\(A=\dfrac{ab+10b+25}{ab+5a+5b+25}+\dfrac{bc+10c+25}{bc+5b+5c+25}+\dfrac{ca+10a+25}{ac+5a+5c+25}\)

\(=\dfrac{\left(ab+5b\right)+\left(5b+25\right)}{\left(ab+5a\right)+\left(5b+25\right)}+\dfrac{\left(bc+5c\right)+\left(5c+25\right)}{\left(bc+5b\right)+\left(5c+25\right)}+\dfrac{\left(ca+5a\right)+\left(5a+25\right)}{\left(ac+5a\right)+\left(5c+25\right)}\)

\(=\dfrac{b\left(a+5\right)+5\left(b+5\right)}{a\left(b+5\right)+5\left(b+5\right)}+\dfrac{c\left(b+5\right)+5\left(c+5\right)}{b\left(c+5\right)+5\left(c+5\right)}+\dfrac{a\left(c+5\right)+5\left(a+5\right)}{a\left(c+5\right)+5\left(c+5\right)}\)

\(=\dfrac{b\left(a+5\right)+5\left(b+5\right)}{\left(a+5\right)\left(b+5\right)}+\dfrac{c\left(b+5\right)+5\left(c+5\right)}{\left(b+5\right)\left(c+5\right)}+\dfrac{a\left(c+5\right)+5\left(a+5\right)}{\left(a+5\right)\left(c+5\right)}\)

\(=\dfrac{b}{b+5}+\dfrac{5}{a+5}+\dfrac{c}{c+5}+\dfrac{5}{b+5}+\dfrac{a}{a+5}+\dfrac{5}{c+5}\)

\(=\left(\dfrac{b}{b+5}+\dfrac{5}{b+5}\right)+\left(\dfrac{a}{a+5}+\dfrac{5}{a+5}\right)+\left(\dfrac{c}{c+5}+\dfrac{5}{c+5}\right)\)

\(=1+1+1=3\) (\(a;b;c\ne-5\))

\(A=\dfrac{ab+5b+5b+25}{a\left(b+5\right)+5\left(b+5\right)}+\dfrac{bc+5c+5c+25}{b\left(c+5\right)+5\left(c+5\right)}+\dfrac{ca+5a+5a+25}{a\left(c+5\right)+5\left(c+5\right)}\)

\(A=\dfrac{b\left(a+5\right)+5\left(b+5\right)}{\left(a+5\right)\left(b+5\right)}+\dfrac{c\left(b+5\right)+5\left(c+5\right)}{\left(b+5\right)\left(c+5\right)}+\dfrac{a\left(c+5\right)+5\left(a+5\right)}{\left(a+5\right)\left(c+5\right)}\)

\(A=\dfrac{b}{b+5}+\dfrac{5}{a+5}+\dfrac{c}{c+5}+\dfrac{5}{b+5}+\dfrac{a}{a+5}+\dfrac{5}{c+5}\)

\(A=\dfrac{a+5}{a+5}+\dfrac{b+5}{b+5}+\dfrac{c+5}{c+5}=1+1+1=3\)

a) 25 + a - (b + c) - a

= 25 - (b + c) = 25 - [25 + (-30)]

= 25 - (-5)

= 30

b) a + b - (c + b)

= a + b - c - b

= a - c

= -13 - 25

= -38

a) 25 + a - (b + c) - a

= 25 - (b + c) = 25 - [25 + (-30)]

= 25 - (-5)

= 30

b) a + b - (c + b)

= a + b - c - b

= a - c

= -13 - 25

= -38

\(A=\left(\dfrac{1}{a^2+b^2}+\dfrac{1}{2ab}\right)+\left(ab+\dfrac{16}{ab}\right)+\dfrac{17}{2ab}\)

\(A\ge\dfrac{4}{a^2+b^2+2ab}+2\sqrt{\dfrac{16ab}{ab}}+\dfrac{17}{\dfrac{2\left(a+b\right)^2}{4}}\)

\(A\ge\dfrac{4}{\left(a+b\right)^2}+8+\dfrac{34}{\left(a+b\right)^2}\ge\dfrac{4}{4^2}+8+\dfrac{34}{4^2}=\dfrac{83}{8}\)

Dấu "=" xảy ra khi \(a=b=2\)

Ta có bất đẳng thức phụ sau (bđt Mincopski)

\(\sqrt{x^2+y^2}+\sqrt{z^2+t^2}\ge\sqrt{\left(x+z\right)^2+\left(y+t\right)^2}\left(x;y;z;t\inℝ\right)\)

Thật vậy :

 \(bđt\Leftrightarrow x^2+y^2+2\sqrt{\left(x^2+y^2\right)\left(z^2+t^2\right)}+z^2+t^2\ge x^2+2xz+z^2+y^2+2yt+t^2\)

\(\Leftrightarrow\sqrt{x^2z^2+x^2t^2+y^2z^2+y^2t^2}\ge xz+yt\)

*Nếu xz + yt < 0 thì bđt hiển nhiên đúng

*Nếu xz + yt > 0 thì bđt trở thành 

\(x^2z^2+x^2t^2+y^2z^2+y^2t^2\ge x^2z^2+2xyzt+y^2t^2\)

\(\Leftrightarrow x^2t^2-2xyzt+y^2z^2\ge0\)

\(\Leftrightarrow\left(xt-yz\right)^2\ge0\)(ĐÚng)

Vậy bđt được chứng minh

Áp dụng bđt trên 2 lần ta được

\(P\ge\sqrt{\left(5+5\right)^2+\left(a^2+b^2\right)^2}+\sqrt{25+c^4}\)

   \(\ge\sqrt{\left(5+5+5\right)^2+\left(a^2+b^2+c^2\right)^2}\)

   \(=\sqrt{225+\left(a^2+b^2+c^2\right)^2}\)

Bài toán quay về tìm \(min\left(a^2+b^2+c^2\right)\)biết \(2\left(a+b+c\right)+ab+bc+ca=18\)

Ta có bđt phụ sau \(a^2+b^2+c^2\ge\frac{\left(a+b+c\right)^2}{3}\)(Tự chứng minh bằng biến đổi tương đương nhé)

        \(\Rightarrow a+b+c\le\sqrt{3\left(a^2+b^2+c^2\right)}\)

Đặt \(3\left(a^2+b^2+c^2\right)=t\left(t\ge0\right)\)

\(\Rightarrow a+b+c\le\sqrt{3t}\)

Lại có bđt phụ sau \(ab+bc+ca\le a^2+b^2+c^2=\frac{t}{3}\)

Tóm lại ta thu được 2 bđt sau \(\hept{\begin{cases}a+b+c\le\sqrt{3t}\\ab+bc+ca\le\frac{t}{3}\end{cases}}\)

Ta có \(18=2\left(a+b+c\right)+ab+bc+ca\le2\sqrt{3t}+\frac{t}{3}\)

\(\Leftrightarrow\frac{t}{3}+2\sqrt{3t}-18\ge0\)

\(\Leftrightarrow t+6\sqrt{3t}-54\ge0\)

\(\Leftrightarrow\orbr{\begin{cases}\sqrt{t}\le-9-3\sqrt{3}\left(Loa_.i\cdot do\cdot\sqrt{t}\ge0\right)\\\sqrt{t}\ge9-3\sqrt{3}\left(Tm\right)\end{cases}}\)

Có \(\sqrt{t}\ge9-3\sqrt{3}\)

\(\Leftrightarrow\sqrt{3\left(a^2+b^2+c^2\right)}\ge9-3\sqrt{3}\)

\(\Leftrightarrow3\left(a^2+b^2+c^2\right)\ge108-54\sqrt{3}\)

\(\Leftrightarrow a^2+b^2+c^2\ge36-18\sqrt{3}\)

Quay trở lại bài toán \(P\ge\sqrt{225+\left(a^2+b^2+c^2\right)^2}\ge\sqrt{225+\left(36-18\sqrt{3}\right)^2}\)

Dấu "=" xảy ra tại a = b = c

P/S: sai đâu thì thôi nha :v a lười ktra lại lắm

Vì ƯCLN(a,b)=25

nên ta đặt:a=25.a'

b=25.b'

với (a',b')=1 ta có 25a'.25b=3750=>625a'b'=3750

=>a'b'=6

mà (a',b')=1 ta có bảng sau :

Vậy (a,b)=(25;150);(50;75);(75;50);(150;25).

toi ko biet

ab=75

kho qua di