\(\dfrac{7}{P}\) chỉ có GTLN chứ ko có GTNN

Với \(x>0;x\ne1\) thì biểu thức này ko tồn tại cả GTNN lẫn GTLN

\(P=\dfrac{\sqrt{x}-2}{\sqrt{x}}=1-\dfrac{2}{\sqrt{x}}\)

Vì \(x\le3\Rightarrow\dfrac{2}{\sqrt{x}}\ge\dfrac{2}{\sqrt{3}}\)\(\Leftrightarrow-\dfrac{2}{\sqrt{x}}\le-\dfrac{2}{\sqrt{3}}\)\(\Leftrightarrow1-\dfrac{2}{\sqrt{3}}\le1-\dfrac{2}{\sqrt{3}}\)

\(\Rightarrow\)\(P\le\dfrac{3-2\sqrt{3}}{3}\)

Dấu = xra khi x=3

Vậy \(P_{max}=\dfrac{3-2\sqrt{3}}{3}\)

\(T=\sqrt{\dfrac{3\sqrt{x}}{\sqrt{x}-6}\cdot\dfrac{x-6\sqrt{x}}{\sqrt{x}-1}}=\sqrt{\dfrac{3\sqrt{x}}{\sqrt{x}-6}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-6\right)}{\sqrt{x}-1}}\\ =\sqrt{\dfrac{3\sqrt{x}\cdot\sqrt{x}}{\sqrt{x}-1}}=\sqrt{\dfrac{3x}{\sqrt{x}-1}}\\ =\sqrt{\dfrac{3\left(x-1\right)+3}{\sqrt{x}-1}}=\sqrt{3\left(\sqrt{x}+1\right)+\dfrac{3}{\sqrt{x}-1}}\\ =\sqrt{3\left(\sqrt{x}-1+\dfrac{1}{\sqrt{x}-1}\right)+6}\)

Áp dụng bất đẳng thức Cosi ta có:

\(\sqrt{x}-1+\dfrac{1}{\sqrt{x}-1}\ge2\)

\(\Rightarrow T\ge\sqrt{3\cdot2+6}=2\sqrt{3}\)

Dấu = xảy ra khi x=4

Bài 1: 

Ta có: \(D=\sqrt{16x^4}-2x^2+1\)

\(=4x^2-2x^2+1\)

\(=2x^2+1\)

\(P=\frac{x^2+\sqrt{x}}{x-\sqrt{x}+1}+1-\frac{2x+\sqrt{x}}{\sqrt{x}}\left(đkxđ\Leftrightarrow x\ge0\right).\)

\(=\frac{\sqrt{x}\left(\sqrt{x}^3+1\right)}{x-\sqrt{x}+1}+1-\frac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}\)

\(=\frac{\sqrt{x}\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{x-\sqrt{x}+1}-\left(2\sqrt{x}+1\right)\)

\(=x+\sqrt{x}-2\sqrt{x}-1=x-\sqrt{x}-1\)

\(P=x-\sqrt{x}-1=\sqrt{x}^2-2.\sqrt{x}.\frac{1}{2}+\frac{1}{4}-\frac{1}{4}-1\)

\(=\left(\sqrt{x}-\frac{1}{2}\right)^2-\frac{5}{4}\)

\(\Rightarrow P_{min}=-\frac{5}{4}\Leftrightarrow\left(\sqrt{x}-\frac{1}{2}\right)^2=0\)

\(\Rightarrow\sqrt{x}=\frac{1}{2}\Rightarrow x=\frac{1}{4}\)

a) ĐKXĐ: \(x\ne9\)

\(P=\frac{x\sqrt{x}+5\sqrt{x}-12-2\left(\sqrt{x}-3\right)^2-\left(\sqrt{x}+3\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}\)

\(P=\frac{x\sqrt{x}+5\sqrt{x}-12-2x+12\sqrt{x}-18-x-5\sqrt{x}-6}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}\)

\(P=\frac{x\sqrt{x}-3x+12\sqrt{x}-36}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}\)

\(P=\frac{\left(\sqrt{x}-3\right)\left(x+12\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}\)

\(P=\frac{x+12}{\sqrt{x}+2}\)

b) Ta có: \(P=\frac{x+12}{\sqrt{x}+2}=\frac{x-4+16}{\sqrt{x}+2}=\sqrt{x}-2+\frac{16}{\sqrt{x}+2}\)

\(=\left(\sqrt{x}+2\right)+\frac{16}{\sqrt{x}+2}-4\)

\(\ge2\sqrt{\left(\sqrt{x}+2\right).\frac{16}{\sqrt{x}+2}}-4=4\)

P = 4 thì \(\left(\sqrt{x}+2\right)^2=16\Rightarrow\sqrt{x}=2\Rightarrow x=4\)

Vậy GTNN của P là 4 khi x = 4.

\(A=\left(\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{2}{x-\sqrt{x}}\right):\frac{1}{\sqrt{x}-1}\)ĐKXĐ : \(x>1\)

\(A=\left(\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{2}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\frac{1}{\sqrt{x}-1}\)

\(A=\frac{x+2}{\sqrt{x}\left(\sqrt{x}-1\right)}.\left(\sqrt{x}-1\right)\)

\(A=\frac{x+2}{\sqrt{x}}\)

\(a)\)\(A=\left(\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{2}{x-\sqrt{x}}\right):\frac{1}{\sqrt{x}-1}\)

\(A=\left(\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{2}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\frac{1}{\sqrt{x}-1}\)

\(A=\left(\frac{\sqrt{x}}{\sqrt{x}-1}:\frac{1}{\sqrt{x}-1}\right)+\left(\frac{2}{\sqrt{x}\left(\sqrt{x}-1\right)}:\frac{1}{\sqrt{x}-1}\right)\)

\(A=\sqrt{x}+\frac{2}{\sqrt{x}}\)

\(b)\) Áp dụng Cosi với hai số dương ta có : 

\(A=\sqrt{x}+\frac{2}{\sqrt{x}}\ge2\sqrt{\sqrt{x}.\frac{2}{\sqrt{x}}}=2\sqrt{2}\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(\sqrt{x}=\frac{2}{\sqrt{x}}\)

\(\Leftrightarrow\)\(x=2\)

Vậy GTNN của \(A\) là \(2\sqrt{2}\) khi \(x=2\)

Chúc bạn học tốt ~ 

PS : mới lớp 8 ko chắc nhé :v 

Theo câu a) thì  \(A=\frac{x+2}{\sqrt{x}}\)

Áp dụng BĐT AM-GM cho 2 số không âm\(x+2\ge2\sqrt{2x}\)

\(\Rightarrow A\ge\frac{2\sqrt{2x}}{\sqrt{x}}=2\sqrt{2}\). Vậy Min \(A=2\sqrt{2}\). Dấu "=" xảy ra <=> \(x=2\).