Câu 1: 

a: \(P=\dfrac{a-4-5-\sqrt{a}-3}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-2\right)}\)

\(=\dfrac{a-\sqrt{a}-12}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-2\right)}=\dfrac{\sqrt{a}-4}{\sqrt{a}-2}\)

b: Để P<1 thì \(\dfrac{\sqrt{a}-4-\sqrt{a}+2}{\sqrt{a}-2}< 0\)

\(\Leftrightarrow\sqrt{a}-2< 0\)

hay 0<a<4

mk giải 1 bài lm mẩu nha .

+) ta có : \(A=x-12\sqrt{x}\Leftrightarrow x-12\sqrt{x}-A=0\)

vì phương trình này luôn có nghiệm \(\Leftrightarrow\Delta'\ge0\)

\(\Leftrightarrow6^2+A\ge0\Leftrightarrow A\ge-36\)

vậy giá trị nhỏ nhất của \(A\) là \(-36\) dấu "=" xảy ra khi \(\sqrt{x}=\dfrac{-b'}{a}=\dfrac{6}{1}=6\Leftrightarrow x=36\)

mấy câu còn lại bn chuyển quế đưa về phương trình bật 2 theo \(x\) rồi giải như trên là đc :

\(A=x-12\sqrt{x}\\ =x-12\sqrt{x}+36-36\\ =\left(\sqrt{x}-6\right)^2-36\ge-36\text{ }\forall x\ge0\)

Vậy \(A_{Min}=-36\text{ }khi\text{ }x=36\)

B tương tự

\(C=\dfrac{\sqrt{x}-5}{\sqrt{x}+3}=\dfrac{\sqrt{x}+3-8}{\sqrt{x}+3}=1-\dfrac{8}{\sqrt{x}+3}\)

\(Do\text{ }\sqrt{x}\ge0\forall x\\ \Rightarrow\sqrt{x}+3\ge3\forall x\\ \Rightarrow\dfrac{8}{\sqrt{x}+3}\le\dfrac{8}{3}\forall x\\ \Rightarrow C=1-\dfrac{8}{\sqrt{x}+3}\ge-\dfrac{5}{3}\forall x\)

Vậy \(C_{Min}=-\dfrac{5}{3}\text{ }khi\text{ }x=0\)

D tương tự

bạn chỉ cần cố gắng là làm được

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

b) \(\dfrac{x-1}{\sqrt{y}-1}\)\(\sqrt{\dfrac{y-2\sqrt{y+1}}{\left(x-1\right)^4}}\)

= \(\dfrac{x-1}{\sqrt{y}-1}\) \(\sqrt{\dfrac{\left(y-1\right)^4}{\left(x-1\right)^4}}\)

= \(\dfrac{x-1}{\sqrt{y}-1}\)\(\dfrac{\left(\sqrt{y}-1\right)^4}{\left(x-1\right)^2}\)

= \(\dfrac{\sqrt{y-1}}{x-1}\)

Chúc bạn học tốt :3

\(F=\left(\dfrac{1}{3-\sqrt{5}}+\dfrac{1}{3+\sqrt{5}}\right):\dfrac{5-\sqrt{5}}{\sqrt{5}-1}=\dfrac{6}{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}:\dfrac{\sqrt{5}\left(\sqrt{5}-1\right)}{\sqrt{5}-1}=\dfrac{3}{2}.\dfrac{1}{\sqrt{5}}=\dfrac{3}{2\sqrt{5}}\)

\(G=\sqrt{3+\sqrt{5}}+\sqrt{7-3\sqrt{5}}-\sqrt{2}=\dfrac{\sqrt{5+2\sqrt{5}+1}+\sqrt{9-2.3.\sqrt{5}+5}-2}{\sqrt{2}}=\dfrac{\sqrt{5}+1+3-\sqrt{5}-2}{\sqrt{2}}=\dfrac{2}{\sqrt{2}}=\sqrt{2}\)

\(H=\sqrt{x+2\sqrt{2x-4}}+\sqrt{x-2\sqrt{2x-4}}=\sqrt{x-2+2\sqrt{2}.\sqrt{x-2}+2}+\sqrt{x-2-2\sqrt{2}.\sqrt{x-2}+2}=\sqrt{\left(\sqrt{x-2}+\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{x-2}-\sqrt{2}\right)^2}=\sqrt{x-2}+\sqrt{2}+\left|\sqrt{x-2}-\sqrt{2}\right|\left(x\ge2\right)\)

+) ta có : \(F=\left(\dfrac{1}{3-\sqrt{5}}+\dfrac{1}{3+\sqrt{5}}\right):\dfrac{5-\sqrt{5}}{\sqrt{5}-1}\)

\(\Leftrightarrow F=\left(\dfrac{3+\sqrt{5}+3-\sqrt{5}}{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}\right):\dfrac{\sqrt{5}\left(\sqrt{5}-1\right)}{\sqrt{5}-1}\)

\(\Leftrightarrow F=\dfrac{6}{4}.\dfrac{1}{\sqrt{5}}=\dfrac{6}{4\sqrt{5}}\)

+) ta có : \(G=\sqrt{3+\sqrt{5}}+\sqrt{7-3\sqrt{5}}-\sqrt{2}\)

\(\Leftrightarrow G=\dfrac{\sqrt{6+2\sqrt{5}}+\sqrt{14-6\sqrt{5}}-2}{\sqrt{2}}\)

\(\Leftrightarrow G=\dfrac{\sqrt{\left(\sqrt{5}+1\right)^2}+\sqrt{\left(3-\sqrt{5}\right)^2}-2}{\sqrt{2}}\)

\(\Leftrightarrow G=\dfrac{\sqrt{5}+1+3-\sqrt{5}-2}{\sqrt{2}}=\dfrac{2}{\sqrt{2}}=\sqrt{2}\)

+) ta có : \(H=\sqrt{x+2\sqrt{2x-4}}+\sqrt{x-2\sqrt{2x-4}}\)

\(\Rightarrow H^2=2x+2\sqrt{x^2-4\left(2x-4\right)}=2x+2\sqrt{\left(x-4\right)^2}\)

\(\Leftrightarrow\left[{}\begin{matrix}H^2=4x-8\\H^2=8\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}H=\sqrt{4x-8}\\H=-\sqrt{4x-8}\left(L\right)\end{matrix}\right.\\\left[{}\begin{matrix}H=\sqrt{8}\\H=-\sqrt{8}\left(L\right)\end{matrix}\right.\end{matrix}\right.\)

vậy \(H=\sqrt{4x-8};H=\sqrt{8}\)