Ta có \(\left(x+y\right)\left(x+2y\right)\left(x+3y\right)\left(x+4y\right)+y^4\)

\(=\left(x+y\right)\left(x+4y\right)\left(x+2y\right)\left(x+3y\right)+y^4\)

\(=\left(x^2+5xy+4y^2\right)\left(x^2+5xy+6y^2\right)+y^4\)

\(=\left(x^2+5xy+5y^2-y^2\right)\left(x^2+5xy+5y^2+y^2\right)+y^4\)

\(=\left(x^2+5xy+5y^2\right)^2\) là số chính phương. \(\Rightarrowđpcm\)

Ta có:

\(A=\left(x+y\right)\left(x+2y\right)\left(x+3y\right)\left(x+4y\right)+y^4\)

\(=\left(x^2+5xy+4y^2\right)\left(x^2+5xy+6y^2\right)+y^4\)

Đặt \(x^2+5xy+5y^2=t\left(t\in Z\right)\) thì:

\(A=\left(t-y^2\right)\left(t+y^2\right)+y^4\)

\(=t^2-y^4+y^4=t^2\)

\(=\left(x^2+5xy+5y^2\right)^2\)

Vì \(x,y,z\in Z\) nên:

\(x^2\in Z,5xy\in Z,5y^2\in Z\)

\(\Leftrightarrow x^2+5xy+5y^2\in Z\)

Vậy \(A\) là số chính phương (Đpcm)

ồ bài này khá dễ

Ta có

\(A=\left(x+y\right)\left(x+2y\right)\left(x+3y\right)\left(x+4y\right)+y^4\)

\(=\left(x^2+5xy+4y^2\right)\left(x^2+5xy+6y^2\right)+y^4\)

Đặt \(x^2+5xy+5y^2=t\left(t\in Z\right)\)

\(\)\(A=\left(t-y^2\right)\left(t+y^2\right)+y^4=t^2-y^4+y^4\)

\(=t^2=\left(x^2+5xy+5y^2\right)^2\)

Vì \(x,y,z\in Z\) nên \(\hept{\begin{cases}x^2\in Z\\5xy\in Z\\5y^2\in Z\end{cases}\Rightarrow x^2+5xy+y^2\in Z}\)

Vậy A là số chính phương

\(A=\left[\left(x+y\right)\left(x+4y\right)\right]\left[\left(x+2y\right)\left(x+3y\right)\right]+y^4.\)

\(=\left(x^2+5xy+4y^2\right)\left(x^2+5xy+6y^2\right)+y^4.\)

\(=\left[\left(x^2+5xy+5y^2\right)-y^2\right]\left[\left(x^2+5xy+5y^2\right)+y^2\right]+y^4.\)

\(=\left(x^2+5xy+5y^2\right)^2-y^4+y^4\)

\(=\left(x^2+5xy+5y^2\right)^2\)

Đến đây ta có điều phải chứng minh rồi :>

\(\left(1+x^2\right)\left(1+y^2\right)+4xy+2\left(x+y\right)\left(1+xy\right)\)\(\)

\(=1+y^2+x^2+x^2y^2+4xy+2\left(x+y\right)\left(1+xy\right)\)

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

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

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

Theo tính chất của cấp số cộng, ta có \(u_1+u_4=u_2+u_3\)

Do đó : \(\Leftrightarrow\left(x-u_1\right)\left(x-u_2\right)\left(x-u_3\right)\left(x-u_4\right)=\left[x^2-\left(u_1-u_4\right)x+u_1u_4\right]\left[x^2-\left(u_2-u_3\right)x+u_2u_3\right]\)(*)

Đặt \(t=x^2-\left(u_1+u_4\right)x=x^2-\left(u_2+u_3\right)x\)

Khi đó (*) \(\Leftrightarrow f\left(t\right)=\left(t+u_1u_4\right)\left(t+u_2u_3\right)+9=t^2+\left(u_1u_4+u_2u_3\right)t+u_1u_4u_2u_3+9\)

Với \(\Delta_t=\left(u_1u_4+u_2u_3\right)^2-4u_1u_4u_2u_3-36=\left(u_1u_4+u_2u_3\right)^2-36\)

Rõ ràng \(\left|u_1u_4-u_2u_3\right|\le6\Rightarrow\Delta_t<0\leftrightarrow f\left(t\right)>0\)với mọi t

<=> A có nghĩa với mọi x

Câu 8 :

\(N=\left(\frac{x-1}{\left(x-1\right)^2+x}-\frac{2}{x-2}\right):\left(\frac{\left(x-1\right)^4+2}{\left(x-1\right)^3-1}-x+1\right)\)

Đặt \(x-1=a\)

\(N=\left(\frac{a}{a^2+x}-\frac{2}{a-1}\right):\left(\frac{a^4+2}{a^3-1}-a\right)\)

\(N=\frac{a\left(a-1\right)-2\left(a^2+x\right)}{\left(a^2+x\right)\left(a-1\right)}:\frac{a^4+2-a\left(a^3-1\right)}{a^3-1}\)

\(N=\frac{a^2-a-2a^2-2x}{\left(a^2+x\right)\left(a-1\right)}:\frac{a^4+2-a^4+a}{a^3-1}\)

\(N=\frac{-a^2-a-2x}{\left(a^2+x\right)\left(a-1\right)}\cdot\frac{\left(a-1\right)\left(a^2+a+1\right)}{2+a}\)

\(N=\frac{-\left(a^2+a+2x\right)\left(a^2+a+1\right)}{\left(a^2+x\right)\left(2+a\right)}\)

\(N=\frac{-\left[\left(x-1\right)^2+x-1+2x\right]\left[\left(x-1\right)^2+x-1+1\right]}{\left[\left(x-1\right)^2+x\right]\left(2+x-1\right)}\)

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

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

\(N=-x\)( đpcm )

Câu 9 : Tìm giá trị nhỏ nhất của biểu thức :

\(P=\frac{x^2}{x+4}\cdot\left(\frac{x^2+16}{x}+8\right)+9\)

Bài làm :

\(P=\frac{x^2}{x+4}\cdot\frac{x^2+8x+16}{x}+9\)

\(P=\frac{x^2\left(x+4\right)^2}{x\left(x+4\right)}+9\)

\(P=x\left(x+4\right)+9\)

\(P=x^2+4x+9\)

\(P=\left(x+2\right)^2+5\ge5\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow x=-2\)

Bài 10 : Tìm GTLN

\(Q=\left(\frac{x^3+8}{x^3-8}\cdot\frac{4x^2+8x+16}{x^2-4}-\frac{4x}{x-2}\right):\frac{-16}{x^4-6x^3+12x^2-8x}\)

\(Q=\left[\frac{\left(x+2\right)\left(x^2-2x+4\right)}{\left(x-2\right)\left(x^2+2x+4\right)}\cdot\frac{4\left(x^2+2x+4\right)}{\left(x-2\right)\left(x+2\right)}-\frac{4x}{x-2}\right]:\frac{-16}{x\left(x^3-6x^2+12x-8\right)}\)

\(Q=\left(\frac{4\left(x^2-2x+4\right)}{\left(x-2\right)^2}-\frac{4x\left(x-2\right)}{\left(x-2\right)^2}\right):\frac{-16}{x\left[x^2\left(x-2\right)-4x\left(x-2\right)+4\left(x-2\right)\right]}\)

\(Q=\frac{4x^2-8x+16-4x^2+8x}{\left(x-2\right)^2}:\frac{-16}{x\left(x-2\right)\left(x^2-4x+4\right)}\)

\(Q=\frac{16}{\left(x-2\right)^2}\cdot\frac{-x\left(x-2\right)\left(x-2\right)^2}{16}\)

\(Q=-x\left(x-2\right)\)

\(Q=-x^2+2x\)

\(Q=-x^2+2x-1+1\)

\(Q=1-\left(x-1\right)^2\le1\forall x\)

Dấu "=" \(\Leftrightarrow x=1\)

Vậy....

\(P=x\left(x+a\right)\left(x-a\right)\left(x+2a\right)+a^4\)        

   \(=\left(x^2+ax\right)\left(x^2+ax-2a^2\right)+a^4\)

Đặt \(x^2+ax=t\)

Khi đó: \(P=t\left(t-2a^2\right)+a^4\)

              \(=t^2-2ta^2+\left(a^2\right)^2=\left(t-a^2\right)^2=\left(x^2+ax-a^2\right)^2\)

Chúc bạn học tốt.