Ta có:
(1 + b/a)(1 + c/b)(1 + a/c) = 8
<=> (a + b)/a.(b + c)/b.(c + a)/c = 8
<=> (a + b)(b + c)(c + a) = 8abc
Áp dụng bất đẳng thức Cauchy cho các số dương a, b, c ta được:
a + b ≥ 2√(ab)
b + c ≥ 2√(bc)
c + a ≥ 2√(ca)
=> (a + b)(b + c)(c + a) ≥ 8√(a^2.b^2.c^2) = 8|abc| = 8abc (vì a, b,c > 0)
Dấu "=" xảy ra <=> a = b; b = c; c = a <=> a = b = c <=> ΔABC đều

https://olm.vn/hoi-dap/detail/2293581520.html cậu tham khảo nhé !

\(\frac{3x}{x-2}-\frac{x}{x-5}+\frac{9x}{x^2-7x+10}=10\)

\(\Rightarrow\frac{3x^2-15x-x^2+2x+9x}{x^2-7x+10}=10\)

\(\Rightarrow\frac{2x^2-4x}{x^2-7x+10}=10\)

\(\Rightarrow2x^2-4x=10x^2-70x+100\)

\(\Rightarrow8x^2-66x+100=0\)

Ta có \(\Delta=66^2-4.8.100=1156,\sqrt{\Delta}=34\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{66+34}{16}=\frac{25}{4}\\x=\frac{66-34}{16}=2\end{cases}}\)

a) \(\frac{3x}{x-2}-\frac{x}{x-5}+\frac{9x}{x^2-7x+10}=10\)

<=> \(\frac{3x\left(x-5\right)}{\left(x-2\right)\left(x-5\right)}-\frac{x\left(x-2\right)}{\left(x-5\right)\left(x-2\right)}+\frac{9x}{\left(x-2\right)\left(x-5\right)}=10\)

<=> \(\frac{3x^2-15x-x^2+2x+9x}{\left(x-5\right)\left(x-2\right)}=10\)

<=> \(\frac{2x^2-4x}{\left(x-5\right)\left(x-2\right)}=10\)

<=> \(\frac{2x\left(x-2\right)}{\left(x-5\right)\left(x-2\right)}=10\)

<=> \(2x=10\left(x-5\right)\)

<=> 2x - 10x = -50

<=> -8x = -50

<=>x = 6,25

Vậy S = {6,25}

b) (x - 7)(x - 2)(x - 4)(x - 5) = 72

<=> (x² - 9x + 14)(x² - 9x + 20) = 72

Đặt x² - 9x + 14 = t <=> t(t + 6) = 72

<=> t² + 6t - 72 = 0

<=> t² + 12t - 6t - 72 = 0

<=> (t + 12)(t - 6) = 0

<=> \(\orbr{\begin{cases}t+12=0\\t-6=0\end{cases}}\)

<=> \(\orbr{\begin{cases}x^2-9x+14+12=0\\x^2-9x+14-6=0\end{cases}}\)

<=> \(\orbr{\begin{cases}\left(x-9x+20,25\right)+5,75=0\\x^2-9x+8=0\end{cases}}\)

<=> \(\orbr{\begin{cases}\left(x-4,5\right)^2+5,75=0\left(vn\right)\\x^2-x-8x+8=0\end{cases}}\)

<=> (x - 1)(x - 8) = 0

<=> \(\orbr{\begin{cases}x-1=0\\x-8=0\end{cases}}\)

<=> \(\orbr{\begin{cases}x=1\\x=8\end{cases}}\)

Vậy S = {1; 8}

\(\frac{3x-7}{5}=\frac{2x-1}{3}\)

\(\Leftrightarrow9x-21=10x-5\)

\(\Leftrightarrow-x=16\Leftrightarrow x=-16\)

\(\frac{4x-7}{12}-x=\frac{3x}{8}\)

\(\Leftrightarrow\frac{4x-7-12x}{12}=\frac{3x}{8}\)

\(\Leftrightarrow\frac{-7-8x}{12}=\frac{3x}{8}\)

\(\Leftrightarrow-56-64x=36x\)

\(\Leftrightarrow-56=100x\Leftrightarrow x=\frac{-14}{25}\)

\(\frac{x-2009}{1234}+\frac{x-2009}{5678}-\frac{x-2009}{197}=0\)

\(\Leftrightarrow\left(x-2019\right)\left(\frac{1}{1234}+\frac{1}{5678}-\frac{1}{197}\right)=0\)

Vì \(\left(\frac{1}{1234}+\frac{1}{5678}-\frac{1}{197}\right)\ne0\)nên x - 2019 = 0

Vậy x = 2019

\(\frac{5x-8}{3}=\frac{1-3x}{2}\)

\(\Leftrightarrow10x-16=3-9x\)

\(\Leftrightarrow19x=19\Leftrightarrow x=1\)

\(\left(2.x+1\right)^2=\left(3.x-5\right)^2\)

\(\Leftrightarrow4.x+4.x+1=9.x^2-30.x+25\)

\(\Leftrightarrow4.x+4.x+1-9.x^2+30.x-25=0\)

\(\Leftrightarrow-5.x^2+34-24=0\)

\(\Leftrightarrow5.x^2-34.x+24=0\)

\(\Leftrightarrow5.x^2-4.x-30.x+24=0\)

\(\Leftrightarrow x.\left(5.x-4\right)-6.\left(5.x-4\right)=0\)

\(\Leftrightarrow\left(5.x-4\right).\left(x-6\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}5.x-4=0\\x-6=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{4}{5}\\x=6\end{cases}}}\)

Vậy \(x=\left\{\frac{4}{5};6\right\}\)

\(\left(2x+1\right)^2=\left(3x-5\right)^2\)

\(\Rightarrow2x+1=3x-5\)

\(\Rightarrow1+5=3x-2x\)

\(\Rightarrow6=x\)

Vậy x=6

Chúc bn học tốt

Giúp mk vs mk cảm ơn 

\(A=2x^2+3x\)

\(=\left(\sqrt{2}x\right)^2+2.\sqrt{2}x.\frac{3}{2\sqrt{2}}+\frac{9}{8}-\frac{9}{8}\)

\(=\left(\sqrt{2}x+\frac{3}{2\sqrt{2}}\right)^2-\frac{9}{8}\ge-\frac{9}{8}\)

Dấu "=" khi \(x=\frac{-3}{4}\)

\(B=-x^2-3x+2=-\left(x^2+3x+\frac{9}{4}\right)+\frac{17}{4}\)

\(=-\left(x+\frac{3}{2}\right)^2+\frac{17}{4}\le\frac{17}{4}\)

Dấu "=" khi \(x=-\frac{3}{2}\)

\(\left(x-4\right)\left(3x+2\right)=\left(5x-3\right)\left(x-4\right)\)

\(\Leftrightarrow\left(x-4\right)\left(3x+2\right)-\left(5x-3\right)\left(x-4\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(-2x+5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-4=0\\-2x+5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=4\\x=\frac{5}{2}\end{cases}}\)

(2x+1)(3x-2)=(5x-8)(2x+1)

<=> (2x+1)(3x-2)-(5x-8)(2x+1)=0

<=>(2x+1)(3x-2-5x+8)=0

<=>(2x+1)(6-2x)=0

<=>2x+1=0 hoặc 6-2x=0

<=>x=-1/2 hoặc x=3

Vậy phương trình có tập nghiệm S={-1/2;3}

       (2x+1)(3x-2)=(5x-8)(2x+1)

<=> (2x+1)(3x-2)-(5x-8)(2x+1)=0

<=>(2x+1)(3x-2-5x+8)=0

<=>(2x+1)(6-2x)=0

<=>2x+1=0 hoặc 6-2x=0

<=>x=\(-\frac{1}{2}\) hoặc x=3

Vậy phương trình có tập nghiệm \(S=\left\{\frac{-1}{2};3\right\}\)