1. Với mọi số thực x;y;z ta có:

\(x^2+y^2+z^2+\dfrac{1}{2}\left(x^2+1\right)+\dfrac{1}{2}\left(y^2+1\right)+\dfrac{1}{2}\left(z^2+1\right)\ge xy+yz+zx+x+y+z\)

\(\Leftrightarrow\dfrac{3}{2}P+\dfrac{3}{2}\ge6\)

\(\Rightarrow P\ge3\)

\(P_{min}=3\) khi \(x=y=z=1\)

1.1

ĐKXĐ: ...

Đặt \(\left\{{}\begin{matrix}\dfrac{1}{\sqrt{x}}=a>0\\\dfrac{1}{\sqrt{y}}=b>0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a+\sqrt{2-b^2}=2\\b+\sqrt{2-a^2}=2\end{matrix}\right.\)

\(\Rightarrow a-b+\sqrt{2-b^2}-\sqrt{2-a^2}=0\)

\(\Leftrightarrow a-b+\dfrac{\left(a-b\right)\left(a+b\right)}{\sqrt{2-b^2}+\sqrt{2-a^2}}=0\)

\(\Leftrightarrow a=b\Leftrightarrow x=y\)

Thay vào pt đầu:

\(a+\sqrt{2-a^2}=2\Rightarrow\sqrt{2-a^2}=2-a\) (\(a\le2\))

\(\Leftrightarrow2-a^2=4-4a+a^2\Leftrightarrow2a^2-4a+2=0\)

\(\Rightarrow a=1\Rightarrow x=y=1\)

2.

\(\left\{{}\begin{matrix}x^2+xy+y^2=7\\\left(x^2+y^2\right)^2-x^2y^2=21\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2+xy+y^2=7\\\left(x^2+xy+y^2\right)\left(x^2-xy+y^2\right)=21\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2+xy+y^2=7\\x^2-xy+y^2=3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}3x^2+3xy+3y^2=21\\7x^2-7xy+7y^2=21\end{matrix}\right.\)

\(\Rightarrow4x^2-10xy+4y^2=0\)

\(\Leftrightarrow2\left(2x-y\right)\left(x-2y\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}y=2x\\y=\dfrac{1}{2}x\end{matrix}\right.\)

Thế vào pt đầu

...

Bài 1)

Ta biết ĐKXĐ:

\(\left\{\begin{matrix}4-x^2\ge0\\x^4-16\ge0\end{matrix}\right.\Leftrightarrow\left\{\begin{matrix}4-x^2\ge0\\\left(x^2-4\right)\left(x^2+4\right)\ge0\end{matrix}\right.\Rightarrow\left\{\begin{matrix}4-x^2\ge0\\x^2-4\ge0\end{matrix}\right.\Rightarrow x^2-4=0\rightarrow x=\pm2\)

Mặt khác \(4x+1\geq 0\Rightarrow x=2\)

Thay vào PT ban đầu : \(\Rightarrow 3+|y-1|=-y+5\Leftrightarrow |y-1|=2-y\)

Xét TH \(y-1\geq 0\) và \(y-1<0\) ta thu được \(y=\frac{3}{2}\)

Thu được cặp nghiệm \((x,y)=\left (2,\frac{3}{2}\right)\)

Bài 2)

BĐT cần chứng minh tương đương với:

\(\sqrt{\frac{z(x-z)}{xy}}+\sqrt{\frac{z(y-z)}{xy}}\leq 1\Leftrightarrow A=\left(\sqrt{\frac{z(x-z)}{xy}}+\sqrt{\frac{z(y-z)}{xy}}\right)^2\leq 1\)

Áp dụng BĐT Cauchy - Schwarz kết hợp AM-GM:

\(A\leq \left ( \frac{z}{y}+\frac{z}{x} \right )\left ( \frac{x-z}{x}+\frac{y-z}{y} \right )=\left ( \frac{z}{x}+\frac{z}{y} \right )\left ( 2-\frac{z}{x}-\frac{z}{y} \right )\)

\(\leq \left ( \frac{\frac{z}{x}+\frac{z}{y}+2-\frac{z}{x}-\frac{z}{y}}{2} \right )^2=1\)

Do đó ta có đpcm.

đề sai à

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:

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

-----------------------------------------------

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.

\(a^2+b^2=\left(a+b-c\right)^2=a^2+\left(b-c\right)^2+2a\left(b-c\right)=b^2+\left(a-c\right)^2+2b\left(a-c\right)\)

\(\Rightarrow\left\{{}\begin{matrix}b^2=\left(b-c\right)^2+2a\left(b-c\right)\\a^2=\left(a-c\right)^2+2b\left(a-c\right)\end{matrix}\right.\)

\(\Rightarrow\dfrac{a^2+\left(a-c\right)^2}{b^2+\left(b-c\right)^2}=\dfrac{\left(a-c\right)^2+2b\left(a-c\right)+\left(a-c\right)^2}{\left(b-c\right)^2+2a\left(b-c\right)+\left(b-c\right)^2}\)

\(=\dfrac{\left(a-c\right)\left(a+b-c\right)}{\left(b-c\right)\left(b+a-c\right)}=\dfrac{a-c}{b-c}\) (đpcm)

\(\sqrt{2x\left(y+z\right)}< =\dfrac{2x+y+z}{2}\)

=>\(\dfrac{1}{\sqrt{x\left(y+z\right)}}>=\dfrac{2\sqrt{2}}{2x+y+z}\)

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

\(\Leftrightarrow P>=2\sqrt{2}\cdot\dfrac{\left(1+1+1\right)^2}{\left(2x+y+z\right)+x+2y+z+x+y+2z}=\dfrac{18\sqrt{2}}{4\cdot18\sqrt{2}}=\dfrac{1}{4}\)

Dấu = xảy ra khi x=y=z=6căn 2

Ai giúp mình với, mình cần sự giúp đỡ, mai nộp bài rồi