có k mk đi

Đ/A:5

24-4-20+5=5

Đọc rồi thì đăng linh tinh ít thoy nha bn.

Học tốt!

\(x^4+1997x^2+1996x+1997=0\)

\(\Leftrightarrow\left(x^4-x\right)+1997\left(x^2+x+1\right)=0\)

\(\Leftrightarrow x\left(x^3-1\right)+1997\left(x^2+x+1\right)=0\)

\(\Leftrightarrow x\left(x-1\right)\left(x^2+x+1\right)+1997\left(x^2+x+1\right)=0\)

\(\Leftrightarrow\left(x^2-x+1997\right)\left(x^2+x+1\right)=0\)

\(\hept{\begin{cases}x^2-x+1997>0\\x^2+x+1>0\end{cases}}\Rightarrow ptvn\)

\(x^2-x+2011.2012=0\)

\(\Leftrightarrow x^2+2011x-2012x+2011.2012=0\)

\(\Leftrightarrow x\left(x+2011\right)-2012\left(x+2011\right)=0\Leftrightarrow\left(x-2012\right)\left(x+2011\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-2012=0\\x+2011=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=2012\\x=-2011\end{cases}}\)

câu b) đề sai nhé,ở trên mk nhầm

c)

\(x^5=x^4+x^3+x^2+x+2\)

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

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

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

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

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

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

\(\Leftrightarrow5x^2-2x+2=x^2+2x+1\)

\(\Leftrightarrow5x^2-x^2-2x-2x=1-2\)

\(\Leftrightarrow4x^2-4x+1=0\)

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

\(\Leftrightarrow2x-1=0\)

\(\Leftrightarrow x=\dfrac{1}{2}\)

Vậy \(S=\left\{\dfrac{1}{2}\right\}\)

\(2,\sqrt{4x^2-x+1}-2x=3\)

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

\(\Leftrightarrow4x^2-x+1=9+12x+4x^2\)

\(\Leftrightarrow4x^2-4x^2-x-12x=9-1\)

\(\Leftrightarrow-13x=8\)

\(\Leftrightarrow x=-\dfrac{8}{13}\)

Vậy \(S=\left\{-\dfrac{8}{13}\right\}\)

1: =>x>=-1 và 5x^2-2x+2=x^2+2x+1

=>x>=-1 và 4x^2-4x+1=0

=>x=1/2

2: =>\(\sqrt{4x^2-x+1}=2x+3\)

=>x>=-3/2 và 4x^2-x+1=4x^2+12x+9

=>x>=-3/2 và -11x=8

=>x=-8/11(nhận)

ĐKXĐ:

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

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

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

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

ý 1: Để pt (1) có 1 nghiệm duy nhất thì \(\Delta=0\)

\(\Delta=\left(-5\right)^2-4m+8=-4m+33\)

\(\Rightarrow33-4m=0\Rightarrow m=\dfrac{33}{4}\)

ý 2: Khi \(m=4\Rightarrow x^2-5x+2=0\)

\(\Delta=\left(-5\right)^2-8=17\Rightarrow\left[{}\begin{matrix}x=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{5-\sqrt{17}}{2}\\x=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{5+\sqrt{17}}{2}\end{matrix}\right.\)

Vậy...

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