\(M=\dfrac{x+6\sqrt{x}+9+25}{\sqrt{x}+3}=\dfrac{\left(\sqrt{x}+3\right)^2+25}{\sqrt{x}+3}=\sqrt{x}+3+\dfrac{25}{\sqrt{x}+3}\)Áp dụng Cô si có

\(M\ge2\sqrt{\left(\sqrt{x}+3\right).\dfrac{25}{\sqrt{x}+3}}=10\)

Dấu "=" \(\sqrt{x}+3=\dfrac{25}{\sqrt{x}+3}\leftrightarrow x=4\)

Vậy GTNN của M = 10 <=> x = 4

\(M=\dfrac{\left(x+6\sqrt{x}+9\right)+25}{\sqrt{x}+3}=\dfrac{\left(\sqrt{x}+3\right)^2+25}{\sqrt{x}+3}=\sqrt{x}+3+\dfrac{25}{\sqrt{x}+3}\)

Do \(\sqrt{x}\ge0\Rightarrow\left\{{}\begin{matrix}\sqrt{x}+3>0\\\dfrac{25}{\sqrt{x}+3}>0\end{matrix}\right.\)

Áp dụng bđt cô-si ta có: 

\(\sqrt{x}+3+\dfrac{25}{\sqrt{x}+3}\ge2\sqrt{\left(\sqrt{x}+3\right)\cdot\dfrac{25}{\sqrt{x}+3}}=2\sqrt{25}=10\)

hay \(M\ge10\)

Dấu "=" xảy ra \(\Leftrightarrow\sqrt{x}+3=\dfrac{25}{\sqrt{x}+3}\Leftrightarrow x=4\)

Vậy GTNN của M = 10 khi x = 4

\(\dfrac{x+6\sqrt{x}+9+25}{\sqrt{x+3}}\)

=\(\dfrac{\sqrt{x}+2.3.\sqrt{x}+3^2+25}{\sqrt{x}+3}\)

=\(\dfrac{\left(\sqrt{x}+3\right)^2+25}{\sqrt{x}+3}\)

=\(\sqrt{x}+3+\dfrac{25}{\sqrt{x}+3}\)

áp dụng cosi

M≥\(^2\sqrt{\left(\sqrt{x}+3\right).\dfrac{25}{\sqrt{x}+3}}\)=10

\(\sqrt{x}+3\)=\(\dfrac{25}{\sqrt{x}+3}\)⇔x=4

vậy...

khó vậy

bai nay mk thay rat kho vi mk ko thay co 1 quy luat nao ca

\(B=\dfrac{x+6\sqrt{x}+34}{\sqrt{x}+3}=\dfrac{x+6\sqrt{x}+9+25}{\sqrt{x}+3}=\dfrac{\left(\sqrt{x}+3\right)^2}{\sqrt{x}+3}+\dfrac{25}{\sqrt{x}+3}=\sqrt{x}+3+\dfrac{25}{\sqrt{x}+3}\) Áp dụng BĐT Cauchy cho các số dương , ta có :

\(\sqrt{x}+3+\dfrac{25}{\sqrt{x}+3}\) ≥ \(2\sqrt{\left(\sqrt{x}+3\right).\dfrac{25}{\sqrt{x}+3}}=2.5=10\)

⇒ \(B_{MIN}=10."="\) ⇔ \(x=4\)

@Akai Haruma

\(M=\dfrac{x+6\sqrt{x}+9+25}{\sqrt{x}+3}=\sqrt{x}+3+\dfrac{25}{\sqrt{x}+3}>=2\cdot\sqrt{25}=10\)

Dấu = xảy ra khi x=4

1.(√x -2)^2 ≥ 0 --> x -4√x +4 ≥ 0 --> x+16 ≥ 12 +4√x --> (x+16)/(3+√x) ≥4 
--> Pmin=4 khi x=4

2. Đặt \(\sqrt{x^2-4x+5}=t\ge1\)1

=> M=2x²-8x+\(\sqrt{x^2-4x+5}\)+6=2(t²-5)+t+6

<=> M=2t²+t-4\(\ge\)2.1²+1-4=-1

M_min=-1 khi t=1 hay x=2

b/ Ko biết yêu cầu

4/ \(E=\frac{x^2}{3}+\frac{x^2}{3}+\frac{x^2}{3}+\frac{1}{x^3}+\frac{1}{x^3}\ge5\sqrt[5]{\frac{x^6}{27x^6}}=\frac{5}{\sqrt[5]{27}}\)

Dấu "=" xảy ra khi \(\frac{x^2}{3}=\frac{1}{x^3}\Leftrightarrow x=\sqrt[5]{3}\)

\(F=x+\frac{1}{x^2}=\frac{x}{2}+\frac{x}{2}+\frac{1}{x^2}\ge3\sqrt[3]{\frac{x^2}{4x^2}}=\frac{3}{\sqrt[3]{4}}\)

Dấu "=" xảy ra khi \(\frac{x}{2}=\frac{1}{x^2}\Rightarrow x=\sqrt[3]{2}\)

6/ \(Q=\frac{\left(x+1\right)^2+16}{2\left(x+1\right)}=\frac{x+1}{2}+\frac{8}{x+1}\ge2\sqrt{\frac{8\left(x+1\right)}{2\left(x+1\right)}}=4\)

Dấu "=" xảy ra khi \(\frac{x+1}{2}=\frac{8}{x+1}\Leftrightarrow x=3\)

7/

\(R=\frac{\left(\sqrt{x}+3\right)^2+25}{\sqrt{x}+3}=\sqrt{x}+3+\frac{25}{\sqrt{x}+3}\ge2\sqrt{\frac{25\left(\sqrt{x}+3\right)}{\sqrt{x}+3}}=10\)

Dấu "=" xảy ra khi \(\sqrt{x}+3=\frac{25}{\sqrt{x}+3}\Leftrightarrow x=4\)

8/

\(S=x^2+\frac{2000}{x}=x^2+\frac{1000}{x}+\frac{1000}{x}\ge3\sqrt[3]{\frac{1000^2x^2}{x^2}}=300\)

Dấu "=" xảy ra khi \(x^2=\frac{1000}{x}\Leftrightarrow x=10\)