đkxđ:x>=0

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

vì \(\left(\sqrt{x}-1\right)^2=x-2\sqrt{x}+1>=0\Rightarrow x+1>=2\sqrt{x}\)

\(\Rightarrow\frac{2\sqrt{x}}{x+1}< =\frac{x+1}{x+1}=1\Rightarrow1-\frac{2\sqrt{x}}{x+1}>=1-1=0\)

dấu = xảy ra khi x=1

vậy min A là 0 khi x-=1

\(A^2>=0\Rightarrow A>=0\)nhá

ĐKXĐ: \(\hept{\begin{cases}x\ge0\\\sqrt{x}-2\ne0\end{cases}}\)<=>\(\hept{\begin{cases}x\ge0\\x\ne4\end{cases}}\)

Ta có \(C=\left(x-1\right)-\frac{2x-3\sqrt{x}-2}{\sqrt{x}-2}\)

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

<=>\(C=x-1-\left(2\sqrt{x}+1\right)\)

<=>\(C=x-2\sqrt{x}-2\)

<=>\(C=\left(\sqrt{x}-1\right)^2-3\ge-3\)

Vậy GTNN của C là -3. Dấu "=" xảy ra <=> x=1 (tm ĐKXĐ)

Đặt \(\sqrt[3]{x^2+1}=t\left(t\ge1\right)\)

\(y=f\left(t\right)=t^2-t+1\)

\(minf\left(t\right)=f\left(1\right)=1\)

\(minf\left(t\right)=1\Leftrightarrow t=1\Leftrightarrow\sqrt[3]{x^2+1}=1\Leftrightarrow x=0\)

Đặt \(\hept{\begin{cases}\sqrt{2x+3}=a\left(a>0\right)\\\sqrt{y}=b\left(b\ge0\right)\end{cases}}\)

Thì ta có

\(\frac{b^2}{a^2}=\frac{a+1}{b+1}\)

\(\Leftrightarrow b^3+b^2=a^3+a^2\)

\(\Leftrightarrow\left(b-a\right)\left(b^2+ab+a^2\right)+\left(b-a\right)\left(b+a\right)=0\)

\(\Leftrightarrow\left(b-a\right)\left(b^2+ab+a^2+b+a\right)=0\)

Mà \(\left(b^2+ab+a^2+b+a\right)>0\)

\(\Rightarrow a=b\)

\(\Rightarrow2x+3=y\)

Thế vào Q ta được 

\(Q=2x^2-5x-12=\left(2x^2-\frac{2x\times\sqrt{2}\times5}{2\sqrt{2}}+\frac{25}{8}\right)-\frac{121}{8}\)

\(=\left(\sqrt{2}x-\frac{5}{2\sqrt{2}}\right)^2-\frac{121}{8}\ge\frac{-121}{8}\)

Ta có: \(y=\sqrt{3+x}+\sqrt{5-x}\)

ĐKXĐ: \(-3\le x\le5\)

\(y^2=3+x+5-x+2\sqrt{\left(3+x\right)\left(5-x\right)}=8+2\sqrt{\left(3+x\right)\left(5-x\right)}\)\(\ge8\)

\(\Rightarrow y\ge2\sqrt{2}\)

Dấu "=" xảy ra khi và chỉ khi \(\left[{}\begin{matrix}x=-3\\x=5\end{matrix}\right.\)(thỏa mãn)

Vậy min y = \(2\sqrt{2}\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=5\end{matrix}\right.\)

mặt khác \(y^2\) = \(8+2\sqrt{\left(3+x\right)\left(5-x\right)}\le8+3+x+5-x=16\)

\(\Rightarrow y\le4\)

Dấu"=" xảy ra khi và chỉ khi \(3+x=5-x\Leftrightarrow x=1\)(thỏa mãn)

Vậy max y = 4 \(\Leftrightarrow x=1\)

1,

Có \(\sqrt{x}\ge0\)với mọi x

=> 2 + \(\sqrt{x}\ge\)2 với mọi x

=> A \(\ge\)2 với mọi x

Dấu "=" xảy ra <=> \(\sqrt{x}=0\)<=> x = 0

KL: A_min = 2 <=> x = 0

2, (câu này phải là GTLN chứ nhỉ)

Có \(\sqrt{x-1}\ge0\)với mọi x

=> \(2.\sqrt{x-1}\ge0\)với mọi x

=> \(5-2.\sqrt{x-1}\le5\)với mọi x

=> B \(\le\)5 với mọi x

Dấu "=" xảy ra <=> \(\sqrt{x-1}=0\)<=> x - 1 = 0 <=> x = 1

KL B_max = 5 <=> x = 1

\(\sqrt{x}\ge0\)

\(\Rightarrow A=2+\sqrt{x}\ge2+0\ge2\)

\(MinA=2\Leftrightarrow\sqrt{x}=0\Rightarrow x=0\)

2) \(5-2\sqrt{x-1}\le5\)

\(MinA=5\Leftrightarrow x-1=0\Rightarrow x=1\)

ĐKXĐ: ...

Đặt \(\sqrt{2x-1}=t\ge0\Rightarrow x=\frac{t^2+1}{2}\)

\(\Rightarrow A=\frac{2t^2+6t+4}{t^2+4t+3}=\frac{2\left(t+1\right)\left(t+2\right)}{\left(t+1\right)\left(t+3\right)}=\frac{2\left(t+2\right)}{t+3}=2-\frac{2}{t+3}\ge2-\frac{2}{3}=\frac{4}{3}\)

Dấu "=" xảy ra khi \(t=0\Leftrightarrow x=\frac{1}{2}\)

Giúp với mn~~

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

\(P=\frac{x\sqrt{x}-3-2\left(x-6\sqrt{x}+9\right)-\left(x+4\sqrt{x}+3\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}\)

\(P=\frac{x\sqrt{x}-3x+8\sqrt{x}-24}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}\)

\(P=\frac{\left(\sqrt{x}-3\right)\left(x+8\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}\)

\(P=\frac{x+8}{\sqrt{x}+1}\)

b) Ta có \(x=14-6\sqrt{5}=9-2.3.\sqrt{5}+5=\left(3-\sqrt{5}\right)^2\)

Vậy nên \(\sqrt{x}=3-\sqrt{5}\)

Suy ra \(P=\frac{\left(3-\sqrt{5}\right)^2+8}{3-\sqrt{5}+1}=\frac{58-2\sqrt{5}}{11}\)

c) \(P=\frac{x+8}{\sqrt{x}+1}=\frac{\left(x-1\right)+9}{\sqrt{x}+1}=\left(\sqrt{x}-1\right)+\frac{9}{\sqrt{x}+1}\)

\(=\left(\sqrt{x}+1\right)+\frac{9}{\sqrt{x}+1}-2\ge2\sqrt{\left(\sqrt{x}+1\right).\frac{9}{\sqrt{x}+1}}-2=4\)

minP = 4 khi \(\sqrt{x}+1=\frac{9}{\sqrt{x}+1}\Rightarrow\sqrt{x}+1=3\Rightarrow x=4.\)