\(=\left(\frac{x}{\left(x+2\right)\left(x-2\right)}-\frac{2}{x-2}+\frac{1}{x+2}\right):\left(\frac{\left(x-2\right)\left(x+2\right)+10-x}{x+2}\right)\)

\(=\left(\frac{x-2\left(x+2\right)+\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}\right):\left(\frac{x^2-4+10-x}{x+2}\right)\)

Đổi 10-x lại thành\(10-x^2\) nha, mk thiếu! sorry!

\(=\left(\frac{x-2x-4+x-2}{\left(x+2\right)\left(x-2\right)}\right):\left(\frac{x^2-4+10-x^2}{x+2}\right)\)

\(=\frac{-6}{\left(x+2\right)\left(x-2\right)}.\frac{x+2}{6}\)

\(=\frac{-6\left(x+2\right)}{6\left(x-2\right)\left(x+2\right)}=-\frac{1}{x-2}\)

<=> (x^8-2x^4+1)+(x^2-2x+1)=0
<=>(x^4-1)^2+(x-1)^2=0
<=>\(\hept{\begin{cases}x^4-1=0\\x-1=0\end{cases}}\) <=> x=1
Chúc bạn học tốt :">

\(x^8-2x^4+x^2-2x+2=0\)

\(\left[\left(x^4\right)^2-2.x^4.1+1^2\right]+\left(x^2-2x+1\right)=0\)

\(\left(x^4-1\right)^2+\left(x-1\right)^2=0\)

Ta có: \(\hept{\begin{cases}\left(x^4-1\right)^2\ge0\forall x\\\left(x-1\right)^2\ge0\forall x\end{cases}}\Rightarrow\left(x^4-1\right)^2+\left(x-1\right)^2\ge0\forall x\)

Mà \(\left(x^4-1\right)^2+\left(x-1\right)^2=0\)

\(\Rightarrow\hept{\begin{cases}\left(x^4-1\right)^2=0\\\left(x-1\right)^2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x^4-1=0\\x-1=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=\pm1\\x=1\end{cases}\Rightarrow}x=1}\)

Vậy \(x=1\)

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

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

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

\(=\frac{-2\left(xy+yz+zx\right)}{2\left[\left(x+y+z\right)^2-2\left(xy+yz+zx\right)-\left(xy+yz+zx\right)\right]}\)

\(=\frac{-2\left(xy+yz+zx\right)}{2\left[-3\left(xy+yz+zx\right)\right]}=\frac{1}{3}\)

\(Q=\left(x^2\right)^2+2.x^2.x+x^2+2x^2+2x+1\)

\(=\left(x^2+x\right)^2+2\left(x^2+x\right)+1=\left(x^2+x+1\right)^2\)

Mà \(x^2+x+1=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\forall x\)

\(\Rightarrow Q=\left(x^2+x+1\right)^2\ge\left(\frac{3}{4}\right)^2=\frac{9}{16}\)

Dấu "=" xảy ra khi: \(x+\frac{1}{2}=0\Rightarrow x=\frac{-1}{2}\)

Vậy GTNN của Q là \(\frac{9}{16}\) khi \(x=\frac{-1}{2}\)

\(\frac{x+y}{x-y}+\frac{x^2-4y^2}{x^2-y^2}-\frac{x-3y}{x+y}\)

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

\(=\frac{x^2+2xy+y^2+x^2-4y^2-x^2+xy+3xy-3y^2}{\left(x+y\right)\left(x-y\right)}\)

\(=\frac{x^2+6xy-6y^2}{\left(x+y\right)\left(x-y\right)}\)

Tham khảo nhé~

Vào link này tham khảo nhé !Câu hỏi của Hà Thuỷ Tiên - Toán lớp 8 | Học trực tuyến

A B C D H

Ta có :  

\(P=\frac{\left(x+\frac{1}{x}^6\right)-\left(x^6+\frac{1}{x}^6\right)-2}{\left(x+\frac{1}{x}\right)^3+x^3+\frac{1}{x^3}}\)

\(=\left(x+\frac{1}{x}\right)^3-\left(x^3+\frac{1}{x}^3\right)\)

\(=3\left(x+\frac{1}{x}\right)\ge6\left(x>0\right)\)

\(\Rightarrow Pmin=6\Leftrightarrow x=1\)

A=(x^2+2xy+y^2)+(y^2-4y+4)-1
=(x+y)^2+(y-2)^2-1 \(\ge\) -1
Dấu "=" xảy ra <=> y=2,x=-2
Nhớ k nha