Ta có: \(\dfrac{a}{3}=\dfrac{b}{4}\)

          \(\dfrac{b}{2}=\dfrac{c}{5}\Rightarrow\dfrac{b}{4}=\dfrac{c}{10}\)

\(\Rightarrow\dfrac{a}{3}=\dfrac{b}{4}=\dfrac{c}{10}\)

Áp dụng tính chất dãy tỉ số bằng nhau:

\(\dfrac{a}{3}=\dfrac{b}{4}=\dfrac{c}{10}=\dfrac{a-c+b}{3-10+4}=\dfrac{3}{-3}=-1\)

\(\Rightarrow\left\{{}\begin{matrix}a=\left(-1\right).3=-3\\b=\left(-1\right).4=-4\\c=\left(-1\right).10=-10\end{matrix}\right.\)

hay quá

\(\dfrac{b}{2}=\dfrac{c}{5}\Rightarrow\dfrac{b}{4}=\dfrac{c}{10}\)

\(\Rightarrow\dfrac{a}{3}=\dfrac{b}{4}=\dfrac{c}{10}=\dfrac{a-c+b}{3-10+4}=\dfrac{3}{-3}=-1\)

\(\Rightarrow\left\{{}\begin{matrix}a=\left(-1\right).3=-3\\b=\left(-1\right).4=-4\\c=\left(-1\right).10=-10\end{matrix}\right.\)

b) Ta có : \(\dfrac{2a}{3}=\dfrac{3b}{4}=\dfrac{4c}{5}\)

\(\Leftrightarrow\dfrac{a}{\dfrac{3}{2}}=\dfrac{b}{\dfrac{4}{3}}=\dfrac{c}{\dfrac{5}{4}}=\dfrac{a+b+c}{\dfrac{3}{2}+\dfrac{4}{3}+\dfrac{5}{4}}=\dfrac{49}{\dfrac{49}{12}}=12\)

Khi đó \(a=12.\dfrac{3}{2}=18;b=12.\dfrac{4}{3}=16;c=12.\dfrac{5}{4}=15\)

Vậy (a,b,c) = (18,16,15) 

Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{a}{8}=\dfrac{b}{12}=\dfrac{c}{15}=\dfrac{a-b-c}{8-12-15}=\dfrac{28}{-19}=\dfrac{-28}{19}\)

Do đó: \(\left\{{}\begin{matrix}a=\dfrac{-224}{19}\\b=\dfrac{-336}{19}\\c=\dfrac{-420}{19}\end{matrix}\right.\)

\(3a=\dfrac{b}{\dfrac{2}{5}}=\dfrac{8c}{3}\Rightarrow\dfrac{6a}{2}=\dfrac{5b}{\dfrac{2}{5}\cdot5}=\dfrac{16c}{6}\\ \Rightarrow\dfrac{6a}{2}=\dfrac{5b}{2}=\dfrac{16c}{6}\)

Áp dụng t/c dtsbn:

\(3a=\dfrac{b}{\dfrac{2}{5}}=\dfrac{8c}{3}=\dfrac{6a}{2}=\dfrac{5b}{2}=\dfrac{16c}{6}=\dfrac{6a+5b+16c}{2+2+6}=\dfrac{10}{10}=1\\ \Rightarrow\left\{{}\begin{matrix}a=\dfrac{1}{3}\\b=\dfrac{2}{5}\\c=\dfrac{3}{8}\end{matrix}\right.\)

\(\dfrac{a}{3}=\dfrac{b}{2};\dfrac{b}{6}=\dfrac{c}{5}\)
\(\Rightarrow\dfrac{a}{9}=\dfrac{b}{6};\dfrac{b}{6}=\dfrac{c}{5}\)

\(\Rightarrow\dfrac{a}{9}=\dfrac{b}{6}=\dfrac{c}{5}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\dfrac{a}{9}=\dfrac{b}{6}=\dfrac{c}{5}=\dfrac{a+b+c}{9+6+5}=\dfrac{-40}{20}=-2\)
\(\Rightarrow\left\{{}\begin{matrix}a=-2\cdot9=-18\\b=-2\cdot6=-12\\c=-2\cdot5=-10\end{matrix}\right.\)

Đặt \(\left(2\sqrt{a}-5;2\sqrt{b}-5;2\sqrt{c}-5\right)=\left(x;y;z\right)\Rightarrow\left\{{}\begin{matrix}x;y;z>0\\a=\left(\dfrac{x+5}{2}\right)^2\\b=\left(\dfrac{y+5}{2}\right)^2\\c=\left(\dfrac{z+5}{2}\right)^2\end{matrix}\right.\)

\(Q=\dfrac{\left(x+5\right)^2}{4y}+\dfrac{\left(y+5\right)^2}{4z}+\dfrac{\left(z+5\right)^2}{4x}\ge\dfrac{\left(x+y+z+15\right)^2}{4\left(x+y+z\right)}\)

\(Q\ge\dfrac{\left(x+y+z\right)^2+30\left(x+y+z\right)+225}{4\left(x+y+z\right)}\)

\(Q\ge\dfrac{x+y+z}{4}+\dfrac{225}{4\left(x+y+z\right)}+\dfrac{15}{2}\ge2\sqrt{\dfrac{225\left(x+y+z\right)}{16\left(x+y+z\right)}}+\dfrac{15}{2}=15\)

Dấu "=" xảy ra khi \(a=b=c=25\)

Áp dụng bđt hoán vị cho hai bộ số đơn điệu ngược chiều \(\left(a,b,c\right);\left(2\sqrt{a}-5,2\sqrt{b}-5,2\sqrt{c}-5\right)\): \(Q\ge\dfrac{a}{2\sqrt{a}-5}+\dfrac{b}{2\sqrt{b}-5}+\dfrac{c}{2\sqrt{c}-5}\).

Mặt khác ta có \(\dfrac{a}{2\sqrt{a}-5}-5=\dfrac{\left(\sqrt{a}-5\right)^2}{2\sqrt{a}-5}\ge0\).

Do đó \(Q\ge5+5+5=15\).

Dấu bằng xảy ra khi a = b = c = 25.

2:

a: Áp dụng tính chất của DTSBN, ta được:

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=\dfrac{x+y+z}{2+3+4}=\dfrac{24}{9}=\dfrac{8}{3}\)

=>x=16/3; y=8; z=32/3

A=3x+2y-6z

=3*16/3+2*8-6*32/3

=16+16-64

=-32

b: Áp dụng tính chất của DTSBN, ta được:

\(\dfrac{x}{5}=\dfrac{y}{6}=\dfrac{z}{7}=\dfrac{x-y+z}{5-6+7}=\dfrac{6\sqrt{2}}{6}=\sqrt{2}\)

=>x=5căn 2; y=6căn 2; y=7căn 2

B=xy-yz

=y(x-z)

=6căn 2(5căn 2-7căn 2)

=-6căn 2*2căn 2

=-24

bài 1 a)áp dụng dãy tỉ số bằng nhau ta có:\(\dfrac{a+b+c}{3+4+5}\)=\(\dfrac{24}{12}\)=2

a=2.3=6 ; b=2.4=8 ;c=2.5=10

M=ab+bc+ac=6.8+8.10+6.10=48+80+60=188

"nhưng bài còn lại làm tương tự"