\(dkxđ\Leftrightarrow\left\{{}\begin{matrix}-x^2+5x\ge0\\-x^2+3x+18\ge0\end{matrix}\right.\)\(\Rightarrow0\le x\le5\Rightarrow\left\{{}\begin{matrix}x\ge0\\x\le5\end{matrix}\right.\)

\(\Rightarrow A=\sqrt{5x-x^2}+\sqrt{18+3x-x^2}\)

\(\sqrt{5x-x^2}=\sqrt{-\left(x^2-5x+\dfrac{25}{4}-\dfrac{25}{4}\right)}=\sqrt{-\left[\left(x-\dfrac{5}{2}\right)^2-\dfrac{25}{4}\right]}=\sqrt{-\left(x-\dfrac{5}{2}\right)^2+\dfrac{25}{4}}\ge0\left(1\right)\)

\(dấu\) \("="\) \(xảy\) \(ra\Leftrightarrow x=5\)

\(\sqrt{-x^2+3x+18}=\sqrt{-\left(x^2-3x-18\right)}=\sqrt{-\left[x^2-3x+\dfrac{9}{4}-\dfrac{81}{4}\right]}=\sqrt{-\left(x-\dfrac{3}{2}\right)^2+\dfrac{81}{4}}\ge\sqrt{-\left(5-\dfrac{3}{2}\right)^2+\dfrac{81}{4}}=\sqrt{8}\left(2\right)\)

dấu"=" xảy ra \(< =>x=5\)

\(\left(1\right)\left(2\right)\Rightarrow A\ge\sqrt{8}\) \(dấu\) \("="\) \(xảy\) \(ra\Leftrightarrow x=5\)\(\Rightarrow MinA=\sqrt{8}\)

\(\left(maxA=\sqrt{48}\right)dấu\) \("="\) \(xảy\) \(ra\Leftrightarrow x=\dfrac{15}{7}\)

\(\)

câu 1 

ta có .....

lười viết Min - cốp xki nha

DKXD của A, ta có \(x^{2\le5\Rightarrow-\sqrt{5}\le x\le\sqrt{5}}\)

mà \(3x\ge-3\sqrt{5}\)

mặt kkhác \(\sqrt{5-x^2}\ge0\Rightarrow A=3x+x\sqrt{5-x^2}\ge-3\sqrt{5}\)

min A= \(-3\sqrt{5}\)\(\Leftrightarrow x=-\sqrt{5}\)

ta có \(A^2\le25\)và ta cx có \(-5\le A\le5\)

nhưng dễ thấy \(A=-5\)không xảy ra, vô lí nên ...........bạn xem đoạn sau nhé ( tiếp phần kia )

a.\(2x^2+5x+8+\sqrt{x}=x^2+3x+35+x^2+2x-7\)

\(=2x^2+5x+8+\sqrt{x}=2x^2+5x+28\Leftrightarrow\sqrt{x}=20\Leftrightarrow x=400.\)

b.\(3\sqrt{x}+7x+5=\sqrt{x}+4x-6+3x+18\)

\(=3\sqrt{x}+7x+5=\sqrt{x}+7x+12\Leftrightarrow2\sqrt{x}=7\Leftrightarrow x=\frac{49}{4}.\)

c.\(8\sqrt{x}+2x-9=5x+7+6\sqrt{x}-3x-12.\)

\(=8\sqrt{x}+2x-9=2x+6\sqrt{x}-5\Leftrightarrow2\sqrt{x}=4\Leftrightarrow x=4.\)

d.\(2\sqrt{3x}+11x-18=5x+3+6\sqrt{3x}+6x-21\)

\(=2\sqrt{3x}+11x-18=11x+6\sqrt{3x}-19\Leftrightarrow4\sqrt{3x}=1\)

\(\Leftrightarrow\sqrt{3x}=\frac{1}{4}\Leftrightarrow3x=\frac{1}{16}\Leftrightarrow x=\frac{1}{48}.\)

a) \(2x^2+5x+8+\sqrt{x}=x^2+3x+35+x^2+2x-7\)

<=> \(2x^2+5x+8+\sqrt{x}=2x^2+5x+28\)

<=> \(2x^2+5x+8+\sqrt{x}-\left(2x^2+5\right)=28\)

<=> \(\sqrt{x}+8=28\)

<=> \(\sqrt{x}=28-8\)

<=> \(\sqrt{x}=20\)

<=> \(\left(\sqrt{x}\right)^2=20^2\)

<=> x = 400

=> x = 400

b) \(3\sqrt{x}+7x+5=\sqrt{x}+4x-6+3x+18\)

<=> \(3\sqrt{x}+7x+5=7x+\sqrt{x}+12\)

<=> \(3\sqrt{x}+5=7x+\sqrt{x}+12-7x\)

<=> \(3\sqrt{x}+5=\sqrt{x}+12\)

<=> \(3\sqrt{x}=\sqrt{x}+12-5\)

<=> \(3\sqrt{x}=\sqrt{x}+7\)

<=> \(3\sqrt{x}-\sqrt{x}=7\)

<=> \(2\sqrt{x}=7\)

<=> \(\sqrt{x}=\frac{7}{2}\)

<=> \(\left(\sqrt{x}\right)^2=\left(\frac{7}{2}\right)^2\)

<=> \(x=\frac{49}{4}\)

=> \(x=\frac{49}{4}\)

c) \(8\sqrt{x}+2x-9=5x+7+6\sqrt{x}-3x-12\)

<=> \(8\sqrt{x}+2x-9=2x+6\sqrt{x}-5\)

<=> \(8\sqrt{x}-9=2x+6\sqrt{x}-5-2x\)

<=> \(8\sqrt{x}-9=6\sqrt{x}-5\)

<=> \(8\sqrt{x}=6\sqrt{x}-5+9\)

<=> \(8\sqrt{x}=6\sqrt{x}+4\)

<=> \(8\sqrt{x}-6\sqrt{x}=4\)

<=> \(2\sqrt{x}=4\)

<=> \(\sqrt{x}=2\)

<=> \(\left(\sqrt{x}\right)^2=2^2\)

<=> x = 4

=> x = 4

d) \(2\sqrt{3x}+11x-18=5x+3+6\sqrt{3x}+6x-21\)

<=> \(2\sqrt{3x}+11x-18=11x+6\sqrt{3x}-18\)

<=> \(2\sqrt{3x}+11x-18-\left(11x-18\right)=6\sqrt{3x}\)

<=>\(2\sqrt{3x}=6\sqrt{3x}\)

<=> \(2\sqrt{3x}-6\sqrt{3x}=0\)

<=>\(-4\sqrt{3x}=0\)

<=> \(\sqrt{3x}=0\)

<=> \(\left(\sqrt{3x}\right)^2=0^2\)

<=> 3x = 0

<=> x = 0

=> x = 0

a: ĐKXĐ: \(\left[{}\begin{matrix}x\ge3\\x\le2\end{matrix}\right.\)

b: ĐKXĐ: \(\left[{}\begin{matrix}x>\dfrac{2\sqrt{14}}{7}\\x< -\dfrac{2\sqrt{14}}{7}\end{matrix}\right.\)

c: ĐKXĐ: \(x=\dfrac{1}{3}\)

d: ĐKXĐ: \(-\dfrac{2}{3}< x\le\sqrt{3}\)

à, phần a ra x = 400. Nhầm