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

\(=x^2+\frac{1}{2}x+\frac{1}{2}x+\frac{1}{4}+\frac{3}{4}\)

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

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

\(=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}>0\forall x\)

\(\Rightarrow P\left(x\right)\) vô nghiệm (ĐPCM)

chỉ cần cho p(x)=0 rùi cm

[] này là j vậy

5(x-2)-(x-3)=1

5x-10-x+3=1

(5x-x)+(-10+3)=1

4x-7=1

4x=8

x=8/4 

x=2

\(\frac{a}{c}=\frac{c}{b}\Rightarrow\frac{a^2}{c^2}=\frac{c^2}{b^2}==\frac{ac}{cb}=\frac{a}{d}\)

Mà \(\frac{a^2}{b^2}=\frac{c^2}{c^2}=\frac{a^2+c^2}{b^2+c^2}\) 

\(\Rightarrow\frac{a^2+c^2}{b^2+c^2}=\frac{a}{b}\)\(\Rightarrow\frac{b^2+c^2}{a^2+c^2}=\frac{b}{a}\)\(\Rightarrow\frac{b^2+c^2}{a^2+c^2}-1=\frac{b}{a}-1\)

\(\Rightarrow\frac{\left(b^2+c^2\right)-\left(a^2+c^2\right)}{a^2+c^2}=\frac{b-a}{a}\)

\(\Rightarrow\frac{b^2-a^2}{a^2+c^2}=\frac{b-a}{a}\) (ĐPCM)

Ta có: \(\frac{a}{c}=\frac{c}{b}\Rightarrow ab=c^2\Rightarrow ab-c^2=0\)

Ta lại có:\(\frac{b^2-a^2}{a^2+c^2}=\frac{b-a}{a}\)

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

\(\Leftrightarrow ab^2-a^3=a^2b+bc^2-a^3-ac^2\)

\(\Leftrightarrow ab^2-a^2b-bc^2+ac^2=0\)

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

\(\Leftrightarrow\left(b-a\right)\left(ab-c^2\right)=0\)[luôn đúng vì ab-c²=0(cmt)]

Ta có \(B=\left(\frac{2010}{2}+1\right)+\left(\frac{2009}{3}+1\right)+...+\left(\frac{2}{2010}+1\right)+\left(\frac{1}{2011}+1\right)+1\)

\(B=\frac{2012}{2}+\frac{2012}{3}+...+\frac{2012}{2010}+\frac{2012}{2011}+\frac{2012}{2012}\)

\(B=2012.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2010}+\frac{1}{2011}+\frac{1}{2012}\right)\)

B=2012.A

=>A/B=1/2012

a/b= 1/2012 nha bạn 

tích

a) |x + 1| \(\ge0\)

    |x + 3| \(\ge0\)

    |x + 5| \(\ge0\)

=> |x + 1| + |x + 3| + |x + 5| \(\ge0\)

=> 7x \(\ge0\)

Mà 7 \(>0\)

=> x \(\ge0\)

=> x + 1 + x + 3 + x + 5 = 7x 

=> 3x + 9 = 7x

=> 4x = 9

=> x = \(\frac{9}{4}\)

a) Vì \(\left|x+1\right|+\left|x+3\right|+\left|x+5\right|\ge0\forall x\in R\Rightarrow7x\ge0\forall x\in R\Rightarrow x\ge0\)

\(\Rightarrow\left|x+1\right|+\left|x+3\right|+\left|x+5\right|=x+1+x+3+x+5=3x+9\)

\(\Rightarrow3x+9=7x\)

\(\Rightarrow7x-3x=9\)

\(\Rightarrow4x=9\)

\(\Rightarrow x=\frac{9}{4}\)

Ta có

x²-x-2=x(x-1)-2

Vì x thuộc Z nên x(x-1) là số chẵn

Ta có x(x-1) \(⋮2\)

          \(2⋮2\)

=> M(x) luôn là 1 số chẵn

=> M(x) không thể là số nguyên tố

Chú ý rằng ko có trường hợp x²-x-2=2

Khi đó x(x-1)=4, ko có x nào thỏa mãn

thanks