Áp dụng BĐT Bunhiacopxki , ta có :

\(\left(\sqrt{3x-5}+\sqrt{7-3x}\right)^2\le\left(1^2+1^2\right)\left(3x-5+7-3x\right)\left(\dfrac{5}{3}\le x\le\dfrac{7}{3}\right)\)
\(\Leftrightarrow\left(\sqrt{3x-5}+\sqrt{7-3x}\right)^2\le4\)

\(\Leftrightarrow\sqrt{3x-5}+\sqrt{7-3x}\le2\)

\(\Rightarrow A_{Max}=2."="\Leftrightarrow x=2\left(TM\right)\)

    (3x - 5) (7 - 5x) - (5x + 2) (2 - 3x) = 4

<=> 21x - 15x² - 35 + 25x - 10x + 15x² - 4 + 6x = 4

<=>  42x - 39 = 4

<=> 42x = 43

<=> x = 43/42

\(A\le\sqrt{2\left(3x-5+7-3x\right)}=2\)

\(A_{max}=2\) khi \(3x-5=7-3x\Leftrightarrow x=2\)

\(A=\sqrt{3x-5}+\sqrt{7-3x}\)

\(ĐKXĐ:\)\(\frac{5}{3}\le x\le\frac{7}{3}\)

\(A^2=3x-5+7-3x+2\sqrt{\left(3x-5\right)\left(7-3x\right)}\)

Áp dụng BĐT Cô-si ta có :

\(A^2\le2+\left(3x-5+7-3x\right)=4\)

Dấu  =  xảy ra \(\Leftrightarrow\)\(3x-5=7-3x\Leftrightarrow x=2\)

Vậy Max \(A^2=4\)suy ra Max A = 2 khi x = 2 

sửa lại đề 

\(A=\sqrt{3x-5}+\sqrt{7-3x}\)

thông cảm nha

\(A=\sqrt{3x-5}+\sqrt{7-3x}\)

\(A^2=3x-5+7-3x+2\sqrt{\left(3x-5\right)\left(7-3x\right)}\)

\(=2+2\sqrt{\left(3x-5\right)\left(7-3x\right)}\)

\(\le2+\left(3x-5\right)+\left(7-3x\right)\)(Bđt Cô-si)

\(=2+2=4\)

\(\Rightarrow A^2\le4\Rightarrow A\le2\)

Dấu = khi \(\sqrt{3x-5}=\sqrt{7-3x}\Leftrightarrow x=2\)

Vậy....

tks bạn nha

DKXD :\(\frac{5}{3}\)\(\le\)\(x\le\)\(\frac{7}{3}\)

áp dụng bdt phụ : ( a + b )\(^2\)\(\ge\)2( a\(^2\) + b\(^2\))   ta duoc :

( \(\sqrt{3x-5}\)+ \(\sqrt{7-3x}\))\(^2\)\(\le\)2(\(3x-5+7-3x\))  = 4

\(\Rightarrow\)0\(\le\)\(\sqrt{3x-5}\)+\(\sqrt{7-3x}\)\(\le\)2

dau '=' xay ra \(\)\(\Leftrightarrow\)\(3x-5=7-3x\)

                          \(\Leftrightarrow\)\(x=2\)(thỏa mãn DKXD )

 Vay GTLN cua A= 2 \(\Leftrightarrow\)\(x=2\)