nhân cả tử và mẫu với 3 rồi làm tương tự bài dưới

lam cu the ra duoc ko ak

\(B=\frac{9a}{3ab+9a+27}+\frac{9b}{3bc+9b+27}+\frac{9c}{9ac+9c+27}\)

Làm tương tự bài dưới

bn đã hỏi rất nhiều bài dạng này r` chả nhẽ mấy lần trc mk làm bn chép lại r` vứt só à 

\(\frac{3a}{ab+3a+9}=\frac{3abc}{ab^2c+3abc+9bc}=\frac{3.27}{27b+3.27+9bc}\)

=\(\frac{81}{9\left[3b+9+bc\right]}\)₁

\(\frac{3c}{ac+3c+9}=\frac{3cb}{abc+3bc+9b}=\frac{3bc}{3\left[9+bc+3b\right]}\)₂

từ 1,2  ta có B=\(\frac{9}{3b+9+bc}+\frac{bc}{9+bc+3b}+\frac{3b}{bc+3b+9}\)

=1

Bài 1: 

Thấy đường cao là a thì cạnh <=> là 1,5a (a > 0)

\(S_{\text{tam giác}}:\frac{1}{2}.a.1,5a=\frac{3}{4}a^2=27\)

\(\Rightarrow a^2=36\Rightarrow a=36\)

1,5a = 9

=> Cạnh tam giác là 9 cm

      Đường cao <=> 6 cm

Bài 2:

\(2a=3b;5b=7c\Rightarrow\frac{a}{3}=\frac{b}{2};\frac{b}{7}=\frac{c}{5}\Rightarrow\frac{a}{21}=\frac{b}{14};\frac{b}{14}=\frac{c}{10}\)

\(\Rightarrow\frac{a}{21}=\frac{b}{14}=\frac{c}{10}\)

Áp dụng t/c dãy tỉ số = nhau, ta có:

\(\frac{a}{21}=\frac{b}{14}=\frac{c}{10}=\frac{3a+5c-7b}{3.21+5.14+7.10}=\frac{30}{15}=2\)

\(\Rightarrow\hept{\begin{cases}\frac{a}{21}=2\Rightarrow a=2.21=42\\\frac{b}{14}=2\Rightarrow b=2.14=28\\\frac{c}{10}=2\Rightarrow c=2.10=20\end{cases}}\)

Vậy:...

\(VP=\frac{6}{\sqrt{\left(3a+bc\right)\left(3b+ca\right)\left(3c+ab\right)}}\)

\(=\frac{6}{\sqrt{\left[\left(a+b+c\right)a+bc\right]\left[\left(a+b+c\right)b+ca\right]\left[\left(a+b+c\right)c+ab\right]}}\)

\(=\frac{6}{\sqrt{\left(a+b\right)^2\left(b+c\right)^2\left(c+1\right)^2}}=\frac{6}{\left(a+b\right)\left(b+c\right)\left(a+c\right)}\)

\(VT=\frac{1}{3a+bc}+\frac{1}{3b+ca}+\frac{1}{3c+ab}\)

\(=\frac{1}{\left(a+b+c\right)a+bc}+\frac{1}{\left(a+b+c\right)b+ac}+\frac{1}{\left(a+b+c\right)c+ab}\)

\(=\frac{\left(b+c\right)+\left(a+c\right)+\left(a+b\right)}{\left(a+b\right)\left(b+c\right)\left(a+c\right)}=\frac{6}{\left(a+b\right)\left(b+c\right)\left(a+c\right)}\)

Vậy VT = VP, đẳng thức được chứng minh

Có \(ab+bc+ac=abc\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=1\)

Áp dụng các bđt sau:Với x;y;z>0 có: \(\dfrac{1}{x+y+z}\le\dfrac{1}{9}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\) và \(\dfrac{1}{x+y}\le\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\) 

Có \(\dfrac{1}{a+3b+2c}=\dfrac{1}{\left(a+b\right)+\left(b+c\right)+\left(b+c\right)}\le\dfrac{1}{9}\left(\dfrac{1}{a+b}+\dfrac{2}{b+c}\right)\)\(\le\dfrac{1}{9}.\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{2}{b}+\dfrac{2}{c}\right)=\dfrac{1}{36}\left(\dfrac{1}{a}+\dfrac{3}{b}+\dfrac{2}{c}\right)\)

CMTT: \(\dfrac{1}{b+3c+2a}\le\dfrac{1}{36}\left(\dfrac{1}{b}+\dfrac{3}{c}+\dfrac{2}{a}\right)\)

\(\dfrac{1}{c+3a+2b}\le\dfrac{1}{36}\left(\dfrac{1}{c}+\dfrac{3}{a}+\dfrac{2}{b}\right)\)

Cộng vế với vế => \(VT\le\dfrac{1}{36}\left(\dfrac{6}{a}+\dfrac{6}{b}+\dfrac{6}{c}\right)=\dfrac{1}{36}.6\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=\dfrac{1}{6}\)

Dấu = xảy ra khi a=b=c=3

Có \(a+b=2\Leftrightarrow2\ge2\sqrt{ab}\Leftrightarrow ab\le1\)

\(E=\left(3a^2+2b\right)\left(3b^2+2a\right)+5a^2b+5ab^2+2ab\)

\(=9a^2b^2+6\left(a^3+b^3\right)+4ab+5ab\left(a+b\right)+20ab\)

\(=9a^2b^2+6\left(a+b\right)^3-18ab\left(a+b\right)+4ab+5ab\left(a+b\right)+20ab\)

\(=9a^2b^2+48-18ab.2+4ab+5.2.ab+20ab\)

\(=9a^2b^2-2ab+48\)

Đặt \(f\left(ab\right)=9a^2b^2-2ab+48;ab\le1\), đỉnh \(I\left(\dfrac{1}{9};\dfrac{431}{9}\right)\)

Hàm đồng biến trên khoảng \(\left[\dfrac{1}{9};1\right]\backslash\left\{\dfrac{1}{9}\right\}\)

 \(\Rightarrow f\left(ab\right)_{max}=55\Leftrightarrow ab=1\)

\(\Rightarrow E_{max}=55\Leftrightarrow a=b=1\)

Vậy...

2,

\(ab\le\dfrac{1}{4}\left(a+b\right)^2=1\Rightarrow0\le ab\le1\)

\(E=9a^2b^2+6\left(a^3+b^3\right)+5ab\left(a+b\right)+24ab\)

\(=9a^2b^2+6\left(a+b\right)^3-18ab\left(a+b\right)+5ab\left(a+b\right)+24ab\)

\(=9a^2b^2-2ab+48\)

Đặt \(ab=x\Rightarrow0\le x\le1\)

\(E=9x^2-2x+48=\left(x-1\right)\left(9x+7\right)+55\le55\)

\(E_{max}=55\) khi \(x=1\) hay \(a=b=1\)

Ủa bị lỗi hả:v? 

\(3a^2+3b^2=10ab\)

\(\Rightarrow3a^2-10ab+3b^2=0\)

\(\Rightarrow3a^2-ab-9ab+3b^2=0\)

\(\Rightarrow\left(3a^2-ab\right)-\left(9ab-3b^2\right)=0\)

\(\Rightarrow a\left(3a-b\right)-3b\left(3a-b\right)=0\)

\(\Rightarrow\left(3a-b\right)\left(a-3b\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}3a-b=0\\a-3b=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}b=-3a\\b=\dfrac{a}{3}\end{matrix}\right.\)

Với \(b=-3a,\)có :

\(P=\dfrac{-3a-a}{-3a+a}=\dfrac{-4a}{-2a}=2\)

Với \(b=\dfrac{a}{3},\)có :

\(P=\dfrac{\dfrac{a}{3}-a}{\dfrac{a}{3}+a}=\dfrac{\dfrac{a}{3}-\dfrac{3a}{3}}{\dfrac{a}{3}+\dfrac{3a}{3}}=\dfrac{-\dfrac{2a}{3}}{\dfrac{4a}{3}}=-\dfrac{2a}{3}.\dfrac{3}{4a}=-\dfrac{1}{2}\)

( Nếu sai thì cho mk xin lỗi nha bn , tại mk ko chắc lắm )