ĐKXĐ: \(x\ge1;y\ge1\)

Ta có: \(\frac{x^2-4}{x}+\frac{y^2-4}{y}+8=4\left(\sqrt{x-1}+\sqrt{y-1}\right)\)

\(\Leftrightarrow\frac{x^2-4}{x}+\frac{y^2-4}{y}=4\left[\left(\sqrt{x-1}-1\right)+\left(\sqrt{y-1}+1\right)\right]\)

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

\(\Leftrightarrow\left(x-2\right)\left(\frac{x+2}{x}-\frac{4}{\sqrt{x-1}+1}\right)+\left(y-2\right)\left(\frac{y+2}{y}-\frac{4}{\sqrt{y-1}+1}\right)=0\)

\(\Leftrightarrow\left(x-2\right)\frac{x\sqrt{x-1}+2\sqrt{x-1}+2+x-4x}{x\left(\sqrt{x-1}+1\right)}+\left(y-2\right)\frac{y\sqrt{y-1}+2\sqrt{y-1}+y-4y}{y\left(\sqrt{y-1}+1\right)}=0\)

\(\Leftrightarrow\left(x-2\right)\frac{\left( x-1\right)\sqrt{x-1}+3\sqrt{x-1}-3\left(x-1\right)-1}{x\left(\sqrt{x-1}+1\right)}\)

      \(+\left(y-2\right)\frac{\left(y-1\right)\sqrt{y-1}+3\sqrt{y-1}-3\left(y-1\right)-1}{y\left(\sqrt{y-1}+1\right)}=0\)

\(\Leftrightarrow\left(x-2\right)\frac{\left(\sqrt{x-1}-1\right)^3}{x\left(\sqrt{x-1}+1\right)}+\left(y-2\right)\frac{\left(\sqrt{y-1}-1\right)^3}{y\left(\sqrt{y-1}+1\right)}=0\)

\(\Leftrightarrow\left(x-2\right)\frac{\left(\sqrt{x-1}-1\right)^3\left(\sqrt{x-1}+1\right)^3}{x\left(\sqrt{x-1}+1\right)^4}+\left(y-2\right)\frac{\left(\sqrt{y-1}-1\right)^3\left(\sqrt{y-1}+1\right)^3}{y\left(\sqrt{y-1}+1\right)^4}=0\)

\(\Leftrightarrow\frac{\left(x-2\right)^4}{x\left(\sqrt{x-1}+1\right)^4}+\frac{\left(y-2\right)^4}{y\left(\sqrt{y-1}+1\right)^4}=0\)

Vì \(x\ge1;y\ge1\Rightarrow\frac{\left(x-2\right)^4}{x\left(\sqrt{x-1}+1\right)^4}\ge0;\frac{\left(y-2\right)^4}{y\left(\sqrt{y-1}+1\right)^4}\ge0\)\(\Rightarrow\frac{\left(x-2\right)^4}{x\left(\sqrt{x-1}+1\right)^4}+\frac{\left(y-2\right)^4}{y\left(\sqrt{y-1}+1\right)^4}\ge0\)

Do đó dấu ''='' xảy ra khi \(\frac{\left(x-2\right)^4}{x\left(\sqrt{x-1}+1\right)^4}=\frac{\left(y-2\right)^4}{y\left(\sqrt{y-1}+1\right)^4}=0\Leftrightarrow x-2=y-2=0\Leftrightarrow x=y=2\)

Vậy \(x=y=2\).

Anh tham khảo tại đây:

Câu hỏi của jihoon oppa! I'm May - Toán lớp 8 - Học toán với OnlineMath

zZz Cool Kid zZz cái đó cô chi nhầm nên bị ngược dấu rồi nhé! Mình đang nghĩ cách:(

1) Ta có: \(2020^2=\left(2019+1\right)^2=2019^2+2.2019+1.\)

\(\Rightarrow1+2019^2=2020^2-2.2019\)

\(\Rightarrow M=\sqrt{1+2019^2+\frac{2019^2}{2020^2}}+\frac{2019}{2020}=\sqrt{2020^2-2.2019+\frac{2019^2}{2020^2}}+\frac{2019}{2020}\)

\(=\sqrt{2020^2-2.2020.\frac{2019}{2020}+\left(\frac{2019}{2020}\right)^2}+\frac{2019}{2020}\)

\(=\sqrt{\left(2020-\frac{2019}{2020}\right)^2}+\frac{2019}{2020}=2020-\frac{2019}{2020}+\frac{2019}{2020}\)

\(=2020\)

Vậy M=2020.

2) Xét  : \(k\in N;k\ge2\)ta có:

\(\left(1+\frac{1}{k-1}-\frac{1}{k}\right)^2=1+\frac{1}{\left(k-1\right)^2}+\frac{1}{k^2}+\frac{2}{k-1}-\frac{2}{\left(k-1\right)k}-\frac{2}{k}\)

                                          \(=1+\frac{1}{\left(k-1\right)^2}+\frac{1}{k^2}+\frac{2}{k-1}-\frac{2}{k-1}+\frac{2}{k}-\frac{2}{k}\)

\(\Rightarrow\left(1+\frac{1}{k-1}-\frac{1}{k}\right)^2=1+\frac{1}{\left(k-1\right)^2}+\frac{1}{k^2}\)

\(\Rightarrow\sqrt{1+\frac{1}{\left(k-1\right)^2}+\frac{1}{k^2}}=1+\frac{1}{k-1}+\frac{1}{k}\)

Cho \(k=3,4,...,2020.\)Ta có:

\(N=\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+\sqrt{1+\frac{1}{3^2}+\frac{1}{4^2}}+...+\sqrt{1+\frac{1}{2019^2}+\frac{1}{2020^2}}\)

\(=\left(1+\frac{1}{2}-\frac{1}{3}\right)+\left(1+\frac{1}{3}-\frac{1}{4}\right)+...+\left(1+\frac{1}{2018}-\frac{1}{2019}\right)+\left(1+\frac{1}{2019}-\frac{1}{2020}\right)\)

\(=2018+\frac{1}{2}-\frac{1}{2020}=2018\frac{1009}{2020}\)

Vậy \(N=2018\frac{1009}{2020}.\)

Làm mẫu 1 phần :

a) \(|3x-1|+|x-1|=4\left(1\right)\)

Ta có: \(3x-1=0\Leftrightarrow x=\frac{1}{3}\)

             \(x-1=0\Leftrightarrow x=1\)

Lập bảng xét dấu :

+) Với \(x< \frac{1}{3}\Rightarrow\hept{\begin{cases}3x-1< 0\\x-1< 0\end{cases}\Rightarrow\hept{\begin{cases}|3x-1|=1-3x\\|x-1|=1-x\end{cases}\left(2\right)}}\)

Thay (2) vào (1) ta được :

\(\left(1-3x\right)+\left(1-x\right)=4\)

\(2-4x=4\)

\(4x=-2\)

\(x=\frac{-1}{2}\)( chọn )

+) Với \(\frac{1}{3}\le x< 1\Rightarrow\hept{\begin{cases}3x-1>0\\x-1< 0\end{cases}\Rightarrow\hept{\begin{cases}|3x-1|=3x-1\\|x-1|=1-x\end{cases}\left(3\right)}}\)

Thay (3) vào (1) ta được :
\(\left(3x-1\right)+\left(1-x\right)=4\)

\(2x=4\)

\(x=2\)( chọn )

+) Với \(x\ge1\Rightarrow\hept{\begin{cases}3x-1>0\\x-1>0\end{cases}\Rightarrow}\hept{\begin{cases}|3x-1|=3x-1\\|x-1|=x-1\end{cases}\left(4\right)}\)

Thay (4) vào (1) ta được :

\(\left(3x-1\right)+\left(x-1\right)=4\)

\(4x-2=4\)

\(4x=6\)

\(x=\frac{3}{2}\)( chọn )

Vậy \(x\in\left\{\frac{-1}{2};2;\frac{3}{2}\right\}\)

1,44224957+2,080083823=3,522333393 \(\in\)I

Liên quan gì bạn @Tam Mai, chứng minh chứ không phải bấm máy tính