\(A=7-x^2-3x=-\left(x^2+3x+\dfrac{9}{4}\right)+\dfrac{37}{4}=-\left(x+\dfrac{3}{2}\right)^2+\dfrac{37}{4}\le\dfrac{37}{4}\)

\(maxA=\dfrac{37}{4}\Leftrightarrow x=-\dfrac{3}{2}\)

B=\(x^2+3x+7\)

=>B= \(x^2+2\times\frac{3}{2}x+\frac{9}{4}+\frac{19}{4}\)

=>B=\(\left(x+\frac{3}{2}\right)^2+\frac{19}{4}\)

Vì \(\left(x+\frac{3}{2}\right)^2\ge0\)   (Với mọi x)

=>\(\left(x+\frac{3}{2}\right)^2+\frac{19}{4}\ge\frac{19}{4}\)   (Với mọi x )

Dấu "='' xảy ra  <=> \(x+\frac{3}{2}=0=>x=-\frac{3}{2}\)

Vậy min B bằng 19/4 <=>x=-3/2

Phần b thì mk làm đc n phần a hình như sai đề pn ạ !!!
 

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 \(\Leftrightarrow\)\(x=2\)

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

\(P_{max}=2\) khi \(3x-5=7-3x\Rightarrow x=2\)

\(A=2\left(x-1\right)+\dfrac{9}{x-1}+2\ge2\sqrt{\dfrac{18\left(x-1\right)}{x-1}}+2=6\sqrt{2}+2\)

\(A_{min}=6\sqrt{2}+2\) khi \(x=\dfrac{2+3\sqrt{2}}{2}\)

ap dung bdt cauchy-schwarz ta co

\(A=\sqrt{3x-5}+\sqrt{7-3x}\) \(\le\sqrt{\left(1^2+1^2\right)\left(3x-5+7-3x\right)}=\sqrt{4}=2\)

dau = xay ra khi \(\frac{1}{3x-5}=\frac{1}{7-3x}\Leftrightarrow x=2\)

bạn tham khảo nhé

áp dụng BĐt cô si ta có

\(\sqrt{3x-5}+\sqrt{7-3x}\le\frac{3x-5+1}{2}+\frac{7-3x+1}{2}=2\)

Vậy A max=2

bạn trần hữu ngọc minh bạn có thể viết công thức ra cho mk đc k

\(A=\frac{3x^2+9x+17}{3x^2+9x+7}=1+\frac{10}{3x^2+9x+7}\)

Có: \(3x^2+9x+7=3\left(x^2+3x+\frac{9}{4}\right)+\frac{1}{4}=3\left(x+\frac{3}{2}\right)^2+\frac{1}{4}\)

Vì: \(3\left(x+\frac{3}{2}\right)^2\ge0,\forall x\)

=> \(3\left(x+\frac{3}{2}\right)^2+\frac{1}{4}\ge\frac{1}{4}\)

=>\(\frac{10}{3\left(x+\frac{3}{2}\right)^2+\frac{1}{4}}\le40\)

=> \(1+\frac{10}{3\left(x+\frac{3}{2}\right)^2+\frac{41}{4}}\le41\)

Vậy GTLN của A là \(\frac{81}{41}\) khi \(x=-\frac{3}{2}\)

HELP ME !!!