+6a ^ 2b ^2 + 4ab ^ 3 + b^4
và a - b =1
(=) ab(2a^2 + 2b^2 + 3ab) = 0 và a - b = 1
Xét a = 0, tương đương b = +-1
b = 0, tương đương a = +-1
2a^2 + 2b^2 + 3ab = 0 => HPt vo nghiem
vay ta co nghiem: x=2,x=3

Đặt 3-x = a ; 2-x = b

=> 5-2x = a+b

pt <=> a^4+b^4 = (a+b)^4 = a^4+4a^3b+6a^2b^2+4ab^3+b^4

<=> a^4+4a^3b+6a^2b^2+4ab^3+b^4-a^4-b^4 = 0

<=> 4a^3b+6a^2b^2+4ab^3 = 0

<=> 2a^3b+3a^2b^2+2ab^3 = 0

<=> ab.(2a^2+3ab+2b^2) = 0

<=> ab=0 ( vì 2a^2+3ab+2b^2 > 0 )

<=> a=0 hoặc b=0

<=> 3-x=0 hoặc 2-x=0

<=> x=3 hoặc x=2

Vậy .............

Tk mk nha

ĐKXĐ: \(x\notin\left\{-3;1\right\}\)

Ta có: \(\frac{4}{x^2+2x-3}=\frac{2x-5}{x+3}-\frac{2x}{x-1}\)

\(\Leftrightarrow\frac{\left(2x-5\right)\left(x-1\right)}{\left(x+3\right)\left(x-1\right)}-\frac{2x\left(x+3\right)}{\left(x-1\right)\left(x+3\right)}=\frac{4}{\left(x-1\right)\left(x+3\right)}\)

Suy ra: \(\left(2x-5\right)\left(x-1\right)-2x\left(x+3\right)=4\)

\(\Leftrightarrow2x^2-2x-5x+5-2x^2-6x=4\)

\(\Leftrightarrow-13x+5=4\)

\(\Leftrightarrow-13x=4-5=-1\)

hay \(x=\frac{1}{13}\)(nhận)

Vậy: \(S=\left\{\frac{1}{13}\right\}\)

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

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

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

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

\(x^2-2x+2x-4-x^2-5x=0\)

\(-5x-4=0\)

\(-5x=4\)

\(\Rightarrow\)\(x=\frac{-4}{5}\)

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

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

\(\left(x-2\right)\left(x-2-2x-10\right)=0\)

\(\left(x-2\right)\left(-x-12\right)=0\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x-2=0\\-x-12=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=-12\end{cases}}}\)

Bạn tự kết luận 2 câu nhé

làm hệ PT nghĩ ngay đến xét hàm

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 :))

a-b=5 và (a,b)/[a,b]. Tim a:b

giải phaj bỏ ngoặc nhức đầu lắm