a: \(A=\dfrac{x-3\sqrt{x}+2x+6\sqrt{x}-3x-9}{x-9}=\dfrac{-3\sqrt{x}-9}{x-9}\)

\(=\dfrac{-3\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{-3}{\sqrt{x}-3}\)

b: A=1/3

=>\(\dfrac{-3}{\sqrt{x}-3}=\dfrac{1}{3}\)

=>căn x-3=-9

=>căn x=-6(loại)

c: căn x-3>=-3

=>3/căn x-3<=-1

=>-3/căn x-3>=1

Dấu = xảy ra khi x=0

a: \(P=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{x-9}=\dfrac{-3\left(\sqrt{x}+1\right)}{x-9}\)

\(M=\dfrac{-3\left(\sqrt{x}+1\right)}{x-9}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}=\dfrac{-3}{\sqrt{x}+3}\)

b: \(A=\dfrac{-3x+4x+7}{\sqrt{x}+3}=\dfrac{x+7}{\sqrt{x}+3}=\dfrac{x-9+16}{\sqrt{x}+3}\)

=>\(A=\sqrt{x}-3+\dfrac{16}{\sqrt{x}+3}=\sqrt{x}+3+\dfrac{16}{\sqrt{x}+3}-6>=2\sqrt{16}-6=2\)

Dấu = xảy ra khi x=1

đợi tý

a) Để \(A=\dfrac{2022}{\left|x\right|+2023}\) đạt Max thì |x| + 2023 phải đạt Min

Ta có \(\left|x\right|\ge0\forall x\Rightarrow\left|x\right|+2023\ge2023\forall x\)

\(\Rightarrow\dfrac{2022}{\left|x\right|+2023}\le\dfrac{2022}{2023}\forall x\)

Dấu "=" xảy ra khi \(\left|x\right|=0\Rightarrow x=0\)

Vậy Max \(A=\dfrac{2022}{\left|x\right|+2023}=\dfrac{2022}{2023}\) đạt được khi x = 0

b) Để \(B=\left(\sqrt{x}+1\right)^{99}+2022\) đạt Min với \(x\ge0\) thì \(\sqrt{x}+1\) phải đạt Min

Ta có \(\sqrt{x}\ge0\forall x\ge0\Rightarrow\sqrt{x}+1\ge1\forall x\ge0\)

\(\Rightarrow\left(\sqrt{x}+1\right)^{99}+2022\ge1+2022\ge2023\forall x\ge0\)

Dấu "=" xảy ra khi \(\sqrt{x}=0\Rightarrow x=0\)

Vậy Max \(B=\left(\sqrt{x}+1\right)^{99}+2022=2023\) đạt được khi x = 0

Câu c) và d) thì tự làm, ko có rảnh =))))

Đã trả lời rồi còn độ tí đồ ngull

Bài 1: 

Ta có: \(D=\sqrt{16x^4}-2x^2+1\)

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

\(=2x^2+1\)

\(P=\dfrac{3\left(x^2+2x+3\right)+1}{x^2+2x+3}=3+\dfrac{1}{x^2+2x+3}=3+\dfrac{1}{\left(x+1\right)^2+2}\le3+\dfrac{1}{2}=\dfrac{7}{2}\)

\(P_{max}=\dfrac{7}{2}\) khi \(x=-1\)

\(M=\dfrac{2\left(x^2+3x+3\right)+1}{x^2+3x+3}=2+\dfrac{1}{x^2+3x+3}=2+\dfrac{1}{\left(x+\dfrac{3}{2}\right)^2+\dfrac{3}{4}}\le2+\dfrac{1}{\dfrac{3}{4}}=\dfrac{10}{3}\)

\(M_{max}=\dfrac{10}{3}\) khi \(x=-\dfrac{3}{2}\)

a) Ta có: 

\(Q=\sqrt{\left(1-3x\right)\left(x+\dfrac{1}{2}\right)}\) Q có nghĩa khi:

\(\left(1-3x\right)\left(x+\dfrac{1}{2}\right)\ge0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}1-3x\ge0\\x+\dfrac{1}{2}\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}1-3x\le0\\x+\dfrac{1}{2}\le\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}3x\le1\\x\ge-\dfrac{1}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}3x\ge1\\x\le-\dfrac{1}{2}\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\le\dfrac{1}{3}\\x\ge-\dfrac{1}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}x\ge\dfrac{1}{3}\\x\le-\dfrac{1}{2}\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}-\dfrac{1}{2}\le x\le\dfrac{1}{3}\\x\in\varnothing\end{matrix}\right.\)

\(\Leftrightarrow-\dfrac{1}{2}\le x\le\dfrac{1}{3}\)

b) Ta có: \(Q=\sqrt{\left(1-3x\right)\left(x+\dfrac{1}{2}\right)}\)

\(Q=\sqrt{x+\dfrac{1}{2}-3x^2-\dfrac{3}{2}x}\)

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

\(Q=\sqrt{-3\left(x^2+\dfrac{1}{6}x-\dfrac{1}{6}\right)}\)

\(Q=\sqrt{-3\left(x^2+2\cdot\dfrac{1}{12}\cdot x+\dfrac{1}{144}-\dfrac{25}{144}\right)}\)

\(Q=\sqrt{-3\left(x+\dfrac{1}{12}\right)^2+\dfrac{25}{144}}\)

Mà: \(Q=\sqrt{-3\left(x+\dfrac{1}{12}\right)^2+\dfrac{25}{144}}\le\sqrt{\dfrac{25}{144}}=\dfrac{5}{12}\)

Dấu "=" xảy ra khi:

\(\Leftrightarrow-3\left(x+\dfrac{1}{12}\right)^2=0\)

\(\Leftrightarrow x+\dfrac{1}{12}=0\)

\(\Leftrightarrow x=-\dfrac{1}{12}\)

Vậy: \(Q_{max}=\dfrac{5}{12}.khi.x=-\dfrac{1}{12}\)

a) Để \(\sqrt{\dfrac{x}{3}}\) có nghĩa thì \(\dfrac{x}{3}\ge0\Leftrightarrow x\ge0\)

b) Để \(\sqrt{-5x}\) có nghĩa thì \(-5x\ge0\Leftrightarrow x\le0\)

c) Để \(\sqrt{4-x}\) có nghĩa thì \(4-x\ge0\Leftrightarrow x\le4\)

d) Để \(\sqrt{3x+7}\) có nghĩa thì \(3x+7\ge0\Leftrightarrow x\ge-\dfrac{7}{3}\)

e) Để \(\sqrt{-3x+4}\) có nghĩa thì \(-3x+4\ge0\Leftrightarrow x\le\dfrac{4}{3}\)

f) Để \(\sqrt{\dfrac{1}{-1+x}}\) có nghĩa thì \(\left\{{}\begin{matrix}\dfrac{1}{-1+x}\ge0\\-1+x\ne0\end{matrix}\right.\)

\(\Leftrightarrow-1+x>0\Leftrightarrow x>1\)

g) Để \(\sqrt{1+x^2}\) có nghĩa thì \(1+x^2\ge0\left(đúng\forall x\right)\)

h) \(\sqrt{\dfrac{5}{x-2}}\) có nghĩ thì \(\left\{{}\begin{matrix}\dfrac{5}{x-2}\ge0\\x-2\ne0\end{matrix}\right.\)

\(\Leftrightarrow x-2>0\Leftrightarrow x>2\)

a. \(x\ge0\)

b. \(x< 0\)

c. \(x\le4\)

d. \(x\ge\dfrac{-7}{3}\)

e. \(x\le\dfrac{4}{3}\)

f. \(x>1\)

g. Mọi x

h. \(x>2\)