Bạn tham khảo lời giải tại đây:

https://hoc24.vn/cau-hoi/cho-0a1-0b2-0c3tim-gtln-cua-a-dfracsqrt1-aa-dfracsqrt2-bb-dfracsqrt3-ccbai-nay-dung-cauchyminh-suy-nghi.179994478119

vì 0<a<1 ;0<b<2 ;0<c<3

=> 1-a > 0 <=> 0<\(\sqrt{1-a}\) < 1

=> 0 <\(\dfrac{\sqrt{1-a}}{a}\) ≤ 1 (1)

c/m tương tự với b,c

=> 0 < \(\dfrac{\sqrt{2-b}}{b}\) ≤ 2 (2)

và 0 < \(\dfrac{\sqrt{3-c}}{c}\) ≤ 3 (3)

Cộng các vế của bđt với nhau

=> 0 < \(\dfrac{\sqrt{1-a}}{a}+\dfrac{\sqrt{2-b}}{b}+\dfrac{\sqrt{3-c}}{c}\) ≤ 6

Vậy GTLN của A là 6

Câu 1:

\(A=\dfrac{81x}{3-x}+\dfrac{3}{x}=\dfrac{81x}{3-x}+\left(\dfrac{3}{x}-1\right)+1=\dfrac{81x}{3-x}+\dfrac{3-x}{x}+1\ge2\sqrt{\dfrac{81x}{3-x}.\dfrac{3-x}{x}}+1=18+1=19\)

Dấu "=" xảy ra <=> x = 0,3

Câu 2:

\(\dfrac{1}{3x-2\sqrt{6x}+5}=\dfrac{1}{\left(3x-2\sqrt{6x}+2\right)+3}=\dfrac{1}{\left(x\sqrt{3}-\sqrt{2}\right)^2+3}\le\dfrac{1}{3}\)

Dấu "=" xảy ra <=> \(x=\sqrt{\dfrac{2}{3}}\)

Câu 3:

\(A=2014\sqrt{x}+2015\sqrt{1-x}=2014\left(\sqrt{x}+\sqrt{1-x}\right)+\sqrt{1-x}\)

Ta có: \(\left(\sqrt{x}+\sqrt{1-x}\right)^2=x+1-x+2\sqrt{x\left(1-x\right)}=1+2\sqrt{x\left(1-x\right)}\ge1\)

=> \(A=2014\left(\sqrt{x}-\sqrt{1-x}\right)+\sqrt{1-x}\ge2014+\sqrt{1-x}\ge2014\)

Dấu "=" xảy ra <=> x = 1

Thanks bn nhìu

CM:$(b+c)(\frac{1}{b}+\frac{1}{c})< \frac{(a+d)^{2}}{ad}$ - Bất đẳng thức và cực trị - Diễn đàn Toán học

\(1.\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}=\dfrac{\sqrt{8+2\sqrt{7}}-\sqrt{8-2\sqrt{7}}}{\sqrt{2}}=\dfrac{\sqrt{\left(\sqrt{7}+1\right)^2}-\sqrt{\left(\sqrt{7}-1\right)^2}}{\sqrt{2}}=\dfrac{|\sqrt{7}+1|-|\sqrt{7}-1|}{\sqrt{2}}=\dfrac{2}{\sqrt{2}}=\sqrt{2}\)

\(3a.x+1-\dfrac{x-1}{3}< x-\dfrac{2x+3}{2}+\dfrac{x}{3}+5\)

\(\Leftrightarrow\dfrac{6\left(x+1\right)-2\left(x-1\right)}{6}< \dfrac{6x-3\left(2x+3\right)+2x+30}{6}\)

\(\Leftrightarrow6x+6-2x+2< 6x-6x-9+2x+30\)

\(\Leftrightarrow6x-2x-2x+6+2+9-30< 0\)

\(\Leftrightarrow2x-13< 0\)

\(\Leftrightarrow x< \dfrac{13}{2}\)

KL...............

\(b.5+\dfrac{x+4}{5}< x-\dfrac{x-2}{2}+\dfrac{x+3}{3}\)

\(\Leftrightarrow\dfrac{150+6\left(x+4\right)}{30}< \dfrac{30x-15\left(x-2\right)+10\left(x+3\right)}{30}\)

\(\Leftrightarrow150+6x+24< 30x-15x+30+10x+30\)

\(\Leftrightarrow6x-30x+15x-10x+150+24-30-30< 0\)

\(\Leftrightarrow-19x+114< 0\)

\(\Leftrightarrow x>6\)

KL..................

Câu 4 :

Ta có :

\(A=\dfrac{3}{1-x}+\dfrac{4}{x}\)

\(=\left(\dfrac{3}{1-x}+\dfrac{4}{x}\right)\left[\left(1-x\right)+x\right]\)

Theo BĐT Bu - nhi a - cốp xki ta có :

\(\left(a^2+b^2\right)\left(x^2+y^2\right)\ge\left(ax+by\right)^2\)

\(\Leftrightarrow\left(\dfrac{3}{1-x}+\dfrac{4}{x}\right)\left[\left(1-x\right)+x\right]\ge\left(\sqrt{\dfrac{3\left(1-x\right)}{1-x}}+\sqrt{\dfrac{4x}{x}}\right)^2=\left(\sqrt{3}+2\right)^2=7+4\sqrt{3}\)

Dấu \("="\) xảy ra khi \(\dfrac{3}{\left(1-x\right)^2}=\dfrac{4}{x^2}\)

\(\Leftrightarrow3x^2=4x^2-8x+4\)

\(\Leftrightarrow x^2-8x+4=0\)

\(\Delta=64-16=48>0\)

\(\Rightarrow\left\{{}\begin{matrix}x_1=4+2\sqrt{3}\\x_2=4-2\sqrt{3}\end{matrix}\right.\)

Vậy GTNN của\(A=7+4\sqrt{3}\) khi \(\left[{}\begin{matrix}x_1=4+2\sqrt{3}\\x_2=4-2\sqrt{3}\end{matrix}\right.\)