Có thể đặt \(u_n=v_n+2017\) nữa bác nhỉ, bác có công thức tổng quát tìm t không ạ: \(u_n=v_n+t\).

Delta xuất hiện cho mấy người lười đó =))

a) \(\left\{{}\begin{matrix}2x^2-5xy-y^2=1\\y\left(\sqrt{xy-2y^2}+\sqrt{4y^2-xy}\right)=1\end{matrix}\right.\)

ĐKXĐ:...

\(\Rightarrow y\left(\sqrt{xy-2y^2}+\sqrt{4y^2-xy}\right)=2x^2-5xy-y^2\)

Từ giả thiết dễ thấy \(y\ne0\), chia cả 2 vế cho \(y^2\) ta được:

\(\dfrac{\sqrt{xy-2y^2}+\sqrt{4y^2-xy}}{y}=\dfrac{2x^2-5xy-y^2}{y^2}\)

\(\Leftrightarrow\sqrt{\dfrac{xy-2y^2}{y^2}}+\sqrt{\dfrac{4y^2-xy}{y^2}}=2\left(\dfrac{x}{y}\right)^2-\dfrac{5x}{y}-1\)

\(\Leftrightarrow\sqrt{\dfrac{x}{y}-2}+\sqrt{4-\dfrac{x}{y}}=2\left(\dfrac{x}{y}\right)^2-5\dfrac{x}{y}-1\)

Đặt \(\dfrac{x}{y}=t\) \(\left(2\le t\le4\right)\)

\(\Leftrightarrow\sqrt{t-2}+\sqrt{4-t}=2t^2-5t-1\)

\(\Leftrightarrow\sqrt{t-2}-1+\sqrt{4-t}-1=2t^2-5t-3\)

\(\Leftrightarrow\left(t-3\right)\left(2t+1\right)=\dfrac{t-3}{\sqrt{t-2}+1}+\dfrac{3-t}{\sqrt{4-t}+1}\)

\(\Leftrightarrow\left(t-3\right)\left(2t+1-\dfrac{1}{\sqrt{t-2}+1}+\dfrac{1}{\sqrt{4-t}+1}\right)=0\)

Xét \(2t+1-\dfrac{1}{\sqrt{t-2}+1}+\dfrac{1}{\sqrt{4-t}+1}=2t+\dfrac{\sqrt{t-2}}{\sqrt{t-2}+1}+\dfrac{1}{\sqrt{4-t}+1}>0\forall t\)

\(\Rightarrow t-3=0\)

\(\Leftrightarrow t=3\)

\(\Leftrightarrow\dfrac{x}{y}=3\Leftrightarrow x=3y\)

Thế vào phương trình \(\left(1\right):2\cdot9y^2-5y\cdot3y-y^2-1=0\)

\(\Leftrightarrow2y^2-1=0\)

\(\Leftrightarrow y=\dfrac{1}{\sqrt{2}}\) do \(y>0\)

\(\Leftrightarrow x=\dfrac{3}{\sqrt{2}}\)

Vậy tập nghiệm của phương trình \(\left(x;y\right)=\left(\dfrac{3}{\sqrt{2}};\dfrac{1}{\sqrt{2}}\right)\)

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

Trừ theo vế 2 phương trình ta được:

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

\(\Leftrightarrow\left(x-y\right)\left(x^2+xy+y^2\right)-2\left(x-y\right)\left(x+y\right)+4\left(x-y\right)=0\)

\(\Leftrightarrow\left(x-y\right)\left(x^2+xy+y^2-2\left(x+y\right)+4\right)=0\)

Xét phương trình \(x^2+x\left(y-2\right)+y^2-2y+4=0\)

\(\Delta_x=\left(y-2\right)^2-4\left(y^2-2y+4\right)=-3y^2+4y-8< 0\) nên phương trình vô nghiệm.

Do đó \(x=y\)

Thế vào phương trình \(\left(1\right):x^3+1=2x^2\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1+\sqrt{5}}{2}\\x=\dfrac{1-\sqrt{5}}{2}\end{matrix}\right.\)

Vậy...

C545 bị sai đề nên mình sửa luôn, nếu không phải thì thôi...

\(\Sigma\dfrac{b^2c}{a^3\left(b+c\right)}\ge\dfrac{1}{2}\Sigma\left(\dfrac{1}{a}\right)\) \(\forall a,b,c>0\)

Giải: 

Xét \(\dfrac{b^2c}{a^3\left(b+c\right)}=\dfrac{1}{\dfrac{a^3}{b^2c}\left(b+c\right)}=\dfrac{1}{\dfrac{a^3}{b}\left(\dfrac{1}{b}+\dfrac{1}{c}\right)}=\dfrac{\dfrac{1}{a^3}}{\dfrac{1}{b}\left(\dfrac{1}{b}+\dfrac{1}{c}\right)}\)

Đặt \(\left(x;y;z\right)\rightarrow\left(\dfrac{1}{a};\dfrac{1}{b};\dfrac{1}{c}\right)\)

\(\dfrac{b^2c}{a^3\left(b+c\right)}=\dfrac{x^3}{y\left(y+z\right)}\)

Khi đó ta chỉ cần chứng minh \(\Sigma\dfrac{x^3}{y\left(y+z\right)}\ge\dfrac{1}{2}\left(x+y+z\right)\)

Áp dụng BĐT Cauchy: 

\(\dfrac{x^3}{y\left(y+z\right)}+\dfrac{y}{2}+\dfrac{y+z}{4}\ge3\sqrt[3]{\dfrac{x^3\cdot y\left(y+z\right)}{8y\left(y+z\right)}}=\dfrac{3x}{2}\)

\(\Leftrightarrow\dfrac{x^3}{y\left(y+z\right)}\ge\dfrac{3x}{2}-\dfrac{3y}{4}-\dfrac{z}{4}\)

\(\Rightarrow\Sigma\dfrac{x^3}{y\left(y+z\right)}\ge\dfrac{3}{2}\left(x+y+z\right)-\dfrac{3}{4}\left(x+y+z\right)-\dfrac{1}{4}\left(x+y+z\right)=\dfrac{1}{2}\left(x+y+z\right)\)

Ta có đpcm.

Dấu "=" xảy ra \(\Leftrightarrow x=y=z\Leftrightarrow a=b=c>0\)

Sửa đề thành \(\sqrt[3]{x+3}\) thì chắc sẽ giải được :>

\(\sum\limits^{2n+1}_{k=0}C^k_{2n+1}=\left(1+1\right)^{2n+1}=2^{2n+1}\)

Lại có \(C^0_{2n+1}+C^1_{2n+1}+...+C^n_{2n+1}=C^{2n+1}_{2n+1}+C^{2n}_{2n+1}+...+C^{n+1}_{2n+1}\)

\(\Rightarrow C^0_{2n+1}+C^1_{2n+1}+...C^n_{2n+1}=\dfrac{2^{2n+1}}{2}\)

\(\Leftrightarrow2^{20}-1=2^{2n}-C^0_{2n+1}\)

\(\Leftrightarrow2^{20}-1=2^{2n}-1\)

\(\Leftrightarrow2n=20\)

\(\Leftrightarrow n=10\)

\(\dfrac{1}{n^2-1}=\dfrac{1}{2}\cdot\dfrac{2}{\left(n-1\right)\left(n+1\right)}=\dfrac{1}{2}\left(\dfrac{1}{n-1}-\dfrac{1}{n+1}\right)\)

Khi đó:

 \(u_n=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{n-2}-\dfrac{1}{n}+\dfrac{1}{n-1}-\dfrac{1}{n+1}\right)\)

\(=\dfrac{1}{2}\left(1+\dfrac{1}{2}+\dfrac{1}{n}-\dfrac{1}{n+1}\right)=\dfrac{3n^2+3n+2}{4n\left(n+1\right)}\)

\(lim_{u_n}=lim\dfrac{3n^2+3n+2}{4n\left(n+1\right)}=lim\dfrac{3+\dfrac{3}{n}+\dfrac{2}{n^2}}{4\left(1+\dfrac{1}{n}\right)}=\dfrac{3}{4}\)