Năm nay tuổi của mẹ là : \(27:\left(4-1\right).4=36\) ( tuổi )

Đ/s : ... 

Tuổi mẹ 6 năm sau : \(32:\left(5-1\right).5=40\)  ( tuổi ) 

Tuổi mẹ hiện nay : \(40-6=34\)  ( tuổi )

Tuổi con hiện nay : \(34-32=2\)  ( tuổi ) 

Đ/s : ... 

Bài này mik nghĩ là tìm Min P 

Thấy : \(\sqrt{\dfrac{ab}{c+ab}}=\sqrt{\dfrac{ab}{c\left(a+b+c\right)+ab}}=\sqrt{\dfrac{ab}{\left(a+c\right)\left(b+c\right)}}\le\dfrac{1}{2}\left(\dfrac{a}{a+c}+\dfrac{b}{b+c}\right)\) 

CMTT : \(\sqrt{\dfrac{bc}{a+bc}}\le\dfrac{1}{2}\left(\dfrac{b}{a+b}+\dfrac{c}{a+c}\right)\)  ; \(\sqrt{\dfrac{ac}{b+ac}}\le\dfrac{1}{2}\left(\dfrac{a}{b+a}+\dfrac{c}{b+c}\right)\)

Suy ra : \(P\le\dfrac{1}{2}\left(\dfrac{a+b}{a+b}+\dfrac{b+c}{b+c}+\dfrac{c+a}{c+a}\right)=\dfrac{3}{2}\)

" = " \(\Leftrightarrow a=b=c=\dfrac{1}{3}\)

5. H/s x/đ trên R \(\Leftrightarrow x^2-2mx-m+2\ge0\forall x\Leftrightarrow\Delta'\le0\)  

\(\Leftrightarrow m^2-\left(2-m\right)\le0\Leftrightarrow m^2+m-2\le0\)  \(\Leftrightarrow-2\le m\le1\)

6. a. Với a ; b ; c > 0 . AD BĐT Cauchy ta được : \(\dfrac{a}{bc}+\dfrac{b}{ac}\ge2\sqrt{\dfrac{a}{bc}.\dfrac{b}{ac}}=\dfrac{2}{c}\)

CMTT : \(\dfrac{b}{ac}+\dfrac{c}{ab}\ge\dfrac{2}{a};\dfrac{a}{bc}+\dfrac{c}{ab}\ge\dfrac{2}{b}\) 

\(\Rightarrow2\left(\dfrac{a}{bc}+\dfrac{b}{ac}+\dfrac{c}{ab}\right)\ge2\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\Rightarrow\dfrac{a}{bc}+\dfrac{b}{ac}+\dfrac{c}{ab}\ge\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\)  

b. AD BĐT Cauchy ta được : \(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}=\dfrac{a^2}{ab+ac}+\dfrac{b^2}{bc+ba}+\dfrac{c^2}{ca+cb}\ge\dfrac{\left(a+b+c\right)^2}{2\left(ab+bc+ac\right)}\ge\dfrac{3\left(ab+bc+ac\right)}{2\left(ab+bc+ac\right)}=\dfrac{3}{2}\)

c. AD BĐT B.C.S ta được : \(\left(2^2+4^2\right)\left(x^2+y^2\right)\ge\left(2x+4y\right)^2\Rightarrow20\left(x^2+y^2\right)\ge1\) \(\Rightarrow x^2+y^2\ge\dfrac{1}{20}\)

" = " \(\Leftrightarrow x=\dfrac{1}{10};y=\dfrac{1}{5}\)

\(\Delta ABC\) đều cạnh là mấy a ? 

ĐKXĐ : \(x\ge\sqrt[3]{7}\) 

P/t \(\Leftrightarrow\left(x^3-4\right)\left(\sqrt{x^3-7}-1\right)+4x^3-x^2+2x-32=0\)

\(\Leftrightarrow\left(x^3-4\right).\dfrac{x^3-8}{\sqrt{x^3-7}+1}+\left(4x^2+7x+16\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left[\dfrac{\left(x^3-4\right)\left(x^2+2x+4\right)}{\sqrt{x^3-7}+1}+4x^2+7x+16\right]=0\)

Dễ thấy : [...] > 0 \(\forall x\in\) ĐKXĐ  \(\Rightarrow x-2=0\Leftrightarrow x=2\)

Vậy ... 

(...) > 0 \(\forall x\in R\Leftrightarrow\Delta< 0\Leftrightarrow\left(2m-1\right)^2-12\left(4m-11\right)< 0\)  

\(\Leftrightarrow4m^2-52m+133\Leftrightarrow\dfrac{7}{2}< m< \dfrac{19}{2}\)

\(6,4=12,6-x.\dfrac{25}{10}\Rightarrow2,5x=6,2\Rightarrow x=\dfrac{62}{25}\)