Ta có: \(A=\left(\frac{x\sqrt{x}-1}{x-\sqrt{x}}-\frac{x\sqrt{x}+1}{x+\sqrt{x}}\right):\left(1-\frac{3-\sqrt{x}}{\sqrt{x}+1}\right)\)     (   ĐKXĐ: \(x>0,\)\(x\ne0,\)\(x\ne1\))

    \(\Leftrightarrow A=\left(\frac{\left(\sqrt{x}-1\right).\left(x+\sqrt{x}+1\right)}{\sqrt{x}.\left(\sqrt{x}-1\right)}-\frac{\left(\sqrt{x}+1\right).\left(x-\sqrt{x}+1\right)}{\sqrt{x}.\left(\sqrt{x}+1\right)}\right):\left(\frac{\sqrt{x}+1-3+\sqrt{x}}{\sqrt{x}+1}\right)\)

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

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

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

    \(\Leftrightarrow A=\frac{\sqrt{x}+1}{\sqrt{x}-1}\)

Để \(A\ge\frac{3}{2}\)\(\Rightarrow\)\(\frac{\sqrt{x}+1}{\sqrt{x}-1}\ge\frac{3}{2}\)

Ta có: \(\frac{\sqrt{x}+1}{\sqrt{x}-1}\ge\frac{3}{2}\)

    \(\Leftrightarrow\frac{\sqrt{x}+1}{\sqrt{x}-1}-\frac{3}{2}\ge0\)

    \(\Leftrightarrow\frac{2\sqrt{x}+2-3\sqrt{x}+3}{2.\left(\sqrt{x}-1\right)}\ge0\)

    \(\Leftrightarrow\frac{5-\sqrt{x}}{2.\left(\sqrt{x}-1\right)}\ge0\)

+ TH₁: \(\hept{\begin{cases}5-\sqrt{x}\ge0\\2\sqrt{x}-2\ge0\end{cases}}\)\(\Leftrightarrow\)\(\hept{\begin{cases}\sqrt{x}\le5\\\sqrt{x}\ge1\end{cases}}\)\(\Leftrightarrow\)\(\hept{\begin{cases}x\le25\\x\ge1\end{cases}}\)\(\Rightarrow\)\(1\le x\le25\)\(\left(TM\right)\)

+ TH₂: \(\hept{\begin{cases}5-\sqrt{x}\le0\\2\sqrt{x}-2\le0\end{cases}}\)\(\Leftrightarrow\)\(\hept{\begin{cases}\sqrt{x}\ge5\\\sqrt{x}\le1\end{cases}}\)\(\Leftrightarrow\)\(\hept{\begin{cases}x\ge25\\x\le1\end{cases}}\)\(\left(L\right)\)

            \(\Rightarrow\)\(1\le x\le25.\)Kết hợp ĐKXĐ: \(x\ne1\)

                         \(\Rightarrow\)\(1< x\le25\)

Vậy để \(A\ge\frac{3}{2}\)\(\Leftrightarrow\)\(1< x\le25\)

\(\left(\sqrt{b}-\sqrt{c}\right)^2\ge0\Leftrightarrow b-2\sqrt{bc}+c\ge0\Leftrightarrow b+c\ge2\sqrt{bc}\) dấu "="xảy ra khi b=c

\(\left(a+2b\right)\left(a+2c\right)=a^2+2a\left(b+c\right)+4bc\ge a^2+4a\sqrt{bc}+4bc=\left(a+2\sqrt{bc}\right)^2\)

\(\Rightarrow\sqrt{\left(a+2b\right)\left(a+2c\right)}\ge a+2\sqrt{bc}\)

tương tự ta có \(\hept{\begin{cases}\sqrt{\left(b+2c\right)\left(b+2c\right)}\ge b+2\sqrt{bc}\\\sqrt{\left(c+2a\right)\left(a+2b\right)}\ge c+2\sqrt{ab}\end{cases}}\)

dấu "=" xảy ra khi a=b=c

\(\Rightarrow A=\sqrt{\left(a+2b\right)\left(a+2c\right)}+\sqrt{\left(b+2a\right)\left(b+2c\right)}+\sqrt{\left(c+2a\right)\left(c+2b\right)}\)\(\ge a+b+c+2\sqrt{ab}+2\sqrt{bc}+2\sqrt{ac}\)

hay \(A\ge\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2=\left(\sqrt{3}\right)^2=3\)

dấu "="xảy ra khi \(\hept{\begin{cases}a=b=c\\\sqrt{a}+\sqrt{b}+\sqrt{c}=3\end{cases}\Leftrightarrow a=b=c=\frac{\sqrt{3}}{3}}\)

\(M=\left(2\sqrt{a}+3\sqrt{b}-4\sqrt{c}\right)^2=\left(2\sqrt{a}+3\sqrt{a}-4\sqrt{a}\right)^2=\left(\sqrt{a}\right)^2=\frac{\sqrt{3}}{3}\)

bổ sung thêm điều kiện x,y là số thực

với x>=1; y>=1 từ giả thiết ta có \(x\sqrt{x}-y\sqrt{y}=\sqrt{y-1}-\sqrt{x-1}\left(1\right)\)

nếu x=y=1 thì S=6 (*)

nếu x,y không đồng thời bằng 1 thì \(\sqrt{y-1}+\sqrt{x-1}>0\)vì vậy

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

\(\Leftrightarrow\left(\sqrt{x}-\sqrt{y}\right)\left(x+\sqrt{xy}+\frac{\sqrt{x}+\sqrt{y}}{\sqrt{x+1}+\sqrt{y+1}}\right)=0\left(2\right)\)

vì x>=1; y>=1 nên từ (2) => x=y

vì vậy S=2x²-8x+12=2(x-2)²+4>=4 (**) với mọi x

dấu "=" xảy ra khi x=2

vậy minS=4 <=> x=y=2

Số 1 bỏ ik nha.

Giá trị của x thì A có nghĩa là : \(x\ne0\)

dễ mà bạn

\(ĐKXĐ:\hept{\begin{cases}x\ge0\\x\ne1\end{cases}}\)

\(P=\left(\frac{3}{x-1}+\frac{1}{\sqrt{x}+1}\right):\frac{1}{\sqrt{x}-1}\)

\(P=\frac{3+\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\left(\sqrt{x}-1\right)\)

\(P=\frac{\sqrt{x}+2}{\sqrt{x}+1}\)