sử dụng bài toán phụ \(3\left(a^2+b^2+c^2\right)\ge\) \(\left(a+b+c\right)^2\)

dễ thế đéo biết làm

sử dụng bài toán phụ:

\(3\left(a^2+b^2+c^2\right)\ge\left(a+b+c\right)^2\) 

dễ thế mà không biết làm

a)  Xét  \(\Delta OAB\)và   \(\Delta OCD\)có:

      \(\widehat{OAB}=\widehat{OCD}\) (slt)

      \(\widehat{OBA}=\widehat{ODC}\) (slt)

suy ra:   \(\Delta OAB~\Delta OCD\)  (g.g)

\(\Rightarrow\)\(\frac{OA}{OC}=\frac{OB}{OD}\)

\(\Rightarrow\)\(OA.OD=OB.OC\)

b)  \(\Delta OAB~\Delta OCD\)  

\(\Rightarrow\)\(\frac{OA}{AC}=\frac{AB}{CD}\)

\(\Rightarrow\)\(OA=\frac{OC.AB}{CD}=3\)

\(\Rightarrow\)\(AC=OA+OC=9\)

\(\Delta AEO~\Delta ADC\)  ( do OE // DC )

\(\Rightarrow\)\(\frac{OE}{DC}=\frac{OA}{AC}\)  \(\Rightarrow\) \(OE=\frac{OA.DC}{AC}=\frac{10}{3}\)

Ta có a+b+c>(a+b+c):1

=>a>1, b<1, c>1

=>.. dpcm

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{bc+ac+ab}{abc}=\frac{bc+ac+ab}{1}=bc+ac+ab\Rightarrow a+b+c>bc+ac+ab\)

\(\left(a-1\right)\left(b-1\right)\left(c-1\right)=\left(ab-a-b+1\right)\left(c-1\right)=abc-ac-bc+c-ab+a+b-1\)

\(=1-1+a+b+c-ac-bc-ab=a+b+c-\left(ac+bc+ab\right)\)

vì \(a+b+c>bc+ac+ab\)(chứng minh trên)\(\Rightarrow a+b+c-\left(bc+ac+ab\right)>0\)

\(\Rightarrow\left(a-1\right)\left(b-1\right)\left(c-1\right)>0\)

PT \(\Leftrightarrow\left(\frac{1}{1+a^2}-\frac{1}{1+ab}\right)-\left(\frac{1}{1+ab}-\frac{1}{1+b^2}\right)< 0\)

\(\Leftrightarrow\frac{ab-a^2}{\left(1+a^2\right)\left(1+ab\right)}-\frac{b^2-ab}{\left(1+b^2\right)\left(1+ab\right)}< 0\)

\(\Leftrightarrow\frac{a\left(b-a\right)}{\left(1+a^2\right)\left(1+ab\right)}-\frac{b\left(b-a\right)}{\left(1+b^2\right)\left(1+ab\right)}< 0\)

\(\Leftrightarrow\frac{b-a}{1+ab}\left(\frac{a}{1+a^2}-\frac{b}{1+b^2}\right)< 0\)

\(\Leftrightarrow\frac{b-a}{1+ab}.\frac{a+ab^2-b-a^2b}{\left(1+a^2\right)\left(1+b^2\right)}< 0\)

\(\Leftrightarrow\frac{b-a}{ab+a}.\frac{\left(ab-1\right)\left(b-a\right)}{\left(1+a^2\right)\left(1+b^2\right)}< 0\\\)

\(\Leftrightarrow\frac{\left(b-a\right)^2\left(ab-1\right)}{\left(1+a^2\right)\left(1+b^2\right)\left(ab+1\right)}< 0\)

vì \(\left(b-a\right)^2\ge0;\left(1+a^2\right),\left(1+b^2\right)>0\)

\(\Leftrightarrow\frac{ab-1}{ab+1}< 0\left(vớia\ne b\right)\)

vì \(ab-1< ab+1\)

\(\Leftrightarrow\hept{\begin{cases}ab-1< 0\\ab+1>0\end{cases}\Leftrightarrow-1< ab< 1}\)

Vậy nghiệm của PT là \(-1< ab< 1\) và \(a\ne b\)

Áp dụngbdt bunhiacopki (a²+b²)(x²+y²)>=(ax+by)²

1.Nửa chu vi mảnh đất là:

600 : 2 = 300 ( m )

Chiều dài là :

( 300 + 190 ) : 2 = 245 ( m )

Chiều rộng là:

245 - 190 = 55 ( m )

Diện tích là:

55 x 245 = 13475 ( m² )

2.số 59

53900 M²