Chieu nay nhe

troi oi anh oi kho nhu vay lam sao ma lam duoc vay de hay la em len hoi thay giao em nhe thay em chinh la bo cua em day va bo em chinh la hieu pho cua truong thcs doan ket

\(x^6+x^5+x^4+x^3+x^2+x+1=x^4\left(x^2+x\right)+x^2\left(x^2+x\right)+x^2+x+1\)

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

\(Taco:\left(x^2+x\right)\left(x^4+x^2+1\right)\ge0\forall x\Rightarrow\left(x^2+x\right)\left(x^4+x^2+1\right)+1\ge1\)

\(Ma:\left(x^2+x\right)\left(x^4+x^2+1\right)+1=0\left(loai\right)\)

Vay pt vo nghiem

Ta có:\(x^4-x^3+2x^2-x+1=0\)

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

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

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

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

Vì \(x^2-x+1=\left(x^2-x+\frac{1}{4}\right)+\frac{3}{4}=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)

và\(x^2+1\ge1\)

nên \(\left(x^2-x+1\right)\left(x^2+1\right)\ge\frac{3}{4}\)

Vậy Pt trên vô nghiệm

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

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

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

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

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

\(\Rightarrow x^2+1=0\)( do \(x^2-x+1\)là bình phương thiếu nên không thể bằng 0)

\(\Leftrightarrow x^2=-1\)( vô lý )

Do đó : Phương trình đã cho vô nghiệm

a) \(ĐKXĐ:x\inℝ\)

\(\frac{x^2+2x+3}{x^2-x+1}=0\)

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

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

\(\Leftrightarrow\)Phương trình vô nghiệm (ĐPCM)

b) \(ĐKXĐ:x\ne\pm2\)

 \(\frac{x}{x+2}+\frac{4}{x-2}=\frac{4}{x^2-4}\)

\(\Leftrightarrow\frac{x}{x+2}+\frac{4}{x-2}-\frac{4}{\left(x-2\right)\left(x+2\right)}=0\)

\(\Leftrightarrow\frac{x\left(x-2\right)+4\left(x+2\right)-4}{\left(x-2\right)\left(x+2\right)}=0\)

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

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

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

\(\Leftrightarrow\)Phương trình vô nghiệm (ĐPCM)

\(a,x^2-\left(2m-3\right)x+m^2=0-vô-ngo\)

\(\Leftrightarrow\Delta< 0\Leftrightarrow[-\left(2m-3\right)]^2-4m^2< 0\Leftrightarrow m>\dfrac{3}{4}\)

\(b,\left(m-1\right)x^2-2mx+m-2=0\)

\(m-1=0\Leftrightarrow m=1\Rightarrow-2x-1=0\Leftrightarrow x=-0,5\left(ktm\right)\)

\(m-1\ne0\Leftrightarrow m\ne1\Rightarrow\Delta'< 0\Leftrightarrow\left(-m\right)^2-\left(m-2\right)\left(m-1\right)< 0\Leftrightarrow m< \dfrac{2}{3}\)

\(c,\left(2-m\right)x^2-2\left(m+1\right)x+4-m=0\)

\(2-m=0\Leftrightarrow m=2\Rightarrow-6x+2=0\Leftrightarrow x=\dfrac{1}{3}\left(ktm\right)\)

\(2-m\ne0\Leftrightarrow m\ne2\Rightarrow\Delta'< 0\Leftrightarrow[-\left(m+1\right)]^2-\left(4-m\right)\left(2-m\right)< 0\Leftrightarrow m< \dfrac{7}{8}\)

+) Ta có: P(x) = 0 có 3 nghiệm phân biệt 

=> Gọi 3 nghiệm đó là m; n ; p. 

=> P(x) = ( x - m ) ( x - p ) (x - n) 

=> P(Q(x)) = ( x^2 + x + 2013 -m )( x^2 + x + 2013 -n )( x^2 + x + 2013 - p )

Vì P(Q(x)) =0 vô nghiệm nên: x^2 + x + 2013 - m = 0 ;x^2 + x + 2013 - m = 0; x^2 + x + 2013 - m = 0 đều vô nghiệm 

=> \(\Delta_m=1^2-4\left(2013-m\right)< 0;\Delta_n=1^2-4\left(2013-n\right)< 0;\Delta_p=1^2-4\left(2013-p\right)< 0\)​​

=> \(2013-m>\frac{1}{4};2013-n>\frac{1}{4};2013-p>\frac{1}{4}\)

=> P(2013) = ( 2013 - m) (2013 -n ) (2013 - p) >\(\frac{1}{4}.\frac{1}{4}.\frac{1}{4}=\frac{1}{64}\)

\(\text{CM vô nghiệm}\)
\(\text{a) }\left(x-2\right)^3=\left(x-2\right).\left(x^2+2x+4\right)-6\left(x-1\right)^2\)
\(\Leftrightarrow x^3-6x^2+12x-8=x^3-8-6\left(x^2-2x+1\right)\)
\(\Leftrightarrow x^3-6x^2+12x-8=x^3-8-6x^2+12x-6\)
\(\Leftrightarrow x^3-6x^2+12x-x^3+6x-12x=-8+8-6\)
\(\Leftrightarrow0x=-6\text{ (vô lí)}\)
\(\text{Vậy }S=\varnothing\)

\(\text{b) }4x^2-12x+10=0\)
\(\Leftrightarrow\left(4x^2-12x+9\right)+1=0\)
\(\Leftrightarrow\left(2x-3\right)^2+1=0\)
\(\Leftrightarrow\left(2x-3\right)^2=-1\text{ (vô lí)}\)
\(\text{Vậy }S=\varnothing\)

\(\text{CM vô số nghiệm}\)
       \(\left(x+1\right)\left(x^2-x+1\right)=\left(x+1\right)^3-3x\left(x+1\right)\)
\(\Leftrightarrow\left(x+1\right)\left(x^2-x+1\right)=\left(x+1\right)\left[\left(x+1\right)^2-3x\right]\)
\(\Leftrightarrow\left(x+1\right)\left(x^2-x+1\right)=\left(x+1\right)\left(x^2+2x+1-3x\right)\)
\(\Leftrightarrow\left(x+1\right)\left(x^2-x+1\right)=\left(x+1\right)\left(x^2-x+1\right)\text{ (luôn luôn đúng)}\)
\(\text{Vậy }S\inℝ\)

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a) Ta có: \(x^2+2x+3\)

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

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

Vậy pt vô nghiệm

b) Ta có \(x^2+2x+4\)

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

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

Vậy pt vô nghiệm