a) \(A=\left(\sqrt{a}+\sqrt{b}\right)^2\le\left(\sqrt{a}+\sqrt{b}\right)^2+\left(\sqrt{a}-\sqrt{b}\right)^2=2a+2b\le2\)

Vậy GTLN của A là 2 \(\Leftrightarrow\hept{\begin{cases}\sqrt{a}=\sqrt{b}\\a+b=1\end{cases}\Leftrightarrow a=b=\frac{1}{2}}\)

b) Ta có : \(\left(\sqrt{a}+\sqrt{b}\right)^4\le\left(\sqrt{a}+\sqrt{b}\right)^4+\left(\sqrt{a}-\sqrt{b}\right)^4=2\left(a^2+b^2+6ab\right)\)

Tương tự : \(\left(\sqrt{a}+\sqrt{c}\right)^4\le2\left(a^2+c^2+6ac\right)\)

\(\left(\sqrt{a}+\sqrt{d}\right)^4\le2\left(a^2+d^2+6ad\right)\)

\(\left(\sqrt{b}+\sqrt{c}\right)^4\le2\left(b^2+c^2+6bc\right)\)

\(\left(\sqrt{b}+\sqrt{d}\right)^4\le2\left(b^2+d^2+6bd\right)\)

\(\left(\sqrt{c}+\sqrt{d}\right)^4\le2\left(c^2+d^2+6cd\right)\)

Cộng các vế lại, ta được :

\(B\le6\left(a^2+b^2+c^2+d^2+2ab+2ac+2ad+2bd+2cd+2bc\right)=6\left(a+b+c+d\right)^2\)

\(\Rightarrow B\le6\)

Vậy GTLN của B là 6 \(\Leftrightarrow\hept{\begin{cases}\sqrt{a}=\sqrt{b}=\sqrt{c}=\sqrt{d}\\a+b+c+d=1\end{cases}}\Leftrightarrow a=b=c=d=\frac{1}{4}\)

Em tham khảo ở đây:

Cho a,b,c > 0 và ab + bc + ac = 1. Chứng minh rằng :\(\dfrac{a}{\sqrt{a^2+1}}+\dfrac{b}{\sqrt{b^2+1}}+\dfrac{c}{\sqrt{c^... - Hoc24

\(P=\sqrt{a\left(b+1\right)}+\sqrt{b\left(a+1\right)}\)

\(\Rightarrow P\sqrt{2}=\sqrt{2a\left(b+1\right)}+\sqrt{2b\left(a+1\right)}\)

\(\le\frac{1}{2}\left(2a+b+1\right)+\frac{1}{2}\left(2b+a+1\right)\)

\(\le\frac{1}{2}\left(3a+3b+2\right)\le\frac{1}{2}.\left(3.2+2\right)=4\)

\(\Rightarrow p\le2\sqrt{2}\)

Dấu"=" xảy ra \(\Leftrightarrow a=b=1\)

Vậy Max P \(=2\sqrt{2}\)\(\Leftrightarrow a=b=1\)

a: ĐKXĐ: a>=0; b>=0; ab<>1

Ta có: \(\frac{\sqrt{a}+\sqrt{b}}{1-\sqrt{ab}}+\frac{\sqrt{a}-\sqrt{b}}{1+\sqrt{ab}}\)

\(=\frac{\left(\sqrt{a}+\sqrt{b}\right)\left(1+\sqrt{ab}\right)+\left(\sqrt{a}-\sqrt{b}\right)\left(1-\sqrt{ab}\right)}{\left(1-\sqrt{ab}\right)\left(1+\sqrt{ab}\right)}\)

\(=\frac{\sqrt{a}+a\cdot\sqrt{b}+\sqrt{b}+b\cdot\sqrt{a}+\sqrt{a}-a\cdot\sqrt{b}-\sqrt{b}+b\cdot\sqrt{a}}{1-ab}=\frac{2\cdot\sqrt{a}+2b\cdot\sqrt{a}}{1-ab}\)

\(=\frac{2\sqrt{a}\left(b+1\right)}{1-ab}\)

Ta có: \(D=\left(\frac{\sqrt{a}+\sqrt{b}}{1-\sqrt{ab}}+\frac{\sqrt{a}-\sqrt{b}}{1+\sqrt{ab}}\right):\left(1+\frac{a+b+2ab}{1-ab}\right)\)

\(=\frac{2\sqrt{a}\left(b+1\right)}{1-ab}:\frac{1-ab+a+b+2ab}{1-ab}=\frac{2\sqrt{a}\left(b+1\right)}{1-ab}\cdot\frac{1-ab}{ab+a+b+1}\)

\(=\frac{2\sqrt{a}\left(b+1\right)}{ab+a+b+1}=\frac{2\sqrt{a}\left(b+1\right)}{\left(b+1\right)\left(a+1\right)}=\frac{2\sqrt{a}}{a+1}\)

b: \(a=\frac{2}{2+\sqrt3}=\frac{2\left(2-\sqrt3\right)}{\left(2+\sqrt3\right)\left(2-\sqrt3\right)}\)

\(=\frac{4-2\sqrt3}{4-3}=4-2\sqrt3=\left(\sqrt3-1\right)^2\)

Thay \(a=\left(\sqrt3-1\right)^2\) vào D, ta được:

\(D=\frac{2\cdot\sqrt{\left(\sqrt3-1\right)^2}}{\left(\sqrt3-1\right)^2+1}\)

\(=\frac{2\left(\sqrt3-1\right)}{4-2\sqrt3+1}=\frac{2\sqrt3-2}{5-2\sqrt3}=\frac{\left(2\sqrt3-2\right)\left(5+2\sqrt3\right)}{\left(5-2\sqrt3\right)\left(5+2\sqrt3\right)}\)

\(=\frac{10\sqrt3+12-10-4\sqrt3}{25-12}=\frac{6\sqrt3+2}{13}\)

c: \(\frac{1}{D}=\frac{a+1}{2\sqrt{a}}\)

=>\(\frac{1}{D}-1=\frac{a+1-2\sqrt{a}}{2\sqrt{a}}=\frac{\left(\sqrt{a}-1\right)^2}{2\sqrt{a}}\ge0\forall a\) thỏa mãn ĐKXĐ

=>\(\frac{1}{D}\ge1\forall a\) thỏa mãn ĐKXĐ

=>D<=1∀a thỏa mãn ĐKXĐ

Dấu '=' xảy ra khi \(\sqrt{a}-1=0\)

=>a=1(nhận)

2M\(\le\)a(9b+4a+5b)+b(9a+4b+5a)  (AM-GM)

     =4(a²+b²)+28ab\(\le\)4(a²+b²)+14(a²+b²)  (AM-GM)

                                =36 (do a²+b²=2)

=> M \(\le\)18

 Dấu bằng có <=> a=b=1

a: \(A=\dfrac{2\sqrt{x}+6+\sqrt{x}-3}{x-9}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\)

\(=\dfrac{3\left(\sqrt{x}+1\right)}{x-9}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}=\dfrac{3}{\sqrt{x}+3}\)

b: \(\sqrt{x}+3>=3\)

=>A<=1

Dấu = xảy ra khi x=0

c: \(P=A:\left(B-1\right)=\dfrac{3}{\sqrt{x}+3}:\dfrac{2\sqrt{x}+1-\sqrt{x}-3}{\sqrt{x}+3}=\dfrac{3}{\sqrt{x}-2}\)

Để P nguyên thì căn x-2\(\in\left\{1;-1;3;-3\right\}\)

=>\(x\in\left\{1;25\right\}\)

Đặt \(\left(\dfrac{1}{a};\dfrac{1}{b};\dfrac{1}{c}\right)=\left(x;y;z\right)\Rightarrow xy+yz+zx=1\)

\(P=\sqrt{\dfrac{yz}{x^2+1}}+\sqrt{\dfrac{zx}{y^2+1}}+\sqrt{\dfrac{xy}{z^2+1}}\)

\(P=\sqrt{\dfrac{yz}{x^2+xy+yz+zx}}+\sqrt{\dfrac{zx}{y^2+xy+yz+zx}}+\sqrt{\dfrac{xy}{z^2+xy+yz+zx}}\)

\(P=\sqrt{\dfrac{yz}{\left(x+y\right)\left(x+z\right)}}+\sqrt{\dfrac{zx}{\left(y+z\right)\left(x+y\right)}}+\sqrt{\dfrac{xy}{\left(x+z\right)\left(y+z\right)}}\)

\(P\le\dfrac{1}{2}\left(\dfrac{y}{x+y}+\dfrac{z}{x+z}\right)+\dfrac{1}{2}\left(\dfrac{z}{y+z}+\dfrac{x}{x+y}\right)+\dfrac{1}{2}\left(\dfrac{x}{x+z}+\dfrac{y}{y+z}\right)=\dfrac{3}{2}\)

\(P_{max}=\dfrac{3}{2}\) khi \(x=y=z=\dfrac{1}{\sqrt{3}}\) hay \(a=b=c=\sqrt{3}\)

Đề thi HSG Toán 9 Huyện Hoàng mia năm 2019-2020 đó