Trả lời:

sửa đề: \(\frac{x^2+y^2+z^2-2xy+2xz-2yz}{x^2-2xy+y^2-z^2}\)

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

Câu hỏi của Vũ Thảo Vy - Toán lớp 8 - Học toán với OnlineMath tham khảo

\(\frac{1}{1.2}+\frac{1}{3.4}+....+\frac{1}{99.100}=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{99}-\frac{1}{100}\)

\(=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{100}-2.\left(\frac{1}{2}+\frac{1}{4}+.....+\frac{1}{100}\right)\)

\(=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{100}-1-\frac{1}{2}-\frac{1}{3}-....-\frac{1}{50}=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)

=> \(2013x.\left(\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)=2013x.\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\right)\)

=> \(2013x.\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\right)=2012.\left(\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\right)\Rightarrow2013x=2012\Rightarrow x=\frac{2012}{2013}\)

Vậy \(x=\frac{2012}{2013}\)

p/s: --trình bày sai sót mong bỏ qua 

ko hiểu

cho hỏi chị là A.R.M.Y đúng ko ạ ???

uk đúng r

Đặt \(A=k^4-8k^3+23k^2-26k+10\)

\(=k^3\left(k-1\right)-7k^2\left(k-1\right)+16k\left(k-1\right)-10\left(k-1\right)\)

\(=\left(k-1\right)\left(k^3-7k^2+16k-10\right)\)

\(=\left(k-1\right)\left[k^2\left(k-1\right)-6k\left(k-1\right)+10\left(k-1\right)\right]\)

\(=\left(k-1\right)^2\left(k^2-6k+10\right)\)

Để A là số chính phương thì \(k^2-6k+10\) là số chính phương hoặc \(\orbr{\begin{cases}k-1=0\\k^2-6k+10=0\end{cases}}\)

-Nếu k² - 6k + 10 là số chính phương thì ta đặt \(k^2-6k+10=t^2\left(t\in Z\right)\)

\(\Rightarrow\left(k-3\right)^2+1=t^2\)

\(\Rightarrow\left(k-3\right)^2-t^2=-1\)

\(\Rightarrow\left(k-t-3\right)\left(k+t-3\right)=-1\)

Vì k,t là số nguyên nên ta có: 

\(TH1:\hept{\begin{cases}k-t-3=-1\\k+t-3=1\end{cases}\Rightarrow}\hept{\begin{cases}k-t=2\\k+t=4\end{cases}\Rightarrow k=\left(2+4\right):2=3}\)

\(TH2:\hept{\begin{cases}k-t-3=1\\k+t-3=-1\end{cases}\Rightarrow}\hept{\begin{cases}k-t=4\\k+t=2\end{cases}\Rightarrow}k=\left(4+2\right):2=3\)

-Nếu \(\orbr{\begin{cases}k-1=0\\k^2-6k+10=0\end{cases}}\)

Mà \(k^2-6k+10=\left(x-3\right)^2+1>0\forall x\)

\(\Rightarrow k-1=0\Rightarrow k=1\) (thỏa mãn)

Vậy \(k\in\left\{1;3\right\}\)

Đặt \(B=k^4-8k^3+23k^2-26k+10\)

\(=\left(k^4-2k^2+1\right)-8k\left(k^2-2k+1\right)+9k^2-18k+1\)

\(=\left(k^2-1\right)^2-8k\left(k-1\right)^2+9\left(k-1\right)^2\)

\(=\left(k-1\right)^2\left[\left(k-3\right)^2+1\right]\)

Vì B là SCP

\(\Rightarrow\left(k-1\right)^2=0\)hoặc \(\left(k-3\right)^2+1\)là SCP

\(TH1:\left(k-1\right)^2=0\Rightarrow k-1=0\Rightarrow k=1\)

\(TH2:\left(k-3\right)^2+1\)

Đặt \(\left(k-3\right)^2+1=n^2\left(n\inℤ\right)\)

\(\Leftrightarrow n^2-\left(k-3\right)^2=1\)

\(\Leftrightarrow\left(n-k+3\right)\left(n+k-3\right)=1\)

\(\Leftrightarrow\hept{\begin{cases}n-k+3=1\\n+k-3=1\end{cases}}\)

hoặc \(\hept{\begin{cases}n-k+3=-1\\n-k+3=-1\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}n=1;k=3\\n=-1;k=3\end{cases}}\Rightarrow k=3\)

Vậy ....

\(16x^4-8x^2+1=\left(4x^2\right)^2-2.4x^2.1+1=\left(4x^2-1\right)^2\ge0\forall x\)

\(\Rightarrow16x^4+1\ge8x^2\)(1)

\(y^4-2y^2+1=\left(y^2-1\right)^2\ge0\forall y\)

\(\Rightarrow y^4+1\ge2y^2\)(2)

Từ (1) và (2) \(\Rightarrow\left(16x^4+1\right)\left(y^4+1\right)\ge8x^2.2y^2=16x^2y^2\)

Dấu "=" xảy ra khi: \(\hept{\begin{cases}4x^2-1=0\\y^2-1=0\end{cases}}\)

Từ đó tìm được \(x=\pm\frac{1}{2},y=\pm1\)

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

2:1:2

\(A=x^2-2xy+6y^2-12x+2y+54\)

\(A=x^2-2xy+y^2-12x+12y+36+5y^2-10y+5+4\)

\(A=\left(x-y\right)^2-2.6\left(x-y\right)+36+5\left(y^2-2y+1\right)+4\)

\(A=\left(x-y-6\right)^2+5\left(y-1\right)^2+4\)

Do: \(\left(x-y-6\right)^2\ge0\forall xy\); \(5\left(y-1\right)^2\ge0\forall y\)

\(\Rightarrow\left(x-y-6\right)^2+5\left(y-1\right)^2\ge0\)

\(\Leftrightarrow A=\left(x-y-6\right)^2+5\left(y-1\right)^2+4\ge4\)

\(\Rightarrow A_{Min}=4\)

Dấu "=" xảy ra khi \(x=7;y=1\)

\(\frac{4}{x+2}\)và \(\frac{2-x}{x^2+4x+4}\)

Ta có : \(x^2+4x+4=\left(x+2\right)^2\)

\(\Rightarrow\text{MTC}=\left(x+2\right)^2\)

\(\Rightarrow\hept{\begin{cases}\frac{4}{x+2}=\frac{4\left(x+2\right)}{\left(x+2\right)\left(x+2\right)}=\frac{4x+8}{\left(x+2\right)^2}\\\frac{2-x}{x^2+4x+4}=\frac{2-x}{\left(x+2\right)^2}\end{cases}}\)