sai cả rồi

x⁴=x

=>x⁴-x=0

=>x.(x³-1)=0

=>x=0 hoặc x³-1=0

Với x³-1=0=>x³=0=>x³=1=>x=1

Vậy có 2 trường hợp thỏa mãn đó là x=0 hoặc x=1

Giải:

Ta có: \(A=\dfrac{2011+2012}{2012+2013}\)

\(=\dfrac{2011}{2012+2013}+\dfrac{2012}{2012+2013}\)

Áp dụng tính chất \(\dfrac{a}{b}>\dfrac{a}{b+m}\left(m>0\right)\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{2011}{2012}>\dfrac{2011}{2012+2013}\\\dfrac{2012}{2013}>\dfrac{2012}{2012+2013}\end{matrix}\right.\)

Cộng vế theo vế ta được:

\(B=\dfrac{2011}{2012}+\dfrac{2012}{2013}>\dfrac{2011}{2012+2013}\) \(+\dfrac{2012}{2012+2013}\)

\(=\dfrac{2011+2012}{2012+2013}=A\)

Vậy \(A< B\)

Bài 3 không có hứng! Bài 5:

Ta có:

\(A=\dfrac{7}{4}\left(\dfrac{3333}{1212}+\dfrac{3333}{2020}+\dfrac{3333}{3030}+\dfrac{3333}{4242}\right)\)

\(=\dfrac{7}{4}\left(\dfrac{33}{12}+\dfrac{33}{20}+\dfrac{33}{30}+\dfrac{33}{42}\right)\)

\(=\dfrac{7}{4}\left(\dfrac{33}{3.4}+\dfrac{33}{4.5}+\dfrac{33}{5.6}+\dfrac{33}{6.7}\right)\)

\(=33.\dfrac{7}{4}\left(\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}\right)\)

\(=33.\dfrac{7}{4}\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{6}-\dfrac{1}{7}\right)\)

\(=33.\dfrac{7}{4}\left(\dfrac{1}{3}-\dfrac{1}{7}\right)\)

\(=33.\dfrac{7}{4}.\dfrac{4}{21}=11\)

Vậy \(A=11\)

Ta có: \(\left(x-5\right)^2=\left|x-5\right|\)

\(\Rightarrow B=\left(x-5\right)^2+\left|x-5\right|+2014\)

\(=\left|x-5\right|+\left|x-5\right|+2014\)

\(=2\left|x-5\right|+2014\)

Dễ thấy: \(\left|x-5\right|\ge0\Leftrightarrow2\left|x-5\right|\ge0\)

\(\Leftrightarrow B=\left(x-5\right)^2+\left|x-5\right|+2014\ge2014\)

Dấu "=" xảy ra \(\Leftrightarrow\left|x-5\right|=0\) \(\Leftrightarrow x-5=0\Leftrightarrow x=5\)

Vậy \(B_{min}=2014\Leftrightarrow x=5\)

Giải:

Ta xét hiệu \(\dfrac{a^2+b^2+c^2}{3}-\left(\dfrac{a+b+c}{3}\right)^2\)

\(=\dfrac{1}{9}\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\right]\) \(\ge0\) (Đúng)

\(\Leftrightarrow\dfrac{a^2+b^2+c^2}{3}\ge\left(\dfrac{a+b+c}{3}\right)^2\) (Đpcm)

Dấu "=" xảy ra \(\Leftrightarrow a=b=c\)

Ta có: \(x^2+2x+2\)

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

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

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

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

\(=\left(x+1\right)^2+1\ge1>0\) (vô nghiệm)

Vậy đa thức \(x^2+2x+2\) không có nghiệm (Đpcm)