Ta có: \(A=\dfrac{15\sqrt{x}-11}{x+2\sqrt{x}-3}-\dfrac{3\sqrt{x}-2}{\sqrt{x}-1}-\dfrac{2\sqrt{x}+3}{\sqrt{x}+3}\)

\(=\dfrac{15\sqrt{x}-11-3x-9\sqrt{x}+2\sqrt{x}+6-\left(2x-2\sqrt{x}+3\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{-3x+8\sqrt{x}-5-2x-\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{-5\sqrt{x}+2}{\sqrt{x}+3}\)

\(\Leftrightarrow A-\dfrac{2}{3}=\dfrac{-5\sqrt{x}+2}{\sqrt{x}+3}-\dfrac{2}{3}\)

\(\Leftrightarrow A-\dfrac{2}{3}=\dfrac{-15\sqrt{x}+6-2\sqrt{x}-6}{3\left(\sqrt{x}+3\right)}\)

\(\Leftrightarrow A-\dfrac{2}{3}=\dfrac{-17\sqrt{x}}{3\left(\sqrt{x}+3\right)}\le0\)

\(\Leftrightarrow A\le\dfrac{2}{3}\)

Đặt \(\left(\sqrt{a};\sqrt{b};\sqrt{c}\right)=\left(x^2;y^2;z^2\right)\) với \(x;y;z>0\Rightarrow xyz=1\)

Đặt vế trái của BĐT cần chứng minh là P

Ta có: \(P=\dfrac{1}{x^2+2y^2+3}+\dfrac{1}{y^2+2z^2+3}+\dfrac{1}{z^2+2x^2+3}\)

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

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

\(P\le\dfrac{1}{2}\left(\dfrac{1}{xy+y+1}+\dfrac{1}{yz+z+1}+\dfrac{1}{zx+x+1}\right)=\dfrac{1}{2}\left(\dfrac{1}{xy+y+1}+\dfrac{xyz}{yz+z+xyz}+\dfrac{y}{xyz+xy+y}\right)\)

\(P\le\dfrac{1}{2}\left(\dfrac{1}{xy+y+1}+\dfrac{xy}{y+1+xy}+\dfrac{y}{1+xy+y}\right)=\dfrac{1}{2}\) (đpcm)

Dấu "=" xảy ra khi \(x=y=z=1\) hay \(a=b=c=1\)

Đặt \(\left(a;b;c\right)=\left(x^4;y^4;z^4\right)\Rightarrow xyz=1\)

\(VT=\dfrac{1}{x^2+2y^2+3}+\dfrac{1}{y^2+2z^2+3}+\dfrac{1}{z^2+2x^2+3}\)

\(VT=\dfrac{1}{x^2+y^2+y^2+1+2}+\dfrac{1}{y^2+z^2+z^2+1+2}+\dfrac{1}{z^2+x^2+x^2+1+2}\)

\(VT\le\dfrac{1}{2xy+2y+2}+\dfrac{1}{2yz+2z+2}+\dfrac{1}{2zx+2x+2}=\dfrac{1}{2}\)

Dấu "=" xảy ra khi \(a=b=c=1\)

a) = =

b) = = = . ( Với điều kiện b # 1)

c) \(\dfrac{a^{\dfrac{1}{3}}b^{-\dfrac{1}{3}-}a^{-\dfrac{1}{3}}b^{\dfrac{1}{3}}}{\sqrt[3]{a^2}-\sqrt[3]{b^2}}\)= = = ( với điều kiện a#b).

d) \(\dfrac{a^{\dfrac{1}{3}}\sqrt{b}+b^{\dfrac{1}{3}}\sqrt{a}}{\sqrt[6]{a}+\sqrt[6]{b}}\) = = = =

Đặt vế trái là T, ta có:

\(\dfrac{a}{\sqrt{b+1}}=\dfrac{a\sqrt{2}}{\sqrt{2}.\sqrt{b+1}}\ge\dfrac{a\sqrt{2}}{\dfrac{b+1+2}{2}}=\dfrac{a.2\sqrt{2}}{b+3}\)

Tương tự: \(\dfrac{b}{\sqrt{c+1}}\ge\dfrac{b.2\sqrt{2}}{c+3}\)

\(\dfrac{c}{\sqrt{a+1}}\ge\dfrac{c.2\sqrt{2}}{a+3}\)

Cộng vế theo vế các BĐT vừa chứng minh, ta được

\(T\ge2\sqrt{2}\left(\dfrac{a}{b+3}+\dfrac{b}{c+3}+\dfrac{c}{a+3}\right)=2\sqrt{2}\left(\dfrac{a^2}{ab+3a}+\dfrac{b^2}{bc+3b}+\dfrac{c^2}{ac+3c}\right)\)

\(T\ge2\sqrt{2}.\dfrac{\left(a+b+c\right)^2}{ab+bc+ca+3\left(a+b+c\right)}\)

\(T\ge2\sqrt{2}.\dfrac{\left(a+b+c\right)^2}{\dfrac{\left(a+b+c\right)^2}{3}+3\left(a+b+c\right)}\)

\(T\ge2\sqrt{2}.\dfrac{3^2}{\dfrac{3^2}{3}+9}=\dfrac{3\sqrt{2}}{2}\)(đpcm)

Đẳng thức xảy ra khi a=b=c=1

b) Đặt vế trái là N,ta có:

\(\sum\sqrt{\dfrac{a^3}{b+3}}=\sum\sqrt{\dfrac{a^4}{ab+3}}=\sum\dfrac{a^2}{\sqrt{ab+3}}=\sum\dfrac{2a^2}{\sqrt{4a\left(b+3\right)}}\ge\sum\dfrac{2a^2}{\dfrac{4a+b+3}{2}}=\sum\dfrac{4a^2}{4a+b+3}\)

\(\sum\dfrac{4a^2}{4a+b+3}\ge\dfrac{\left(2a+2b+2c\right)^2}{4a+b+3+4b+c+3+4c+a+3}=\dfrac{3}{2}\)(đpcm)

Đẳng thức xảy ra khi a=b=c=1

a) Ta có: \(A=\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}-\dfrac{\sqrt{x}}{3-\sqrt{x}}-\dfrac{3x+3}{x-9}\right):\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)

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

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

\(=\dfrac{-3}{\sqrt{x}+3}\)

b) Ta có: \(x=\sqrt{3+2\sqrt{2}}-\sqrt{3-2\sqrt{2}}\)

\(=\sqrt{2}+1-\sqrt{2}+1\)

=2

Thay x=2 vào A, ta được:

\(A=\dfrac{-3}{3+\sqrt{2}}=\dfrac{-9+3\sqrt{2}}{7}\)

\(a,ĐK:x\ge0;x\ne9\\ A=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}:\dfrac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\\ A=\dfrac{-3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}=\dfrac{-3}{\sqrt{x}+3}\\ b,x=13-4\sqrt{3}=\left(2\sqrt{3}-1\right)^2\\ \Leftrightarrow A=\dfrac{-3}{2\sqrt{3}-1+3}=\dfrac{-3}{2\sqrt{3}+2}=\dfrac{-3\left(2\sqrt{3}-2\right)}{8}\)

\(c,A< -\dfrac{1}{2}\Leftrightarrow\dfrac{-3}{\sqrt{x}+3}+\dfrac{1}{2}< 0\Leftrightarrow\dfrac{\sqrt{x}-3}{2\left(\sqrt{x}+3\right)}< 0\\ \Leftrightarrow\sqrt{x}-3< 0\left(\sqrt{x}+3>0\right)\\ \Leftrightarrow\sqrt{x}< 3\Leftrightarrow0\le x< 9\\ d,A=-\dfrac{2}{3}\Leftrightarrow\dfrac{3}{\sqrt{x}+3}=\dfrac{2}{3}\\ \Leftrightarrow2\sqrt{x}+6=9\\ \Leftrightarrow\sqrt{x}=\dfrac{3}{2}\Leftrightarrow x=\dfrac{9}{4}\left(tm\right)\\ e,\Leftrightarrow\sqrt{x}+3\inƯ\left(-3\right)=\left\{-3;-1;1;3\right\}\\ \Leftrightarrow\sqrt{x}=0\left(\sqrt{x}\ge0\right)\\ \Leftrightarrow x=0\left(tm\right)\\ f,\sqrt{x}+3\ge3\\ \Leftrightarrow A=-\dfrac{3}{\sqrt{x}+3}\ge-\dfrac{3}{3}=-1\\ A_{min}=-1\Leftrightarrow x=0\)

\(a,B=\dfrac{\sqrt{x}+1}{\sqrt{x}-2}+\dfrac{2}{\sqrt{x}+3}-\dfrac{9\sqrt{x}-3}{x+\sqrt{x}-6}\left(x>0;x\ne6\right)\\ =\dfrac{\sqrt{x}+1}{\sqrt{x}-2}+\dfrac{2}{\sqrt{x}+3}-\dfrac{9\sqrt{x}-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\\ =\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}+\dfrac{2\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}-\dfrac{9\sqrt{x}-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\\ =\dfrac{x+3\sqrt{x}+\sqrt{x}+3+2\sqrt{x}-4-9\sqrt{x}+3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\\ =\dfrac{x-3\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\\\)

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

`b,` Tớ tính mãi ko ra, xl cậu nha=')

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