Câu hỏi của Đàm Vũ Đức Anh

Question

1 Giải hệ pt $\left\{{}\begin{matrix}\left(x-1\right)^3=1-\dfrac{27}{y^3}\x^2+\dfrac{9}{y^2}=2x\end{matrix}\right.$

2 CM  $n^4-10n^2+9$ chia hết 384 với mọi n lẻ

3 cho $0\le x\le\dfrac{1}{2}$...

Kuro Kazuya · Accepted Answer

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 $x=y=z=1$Câu trả lời: 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 $x=y=z=1$

Bùi Nhất Duy · Answer

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=1/3Câu trả lời: 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

Neet · Answer

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)

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

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

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