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

Đã từng lm qua nhưng ko chắc á 

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

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

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

Áp dụng BĐT Cô - si Ta có : \(A^2\le2+\left(3x-5+7-3x\right)=4\)

Dấu ''='' xãy ra \(\Leftrightarrow3x-5=7-3x\Leftrightarrow x=2\)

Vậy Max A²=4 => Max A=2 khi x=2 

tui đã hỉu 

cam on Kid 

có dịp giúp á á á 

e lm cách khác nhưng ko bt có đúng ko nữa:(

Ta có:

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

\(\Rightarrow A^2=\left(\sqrt{3x-5}\cdot1+\sqrt{7-3x}\cdot1\right)^2\)

Áp dụng BĐT bu-nhi-a-cốp-ski ta có:

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

\(A^2\le\left(3x-5+7-3x\right)\cdot2\)

\(A^2\le4\)

\(\Rightarrow A\le2\left(because:A\ge0\right)\)

Dấu "=" xảy ra khi \(x=2\)

Với các số thực không âm a; b ta luôn có BĐT sau:

\(\sqrt{a}+\sqrt{b}\ge\sqrt{a+b}\) (bình phương 2 vế được \(2\sqrt{ab}\ge0\) luôn đúng)

Áp dụng:

a. 

\(A\ge\sqrt{x-4+5-x}=1\)

\(\Rightarrow A_{min}=1\) khi \(\left[{}\begin{matrix}x=4\\x=5\end{matrix}\right.\)

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

\(A_{max}=\sqrt{2}\) khi \(x-4=5-x\Leftrightarrow x=\dfrac{9}{2}\)

b.

\(B\ge\sqrt{3-2x+3x+4}=\sqrt{x+7}=\sqrt{\dfrac{1}{3}\left(3x+4\right)+\dfrac{17}{3}}\ge\sqrt{\dfrac{17}{3}}=\dfrac{\sqrt{51}}{3}\)

\(B_{min}=\dfrac{\sqrt{51}}{3}\) khi \(x=-\dfrac{4}{3}\)

\(B=\sqrt{3-2x}+\sqrt{\dfrac{3}{2}}.\sqrt{2x+\dfrac{8}{3}}\le\sqrt{\left(1+\dfrac{3}{2}\right)\left(3-2x+2x+\dfrac{8}{3}\right)}=\dfrac{\sqrt{510}}{6}\)

\(B_{max}=\dfrac{\sqrt{510}}{6}\) khi \(x=\dfrac{11}{30}\)

a)Ta có:A=\(\sqrt{x-4}+\sqrt{5-x}\)

        =>A²=\(x-4+2\sqrt{\left(x-4\right)\left(5-x\right)}+5-x\)

        =>A²= 1+\(2\sqrt{\left(x-4\right)\left(5-x\right)}\ge1\)

        =>A\(\ge\)1

Dấu '=' xảy ra <=> x=4 hoặc x=5

Vậy,Min A=1 <=>x=4 hoặc x=5

Còn câu b tương tự nhé

\(x=\sqrt[3]{5\sqrt{2}+7}-\sqrt[3]{5\sqrt{2}-7}\Rightarrow x^3=5\sqrt{2}+7-\left(5\sqrt{2}-7\right)-3\sqrt[3]{\left(5\sqrt{2}\right)^2-7^2}.x\)

\(=14-3.\sqrt[3]{50-49}.x=14-3x\)

\(\Rightarrow x^3=14-3x\Rightarrow x^3+3x=14\)

Đk: x = \(5+2\sqrt{7}\)> 5

Đặt A = \(\sqrt{3x+\sqrt{6x-1}}-\sqrt{3x-\sqrt{6x-1}}\)

A² = \(\left(\sqrt{3x+\sqrt{6x-1}}-\sqrt{3x-\sqrt{6x-1}}\right)^2\)

A² = \(3x+\sqrt{6x-1}+3x-\sqrt{6x-1}-2\sqrt{\left(3x+\sqrt{6x-1}\right)\left(3x-\sqrt{6x-1}\right)}\)

A² = \(6x-2\sqrt{9x^2-6x+1}\)

A² = \(6x-2\sqrt{\left(3x-1\right)^2}\) (vì x > \(\frac{1}{3}\))

A² = \(6x-2\left(3x-1\right)\)

A² = \(6x-6x+2\)

A² = 2

=> A = \(\sqrt{2}\)

Vậy ....

Đặt:    \(A=\sqrt{3x+\sqrt{6x-1}}-\sqrt{3x-\sqrt{6x-1}}\)

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

=>    \(A^2=6x-2\sqrt{9x^2-6x+1}\)

=>    \(A^2=6x-2\sqrt{\left(3x-1\right)^2}\)

Mà:    \(x=5+2\sqrt{7}\Rightarrow x>\frac{1}{3}\Rightarrow3x>1\Rightarrow3x-1>0\)

=>   \(A^2=6x-2\left(3x-1\right)\)

=>    \(A^2=6x-6x+2=2\)

Mà:    \(\sqrt{3x+\sqrt{6x-1}}>\sqrt{3x-\sqrt{6x-1}}\Rightarrow A>0\)

=>    \(A=\sqrt{2}\)

VẬY    \(A=\sqrt{2}\)