Chọn B

Áp dụng BĐT Cauchy cho cặp số dương \(\dfrac{1}{\left(z+x\right)};\dfrac{1}{\left(z+y\right)}\)

\(\dfrac{1}{\left(z+x\right)}+\dfrac{1}{\left(z+y\right)}\ge\dfrac{1}{2}.\dfrac{1}{\sqrt[]{\left(z+x\right)\left(z+y\right)}}\)

\(\Rightarrow\dfrac{xy}{\sqrt[]{\left(z+x\right)\left(z+y\right)}}\le\dfrac{2xy}{z+x}+\dfrac{2xy}{z+y}\left(1\right)\)

Tương tự ta được

\(\dfrac{zx}{\sqrt[]{\left(y+z\right)\left(y+x\right)}}\le\dfrac{2zx}{y+z}+\dfrac{2zx}{y+x}\left(2\right)\)

\(\dfrac{yz}{\sqrt[]{\left(x+y\right)\left(x+z\right)}}\le\dfrac{2yz}{x+y}+\dfrac{2yz}{x+z}\left(3\right)\)

\(\left(1\right)+\left(2\right)+\left(3\right)\) ta được :

\(P=\dfrac{yz}{\sqrt[]{\left(x+y\right)\left(x+z\right)}}+\dfrac{zx}{\sqrt[]{\left(y+z\right)\left(y+x\right)}}+\dfrac{xy}{\sqrt[]{\left(z+x\right)\left(z+y\right)}}\le\dfrac{2yz}{x+y}+\dfrac{2yz}{x+z}+\dfrac{2zx}{y+z}+\dfrac{2zx}{y+x}+\dfrac{2xy}{z+x}+\dfrac{2xy}{z+y}\)

\(\Rightarrow P\le2\left(x+y+z\right)=2.3=6\)

\(\Rightarrow GTLN\left(P\right)=6\left(tạix=y=z=1\right)\)

\(\dfrac{1}{2}x^2-\left(-2+1\right)x+\dfrac{-2-1}{2}=0\)

\(\Rightarrow\dfrac{1}{2}x^2+x-\dfrac{3}{2}=0\)

Tới đây dùng \(\Delta\) chứ, nếu bn lấy \(\dfrac{1}{2}\) đặt lm nhân tử chung thì ở đây hơi vô lí 

\(\Delta=b^2-4ac=1-4.\dfrac{1}{2}.\left(-\dfrac{3}{2}\right)=4>0\)

\(\Rightarrow\)Pt có 2 nghiệm phân biệt

\(\left\{{}\begin{matrix}x_1=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{-1+2}{1}=1\\x_2=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{-1-2}{1}=-3\end{matrix}\right.\)

Thay \(x_1=1\) vào \(y=\dfrac{1}{2}x^2\Rightarrow y=\dfrac{1}{2}\)

Thay \(x_2=-3\) vào \(y=-x+\dfrac{3}{2}\Rightarrow y=\dfrac{9}{2}\)

Đề bài sai, biểu thức này ko có min

a. \(\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)=x-3\sqrt{x} +2\sqrt{x}-6=x-\sqrt{x}-6\)

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

c. \(\left(\sqrt{\dfrac{25}{3}}-\sqrt{\dfrac{49}{3}}+\sqrt{3}\right).\sqrt{3}\)

\(=\left(\dfrac{5}{\sqrt{3}}-\dfrac{7}{\sqrt{3}}+\sqrt{3}\right).\sqrt{3}=\dfrac{5}{3}-\dfrac{7}{3}+9=\dfrac{25}{3}\)

d. \(\left(1+\sqrt{3}-\sqrt{5}\right)\left(1+\sqrt{3}+\sqrt{5}\right)\)

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

Lời giải:

Từ \(x+y+z=xyz\Rightarrow \frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}=1\)

Đặt \((\frac{1}{a}, \frac{1}{b}, \frac{1}{c})=(x,y,z)\), trong đó $a,b,c>0$ thì ta có:

\(ab+bc+ac=1\) và cần phải CMR:

\(P=\frac{\sqrt{(\frac{1}{b^2}+1)(\frac{1}{c^2}+1})-\sqrt{\frac{1}{b^2}+1}-\sqrt{\frac{1}{c^2}+1}}{\frac{1}{bc}}+\frac{\sqrt{(\frac{1}{c^2}+1)(\frac{1}{a^2}+1})-\sqrt{\frac{1}{c^2}+1}-\sqrt{\frac{1}{a^2}+1}}{\frac{1}{ac}}+\frac{\sqrt{(\frac{1}{a^2}+1)(\frac{1}{b^2}+1})-\sqrt{\frac{1}{a^2}+1}-\sqrt{\frac{1}{b^2}+1}}{\frac{1}{ab}}\)

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Ta có:
\(\frac{\sqrt{(\frac{1}{b^2}+1)(\frac{1}{c^2}+1})-\sqrt{\frac{1}{b^2}+1}-\sqrt{\frac{1}{c^2}+1}}{\frac{1}{bc}}=\sqrt{(b^2+1)(c^2+1)}-b\sqrt{c^2+1}-c\sqrt{b^2+1}\)

\(=\sqrt{(b^2+ab+bc+ac)(c^2+ac+bc+ab)}-b\sqrt{c^2+ac+bc+ab}-c\sqrt{b^2+ab+bc+ac}\)

\(=\sqrt{(b+a)(b+c)(c+a)(c+b)}-b\sqrt{(c+a)(c+b)}-c\sqrt{(b+a)(b+c)}\)

\(=(b+c)\sqrt{(a+b)(a+c)}-b\sqrt{(c+a)(c+b)}-c\sqrt{(b+a)(b+c)}(1)\)

Tương tự:

\(\frac{\sqrt{(\frac{1}{c^2}+1)(\frac{1}{a^2}+1})-\sqrt{\frac{1}{c^2}+1}-\sqrt{\frac{1}{a^2}+1}}{\frac{1}{ac}}=(a+c)\sqrt{(b+a)(b+c)}-a\sqrt{(c+a)(c+b)}-c\sqrt{(a+b)(a+c)}(2)\)

\(\frac{\sqrt{(\frac{1}{a^2}+1)(\frac{1}{b^2}+1})-\sqrt{\frac{1}{a^2}+1}-\sqrt{\frac{1}{b^2}+1}}{\frac{1}{ab}}=(a+b)\sqrt{(c+a)(c+b)}-b\sqrt{(a+b)(a+c)}-a\sqrt{(b+c)(b+a)}(3)\)

Từ \((1);(2);(3)\Rightarrow P=(b+c-c-b)\sqrt{(a+b)(a+c)}+(a+c-c-a)\sqrt{(b+a)(b+c)}+(a+b-b-a)\sqrt{(c+a)(c+b)}\)

\(=0\)

Ta có đpcm.