a, b, c > 0 mà sao abc = 0 được vậy nhỉ:))

#)Góp ý :

Nguyễn Khang chuẩn :v

Rõ bảo mong k muốn ai thấy nick này mak cứ ló mặt ra lm chi ???

Lấy nick tth_new có ph nhanh hơn k ^^

#)Giải :

Bài 1 :

a) \(P=\left(\frac{\sqrt{x}-2}{x-1}-\frac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right)\left(\frac{1-x}{\sqrt{2}}\right)^2\)

\(=\left[\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}-\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}\right]\frac{\left(1-x\right)^2}{2}\)

\(=\frac{x-\sqrt{x}-2-x-\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}.\frac{\left(\sqrt{x}-1\right)^2\left(\sqrt{x+1}\right)^2}{2}\)

\(=\frac{-2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}.\frac{\left(\sqrt{x}-1\right)^2\left(\sqrt{x}+1\right)^2}{2}\)

\(=-\sqrt{x}\left(\sqrt{x}-1\right)\)

b) Để \(P>0\Rightarrow\hept{\begin{cases}\sqrt{x}>0\\1-\sqrt{x}>0\end{cases}\Rightarrow0< x< 1}\)

c) \(P=-x+\sqrt{x}=-\left(x-2\sqrt{x}.\frac{1}{2}+\frac{1}{4}\right)+\frac{1}{4}=-\left(\sqrt{x}-\frac{1}{2}\right)^2+\frac{1}{4}\le\frac{1}{4}\)

Dấu ''='' xảy ra khi \(x=\frac{1}{4}\)

a) \(A=\frac{3-\sqrt{3}}{\sqrt{3}}=\frac{\sqrt{3}\left(\sqrt{3}-1\right)}{\sqrt{3}}=\sqrt{3}-1\)

b) \(B=\frac{\sqrt{6+2\sqrt{5}}}{\sqrt{5}+1}=\frac{\sqrt{\left(\sqrt{5}+1\right)^2}}{\sqrt{5}+1}=\frac{\sqrt{5}+1}{\sqrt{5}+1}=1\)

c) \(C=\frac{2\sqrt{2}+\sqrt{6}}{4+\sqrt{12}}=\frac{2\sqrt{2}+\sqrt{6}}{4+2\sqrt{3}}=\frac{\left(2\sqrt{2}+\sqrt{6}\right)\left(4-2\sqrt{3}\right)}{\left(4+2\sqrt{3}\right)\left(4-2\sqrt{3}\right)}=\frac{2\sqrt{2}}{4}=\frac{\sqrt{2}}{2}\)

d) \(D=\frac{\sqrt{5+2\sqrt{6}}}{\sqrt{2}+\sqrt{3}}=\frac{\sqrt{5+2\sqrt{6}}\left(\sqrt{2}-\sqrt{3}\right)}{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}=-\sqrt{5+2\sqrt{6}}\left(\sqrt{2}-\sqrt{3}\right)\) 

= \(\frac{\sqrt{3}+\sqrt{11+6\sqrt{2}}-\sqrt{5+2\sqrt{6}}}{\sqrt{2}+\sqrt{6+2\sqrt{5}}-\sqrt{7+2\sqrt{10}}}\)

=\(\frac{\sqrt{3}+\sqrt{\left(3+\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{5}+\sqrt{2}\right)^2}}{\sqrt{2}+\sqrt{\left(\sqrt{5}+1\right)^2}-\sqrt{\left(\sqrt{2}+\sqrt{5}\right)^2}}\)

= \(\frac{\sqrt{3}+3+\sqrt{2}-\left(\sqrt{5}+\sqrt{2}\right)}{\sqrt{2}+\sqrt{5}+1-\left(\sqrt{2}+\sqrt{5}\right)}\)

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

= \(\sqrt{3}+\sqrt{5}+3\)

Bạn Khanh đúng r chỉ sai chỗ\(\sqrt{5+2\sqrt{6}}=\sqrt{\left(\sqrt{2}+\sqrt{3}\right)^2}\) mới đúng

\(x^2-2x+3=t\left(t\ge0\right)\)

\(pt\Leftrightarrow\frac{1}{t-1}+\frac{1}{t}=\frac{9}{2\left(t+1\right)}\)

\(\Leftrightarrow\frac{2t\left(t+1\right)}{2t\left(t^2-1\right)}+\frac{2\left(t^2-1\right)}{2t\left(t^2-1\right)}-\frac{9t\left(t-1\right)}{2t\left(t^2-1\right)}=0\)

\(\Leftrightarrow-5t^2+11t-2=0\)

\(\Leftrightarrow\left(5x-1\right)\left(x-2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{5}\\x=2\end{cases}}\)

\(=\left(\frac{\sqrt{y}}{\sqrt{x}\left(\sqrt{x}+\sqrt{y}\right)}-\frac{\sqrt{x}}{\sqrt{y}\left(\sqrt{x}+\sqrt{y}\right)}\right)\cdot\frac{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}{2\sqrt{xy}}\)

\(=\left(\frac{y-x}{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)}\right)\cdot\frac{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}{2\sqrt{xy}}\)

\(=\frac{\left(y-x\right)\left(\sqrt{x}-\sqrt{y}\right)}{2xy}\)

\(\left(\frac{\sqrt{y}}{x+\sqrt{xy}}-\frac{\sqrt{y}}{x-\sqrt{xy}}\right).\frac{x-y}{2\sqrt{xy}}\)

\(=\frac{x-y}{2\sqrt{xy}}.\left[-\frac{2y\sqrt{x}}{\left(x+\sqrt{yx}\right)\left(x-\sqrt{yx}\right)}\right]\)

\(=-\frac{2y\sqrt{x}}{\left(x+\sqrt{xy}\right)\left(x-\sqrt{xy}\right)}.\frac{x-y}{2\sqrt{xy}}\)

\(=-\frac{2y\sqrt{x}.\left(x-y\right)}{\left(x+\sqrt{xy}\right)\left(x-\sqrt{xy}\right).2\sqrt{xy}}\)

\(=-\frac{y\sqrt{x}\left(x-y\right)}{\left(x+\sqrt{xy}\right)\left(x-\sqrt{xy}\right).\sqrt{xy}}\)

\(=-\frac{\sqrt{y}\left(x-y\right)}{\left(x+\sqrt{yx}\right)\left(x-\sqrt{yx}\right)}\)

\(=-\frac{\sqrt{y}}{x}\)

Đặt \(\left(\sqrt{a};\sqrt{b};\sqrt{c}\right)\rightarrow\left(x;y;z\right)\)\(\Rightarrow\)\(x^2+y^2+z^2=4\)

\(P=\frac{x^3}{x+3y}+\frac{y^3}{y+3z}+\frac{z^3}{z+3x}=\frac{x^4}{x^2+3xy}+\frac{y^4}{y^2+3yz}+\frac{z^4}{z^2+3zx}\)

\(\ge\frac{\left(x^2+y^2+z^2\right)^2}{x^2+y^2+z^2+3\left(xy+yz+zx\right)}\ge\frac{\left(x^2+y^2+z^2\right)^2}{x^2+y^2+z^2+3\left(x^2+y^2+z^2\right)}=\frac{4^2}{4+3.4}=1\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(a=b=c=\frac{2}{\sqrt{3}}\)

à nhầm, \(a=b=c=\frac{4}{3}\) nhé