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Pt đầu chắc là sai đề (chắc chắn), bạn kiểm tra lại

Với pt sau:

Nhận thấy một ẩn bằng 0 thì 2 ẩn còn lại cũng bằng 0, do đó \(\left(x;y;z\right)=\left(0;0;0\right)\) là 1 nghiệm

Với \(x;y;z\ne0\)

Từ pt đầu ta suy ra \(y>0\) , từ đó suy ra \(z>0\) từ pt 2 và hiển nhiên \(x>0\) từ pt 3

Do đó:

\(\left\{{}\begin{matrix}y=\dfrac{2x^2}{x^2+1}\le\dfrac{2x^2}{2x}=x\\z=\dfrac{3y^3}{y^4+y^2+1}\le\dfrac{3y^3}{3\sqrt[3]{y^4.y^2.1}}=y\\x=\dfrac{4z^4}{z^6+z^4+z^2+1}\le\dfrac{4z^4}{4\sqrt[4]{z^6z^4z^2}}=z\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}y\le x\\z\le y\\x\le z\end{matrix}\right.\) \(\Rightarrow x=y=z\)

Dấu "=" xảy ra khi và chỉ khi \(x=y=z=1\)

Vậy nghiệm của hệ là \(\left(x;y;z\right)=\left(0;0;0\right);\left(1;1;1\right)\)

a) \(\left\{{}\begin{matrix}5y-5x=xy\\\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{4}{5}\end{matrix}\right.\) \(\Leftrightarrow\)\(\left\{{}\begin{matrix}5y-5x=xy\\\dfrac{x+y}{xy}=\dfrac{4}{5}\end{matrix}\right.\)

\(\Leftrightarrow\)\(\left\{{}\begin{matrix}5y-5x=xy\\5\left(x+y\right)=4xy\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}5y-5x=xy\\5\left(x+y\right)=4\left(5y-5x\right)\end{matrix}\right.\)

\(\Leftrightarrow\)\(\left\{{}\begin{matrix}5y-5x=xy\\5x+5y=20y-20x\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}5y-5x=xy\\5x+5y-20y+20x=0\end{matrix}\right.\)

\(\Leftrightarrow\)\(\left\{{}\begin{matrix}5y-5x=xy\\-15y+25x=0\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}5y-5x=xy\\-5\left(3y-5x\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\)\(\left\{{}\begin{matrix}5y-5x=xy\\3y-5x=0\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}5y-5x=xy\\5x=3y\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}5y-3y=xy\\5x=3y\end{matrix}\right.\)

\(\Leftrightarrow\)\(\left\{{}\begin{matrix}2y=xy\\5x=3y\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x=2\\y=\dfrac{10}{3}\end{matrix}\right.\)

b) \(\left\{{}\begin{matrix}\dfrac{1}{2x-3y}+\dfrac{5}{3x+y}=\dfrac{5}{8}\\\dfrac{2}{2x-3y}-\dfrac{5}{3x+y}=\dfrac{-3}{8}\end{matrix}\right.\)

Đặt \(\dfrac{1}{2x-3y}=a;\dfrac{1}{3x+y}=b\)

=> hpt <=> \(\left\{{}\begin{matrix}a+5b=\dfrac{5}{8}\\2a-5b=\dfrac{-3}{8}\end{matrix}\right.\)

\(\Leftrightarrow\)\(\left\{{}\begin{matrix}a+5b=\dfrac{5}{8}\\2a-5b+a+5b=\dfrac{-3}{8}+\dfrac{5}{8}=0,25\end{matrix}\right.\)

\(\Leftrightarrow\)\(\left\{{}\begin{matrix}a+5b=\dfrac{5}{8}\\3a=0,25\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a+5b=\dfrac{5}{8}\\a=\dfrac{1}{12}\end{matrix}\right.\)

\(\Leftrightarrow\)\(\left\{{}\begin{matrix}a=\dfrac{1}{12}\\b=\dfrac{13}{120}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{2x-3y}=\dfrac{1}{12}\\\dfrac{1}{3x+y}=\dfrac{13}{120}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x-3y=12\\3x+y=\dfrac{120}{13}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{516}{143}\\y=-\dfrac{228}{143}\end{matrix}\right.\)

c) \(\left\{{}\begin{matrix}x-y=2\\y-3z=2\\-3x-2y+z=-2\end{matrix}\right.\)

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

\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x=y+2\\y=3z+2\\-3y-6-6z-4+z=-2\end{matrix}\right.\)

\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x=y+2\\y=3z+2\\-3y-5z=8\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x=y+2\\y=3z+2\\-3\left(3z+2\right)-5z=8\end{matrix}\right.\)

\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x=y+2\\y=3z+2\\-9z-6-5z=8\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x=y+2\\y=3z+2\\-14z=14\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x=\left(-1\right)+2=1\\y=3\left(-1\right)+2=-1\\z=-1\end{matrix}\right.\)

Vậy...

4) Áp dụng bất đẳng thức Bunyakovsky

\(\Rightarrow\left(x^4+yz\right)\left(1+1\right)\ge\left(x^2+\sqrt{yz}\right)^2\)

\(\Rightarrow\dfrac{x^2}{x^4+yz}\le\dfrac{2x^2}{\left(x^2+\sqrt{yz}\right)^2}\)

Tượng tự ta có \(\left\{{}\begin{matrix}\dfrac{y^2}{y^4+xz}\le\dfrac{2y^2}{\left(y^2+\sqrt{xz}\right)^2}\\\dfrac{z^2}{z^4+xy}\le\dfrac{2z^2}{\left(z^2+\sqrt{xy}\right)^2}\end{matrix}\right.\)

\(\Rightarrow VT\le2\left[\dfrac{x^2}{\left(x^2+\sqrt{yz}\right)^2}+\dfrac{y^2}{\left(y^2+\sqrt{xz}\right)^2}+\dfrac{z^2}{\left(z^2+\sqrt{xy}\right)}\right]\)

Chứng minh rằng \(2\left[\dfrac{x^2}{\left(x^2+\sqrt{yz}\right)^2}+\dfrac{y^2}{\left(y^2+\sqrt{xz}\right)^2}+\dfrac{z^2}{\left(z^2+\sqrt{xy}\right)}\right]\le\dfrac{3}{2}\)

\(\Leftrightarrow\dfrac{x^2}{\left(x^2+\sqrt{yz}\right)^2}+\dfrac{y^2}{\left(y^2+\sqrt{xz}\right)^2}+\dfrac{z^2}{\left(z^2+\sqrt{xy}\right)^2}\le\dfrac{3}{4}\)

Áp dụng bất đẳng thức Cauchy

\(\Rightarrow x^2+\sqrt{yz}\ge2\sqrt{x^2\sqrt{yz}}=2x\sqrt{\sqrt{yz}}\)

\(\Rightarrow\left(x^2+\sqrt{yz}\right)^2\ge4x^2\sqrt{yz}\)

\(\Rightarrow\dfrac{x^2}{\left(x^2+\sqrt{yz}\right)^2}\le\dfrac{x^2}{4x^2\sqrt{yz}}=\dfrac{1}{4\sqrt{yz}}\)

Tượng tự ta có \(\left\{{}\begin{matrix}\dfrac{y^2}{\left(y^2+\sqrt{xz}\right)^2}\le\dfrac{1}{4\sqrt{xz}}\\\dfrac{z^2}{\left(z^2+\sqrt{xy}\right)^2}\le\dfrac{1}{4\sqrt{xy}}\end{matrix}\right.\)

\(\Leftrightarrow\dfrac{x^2}{\left(x^2+\sqrt{yz}\right)^2}+\dfrac{y^2}{\left(y^2+\sqrt{xz}\right)^2}+\dfrac{z^2}{\left(z^2+\sqrt{xy}\right)^2}\le\dfrac{1}{4}\left(\dfrac{1}{\sqrt{xy}}+\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{xz}}\right)\)

Chứng minh rằng \(\dfrac{1}{4}\left(\dfrac{1}{\sqrt{xy}}+\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{xz}}\right)\le\dfrac{3}{4}\)

\(\Leftrightarrow\dfrac{1}{\sqrt{xy}}+\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{xz}}\le3\)

Theo đề bài ta có \(x^2+y^2+z^2=3xyz\)

\(\Rightarrow\dfrac{x}{yz}+\dfrac{y}{xz}+\dfrac{z}{xy}=3\)

\(\Rightarrow\dfrac{1}{\sqrt{xy}}+\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{xz}}\le3\)

\(\Leftrightarrow\dfrac{1}{\sqrt{xy}}+\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{xz}}\le\dfrac{x}{yz}+\dfrac{y}{xz}+\dfrac{z}{xy}\)

Áp dụng bất đẳng thức Cauchy

\(\Rightarrow\dfrac{1}{\sqrt{xy}}\le\dfrac{\dfrac{1}{x}+\dfrac{1}{y}}{2}\)

Tượng tự ta có \(\left\{{}\begin{matrix}\dfrac{1}{\sqrt{xz}}\le\dfrac{\dfrac{1}{x}+\dfrac{1}{z}}{2}\\\dfrac{1}{\sqrt{yz}}\le\dfrac{\dfrac{1}{z}+\dfrac{1}{y}}{2}\end{matrix}\right.\)

\(\Rightarrow\dfrac{1}{\sqrt{xy}}+\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{xz}}\le\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\) (1)

Áp dụng bất đẳng thức Cauchy

\(\Rightarrow\dfrac{x}{yz}+\dfrac{y}{xz}\ge2\sqrt{\dfrac{1}{z^2}}=\dfrac{2}{z}\)

Tượng tự ta có \(\left\{{}\begin{matrix}\dfrac{y}{xz}+\dfrac{z}{xy}\ge\dfrac{2}{x}\\\dfrac{x}{zy}+\dfrac{z}{xy}\ge\dfrac{2}{y}\end{matrix}\right.\)

\(\Rightarrow2\left(\dfrac{x}{yz}+\dfrac{y}{xz}+\dfrac{z}{xy}\right)\ge2\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\)

\(\Leftrightarrow\dfrac{x}{yz}+\dfrac{y}{xz}+\dfrac{z}{xy}\ge\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\) (2)

Từ (1) và (2)

\(\Rightarrow\dfrac{1}{\sqrt{xy}}+\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{xz}}\le3\) ( đpcm )

Vậy \(\dfrac{1}{4}\left(\dfrac{1}{\sqrt{xy}}+\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{xz}}\right)\le\dfrac{3}{4}\)

\(\Rightarrow2\left[\dfrac{x^2}{\left(x^2+\sqrt{yz}\right)^2}+\dfrac{y^2}{\left(y^2+\sqrt{xz}\right)^2}+\dfrac{z^2}{\left(z^2+\sqrt{xy}\right)}\right]\le\dfrac{3}{2}\)

Mà \(VT\le2\left[\dfrac{x^2}{\left(x^2+\sqrt{yz}\right)^2}+\dfrac{y^2}{\left(y^2+\sqrt{xz}\right)^2}+\dfrac{z^2}{\left(z^2+\sqrt{xy}\right)}\right]\)

\(\Rightarrow VT\le\dfrac{3}{2}\) ( đpcm )

Dấu " = " xảy ra khi \(x=y=z=1\)

3. Ta có :\(x^2\left(1-2x\right)=x.x.\left(1-2x\right)\le\dfrac{\left(x+x+1-2x\right)^3}{27}=\dfrac{1}{27}\)(bđt cô si)

Dấu "=" xảy ra khi :x=1-2x\(\Leftrightarrow x=\dfrac{1}{3}\)

Vậy max của Qlaf 1/27 khi x=1/3

Bài 1:\(HpT\Leftrightarrow\left\{{}\begin{matrix}\left(x-1\right)^3+\left(\dfrac{3}{y}\right)^3=1\\\left(x-1\right)^2+\left(\dfrac{3}{y}\right)^2=1\end{matrix}\right.\)Đẹp !!

Bài 2:phân tích đc thành (n+1)(n-1)(n+3)(n-3)

đến đây mình tịt ah

Bài 4:

góp thêm 1 cách :(vắn tắt thôi )

\(GT\Leftrightarrow3=\dfrac{x}{yz}+\dfrac{y}{xz}+\dfrac{z}{xy}\ge\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\)(AM-GM)

\(VT\le\sum\dfrac{1}{2\sqrt{yz}}\le\sum\dfrac{1}{4}\left(\dfrac{1}{y}+\dfrac{1}{z}\right)=\dfrac{1}{2}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)=\dfrac{3}{2}\)

Sau vài phút cố gắng thì khẳng định đề bài của em bị sai

Lời giải:
$xy+yz+xz=\frac{1}{2}[(x+y+z)^2-(x^2+y^2+z^2)]=\frac{1}{2}(a^2-b^2)$

$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{c}$

$\Rightarrow xyz=c(xy+yz+xz)=\frac{1}{2}c(a^2-b^2)$

Khi đó:

$P=(x+y+z)^3-3(x+y)(y+z)(x+z)$

$=(x+y+z)^3-3[(x+y+z)(xy+yz+xz)-xyz]=(x+y+z)^3-3(xy+yz+xz)(x+y+z)+3xyz$
$=a^3-\frac{3}{2}a(a^2-b^2)+\frac{3}{2}c(a^2-b^2)$

Đề này còn có lý, lần sau chú ý đọc kĩ đề trước khi đăng lên, tránh làm mất thời gian vô ích:

\(\left|x-2y\right|\le\dfrac{1}{\sqrt{x}}\Rightarrow1\ge\sqrt{x}\left|x-2y\right|\Rightarrow1\ge x\left(x-2y\right)^2\)

\(\Rightarrow1\ge x^3-4x^2y+4xy^2\)

Tương tự: \(\dfrac{1}{\sqrt{y}}\ge\left|y-2x\right|\Rightarrow1\ge y^3-4xy^2+4xy^2\)

Cộng vế:

\(\Rightarrow2\ge x^3+y^3=\dfrac{1}{2}\left(x^3+x^3+1\right)+\left(y^3+1+1\right)-\dfrac{5}{2}\ge\dfrac{1}{2}.3x^2+3y-\dfrac{3}{2}=\dfrac{3}{2}\left(x^2+2y\right)-\dfrac{5}{2}\)

\(\Rightarrow\dfrac{3}{2}\left(x^2+2y\right)\le\dfrac{9}{2}\Rightarrow x^2+2y\le3\)