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.

Nguyễn Việt Lâm giúp mk nhá, thanks bn nhìu :>>>

Nguyễn Việt Lâm DƯƠNG PHAN KHÁNH DƯƠNG Mysterious Person help

Do \(a,b,c>\dfrac{25}{4}\Rightarrow\) các mẫu số đều dương

Áp dụng BĐT Cauchy:

\(M\ge3\sqrt[3]{\dfrac{abc}{\left(2\sqrt{b}-5\right)\left(2\sqrt{c}-5\right)\left(2\sqrt{a}-5\right)}}\)

\(\Rightarrow M\ge3\sqrt[3]{\dfrac{5^3.abc}{5\left(2\sqrt{b}-5\right).5\left(2\sqrt{c}-5\right).5\left(2\sqrt{a}-5\right)}}\)

Ta có: \(\left\{{}\begin{matrix}5\left(2\sqrt{a}-5\right)\le\dfrac{\left(5+2\sqrt{a}-5\right)^2}{4}=a\\5\left(2\sqrt{b}-5\right)\le\dfrac{\left(5+2\sqrt{b}-5\right)^2}{4}=b\\5\left(2\sqrt{c}-5\right)\le\dfrac{\left(5+2\sqrt{c}-5\right)^2}{4}=c\end{matrix}\right.\)

\(\Rightarrow M\ge3\sqrt[3]{\dfrac{5^3.abc}{abc}}=3.5=15\)

\(\Rightarrow M_{min}=15\) khi \(a=b=c=25\)

Bạn tham khảo:

Cho \(a,b,c>\dfrac{25}{4}.\)Tìm GTNN của \(Q=\dfrac{a}{2\sqrt{b}-5}+\dfrac{b}{2\sqrt{c}-5}+\dfrac{c}{2\sqrt{a}-5}\) - Hoc24

`A)đk:x>=0,x ne 25`

`A=9=>A=(3+2)/(3-5)=-5/2`

`B)B=(3sqrtx-15+20-2sqrtx)/(x-25)`

`=(sqrtx+5)/(x-25)`

`=1/(sqrtx-5)`

`A=B.|x-4|`

`<=>A/B=|x-4|`

`<=>\sqrtx+2=|x-4|`

`<=>\sqrtx+2=(sqrtx+2)|sqrtx-2|`

`<=>|sqrtx-2|=1`

`+)sqrtx-2=1<=>x=9(tm)`

`+)sqrtx-2=-1<=>x=1(tm)`

Vậy `S={1,9}`

a, Thay x=9 vào biểu thức A ta có

\(A=\dfrac{\sqrt{9}+2}{\sqrt{9}-5}\)

\(A=\dfrac{3+2}{3-5}=\dfrac{5}{-2}=-2,5\)

Vậy A =-2,5 khi x=9

a. A=\(\dfrac{\sqrt{x}+2}{\sqrt{x}-5}\)

=\(\dfrac{\sqrt{9}+2}{\sqrt{9}-5}=\dfrac{-5}{2}\)

\(a^2+2ab+b^2=\left(a+b\right)^2\ge0\forall a,b\)

\(a^2-2ab+b^2=\left(a-b\right)^2\ge0\forall a,b\)

\(A^{2n}\ge0\forall A\)

\(-A^{2n}\le0\forall A\)

\(\left|A\right|\ge0\forall A\)

\(-\left|A\right|\le0\forall A\)

\(\left|A\right|+\left|B\right|\ge\left|A+B\right|\)

\(\left|A\right|-\left|B\right|\le\left|A-B\right|\)

Mk cho bn cong thức nè

\(=\left(1^2+4^2\right)\left(a^2+\dfrac{1}{b^2}\right)\ge\left(1a+4.\dfrac{1}{b}\right)^2\\ \Rightarrow\sqrt{a^2+\dfrac{1}{vb^2}}\ge\dfrac{1}{\sqrt{17}}\left(a+\dfrac{4}{b}\right)\) 

Tương tự

\(\sqrt{b^2+\dfrac{1}{c^2}}\ge\dfrac{1}{\sqrt{17}}\left(b+\dfrac{4}{c}\right)\\ \sqrt{c^2+\dfrac{1}{a^2}}\ge\dfrac{1}{\sqrt{17}}\left(c+\dfrac{4}{a}\right)\\ Do.đó:\\ Q\ge\dfrac{1}{\sqrt{17}}\left(a+b+c+\dfrac{4}{a}+\dfrac{4}{b}+\dfrac{4}{c}\right)\ge\dfrac{1}{\sqrt{17}}\\ \left(a+b+c+\dfrac{36}{a+b+c}\right)\) 

\(=\dfrac{1}{\sqrt{17}}\\ \left[a+b+c+\dfrac{9}{4\left(a+b+c\right)}+\dfrac{135}{4\left(a+b+c\right)}\right]\\ \ge\dfrac{3\sqrt{17}}{2}\)