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

\(ĐKXĐ:x\ge\frac{1}{2}\)

Phương trình đã cho tương đương :

\(4.\left(x^2+1\right)+3.x.\left(x-2\right).\sqrt{2x-1}=2x^3+10x\)

\(\Leftrightarrow3x\left(x-2\right)\sqrt{2x-1}=2x^3-8x^2+10x-4\)

\(\Leftrightarrow3x.\left(x-2\right).\sqrt{2x-1}=2.\left(x-2\right).\left(x-1\right)^2\) (1)

Dễ thấy \(x=2\) là một nghiệm của (1). Xét \(x\ne2\). Khi đó ta có :

\(3x.\sqrt{2x-1}=2.\left(x-1\right)^2\)(*)

Đặt \(\sqrt{2x-1}=a\left(a\ge0\right)\Rightarrow-a^2=1-2x\)

Khi đó pt (*) có dạng :

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

\(\Leftrightarrow2x^2-3xa-2a^2=0\)

\(\Leftrightarrow2x^2-4ax+xa-2a^2=0\)

\(\Leftrightarrow2x.\left(x-2a\right)+a.\left(x-2a\right)=0\)

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

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

+) Với \(2a=x\Rightarrow2\sqrt{2x-1}=x\left(x\ge0\right)\)

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

\(\Leftrightarrow x^2-8x+4=0\)

\(\Leftrightarrow x=4\pm2\sqrt{3}\) ( Thỏa mãn )

+) Với \(a=-2x\Rightarrow\sqrt{2x-1}=-2x\left(x\le0\right)\)

\(\Leftrightarrow4x^2=2x-1\)

\(\Leftrightarrow4x^2-2x+1=0\) ( Vô nghiệm )

Vậy phương trình đã cho có tập nghiệm \(S=\left\{4\pm2\sqrt{3},2\right\}\)

=>x^3+6x^2+12x+8+1/3(8x^3-24x^2+24x-8)=1/5x+2/5+8

=>x^3+6x^2+12x+8+8/3x^3-8x^2+8x-8/3=1/5x+42/5

=>11/3x^3-2x^2+20x+16/3-1/5x-42/5=0

=>11/3x^3-11/5x^2+20x-46/15=0

=>\(x\simeq0,16\)

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

\(\dfrac{1}{x+3}+\dfrac{8}{\left(x+1\right)\left(x-3\right)}=\dfrac{2x}{x^2-2x-3}\)

* x² - 2x - 3 = x²- 3x + x - 3 = x(x-3 ) + ( x - 3) = ( x - 3 ) (  x + 1 )

\(\Leftrightarrow\dfrac{1}{x+3}+\dfrac{8}{\left(x+1\right)\left(x-3\right)}=\dfrac{2x}{\left(x-3\right)\left(x+1\right)}\left(ĐKXĐ:x\ne\pm3;x\ne-1\right)\)

\(\Leftrightarrow\left(x+1\right)\left(x-3\right)+8\left(x+3\right)=2x\left(x+3\right)\)

\(\Leftrightarrow x^2-2x+1+8x+24=2x^2+6x\)

\(\Leftrightarrow-x^2+25=0\)

\(\Leftrightarrow x^2-25=0\Leftrightarrow\left(x-5\right)\left(x+5\right)=0\)

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

Vậy \(S=\left\{-5;5\right\}\)

a.

ĐKXĐ: \(1\le x\le7\)

\(\Leftrightarrow x-1-2\sqrt{x-1}+2\sqrt{7-x}-\sqrt{\left(x-1\right)\left(7-x\right)}=0\)

\(\Leftrightarrow\sqrt{x-1}\left(\sqrt{x-1}-2\right)-\sqrt{7-x}\left(\sqrt{x-1}-2\right)=0\)

\(\Leftrightarrow\left(\sqrt{x-1}-\sqrt{7-x}\right)\left(\sqrt{x-1}-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-1}=\sqrt{7-x}\\\sqrt{x-1}=2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=7-x\\x-1=4\end{matrix}\right.\)

\(\Leftrightarrow...\)

b. ĐKXĐ: ...

Biến đổi pt đầu:

\(x\left(y-1\right)-\left(y-1\right)^2=\sqrt{y-1}-\sqrt{x}\)

Đặt \(\left\{{}\begin{matrix}\sqrt{x}=a\ge0\\\sqrt{y-1}=b\ge0\end{matrix}\right.\)

\(\Rightarrow a^2b^2-b^4=b-a\)

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

\(\Leftrightarrow\left(a-b\right)\left(b^2\left(a+b\right)+1\right)=0\)

\(\Leftrightarrow a=b\)

\(\Leftrightarrow\sqrt{x}=\sqrt{y-1}\Rightarrow y=x+1\)

Thế vào pt dưới:

\(3\sqrt{5-x}+3\sqrt{5x-4}=2x+7\)

\(\Leftrightarrow3\left(x-\sqrt{5x-4}\right)+7-x-3\sqrt{5-x}=0\)

\(\Leftrightarrow\dfrac{3\left(x^2-5x+4\right)}{x+\sqrt{5x-4}}+\dfrac{x^2-5x+4}{7-x+3\sqrt{5-x}}=0\)

\(\Leftrightarrow\left(x^2-5x+4\right)\left(\dfrac{3}{x+\sqrt{5x-4}}+\dfrac{1}{7-x+3\sqrt{5-x}}\right)=0\)

\(\Leftrightarrow...\)