a) \(A=\frac{3x^2+5}{x^2+1}=\frac{3\left(x^2+1\right)+2}{x^2+1}=3+\frac{2}{x^2+1}\)

để A nguyên =>\(x^2+1\inƯ\left(2\right)\)

\(\Leftrightarrow x^2\in\left\{0;1\right\}\)

\(\Leftrightarrow x\in\left\{0;\pm1\right\}\)

Để A nguyên thì \(3x^2+5⋮x^2+1\)

\(\Rightarrow3\left(x^2+1\right)+2⋮x^2+1\)

\(\Rightarrow2⋮x^2+1\)

\(\Rightarrow x^2+1\in\left\{1;2\right\}\Rightarrow x=0;x=1;x=-1\)

a) \(Q=\frac{x^4-x^2+2x+2}{x^4+x^3+x+1}\)

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

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

\(Q=\frac{\left(x+1\right)\left[x^2\left(x-1\right)+2\right]}{\left(x+1\right)\left(x^3+1\right)}\)

\(Q=\frac{x^3-x^2+2}{x^3+1}\)

b) \(Q=\left|Q\right|=\frac{x^3-x^2+2}{x^3+1}\)

Ta có: \(x^3-x=\left(x-1\right)x\left(x+1\right)\)

Tích 3 số nguyên liên tiếp chia hết cho 3 nên \(\left(x-1\right)x\left(x+1\right)⋮3\)

hay \(x^3-x⋮3\)

Tương tự \(y^3-y⋮3\);\(z^3-z⋮3\)

\(\Rightarrow x^3+y^3+z^3-\left(x+y+z\right)⋮3\)

Mà \(\left(x+y+z\right)⋮3\left(gt\right)\Rightarrow a^3+b^3+c^3⋮3\left(đpcm\right)\)

\(x^2+5y^2+2y-4xy-3=0.\)

\(\Rightarrow x^2-4xy+4y^2+y^2+2y-3=0\)

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

Vậy cặp số x,y nhỏ nhất thỏa mãn là \(\hept{\begin{cases}x-2y=0\\y+1=0\end{cases}\Rightarrow\hept{\begin{cases}x-2y=0\\y=-1\end{cases}\Rightarrow}\hept{\begin{cases}x+2=0\\y=-1\end{cases}}}\)

\(\Rightarrow x=-2;y=-1\)

\(x^2+5y^2+2y-4xy-3=0\)

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

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

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

=> \(\left(x-2y\right)^2+\left(y-1\right)\left(y+3\right)=0\)

Mà   \(\left(x-2y\right)^2 \ge 0 \forall x\) 

=> \(\left(y-1\right)\left(y+3\right)\le0\)   Mặt khác \(y-1 < y+3 \)

=> \(\hept{\begin{cases}y-1\le0\\y+3\ge0\end{cases}}\)=> \(-3\le y\le1\)  mà y nhỏ nhất 

=> \(y=-3\)

Thay vào biểu thức, ta có \(\left(x+6\right)^2+\left(-3-1\right)\left(-3+3\right)=0\) => \(\left(x+6\right)^2=0\)  => \(x+6=0\) => \(x=-6\)

    Vậy x=-6 , y=-3

\(=2x^2y-\frac{1}{2}x^2y^2-xy\left(2x-xy\right)\)

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

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

\(=xy.\frac{1}{2}xy\)

\(=\frac{1}{2}x^2y^2\)

mọi người giải hết giúp em ạ

a) MTC: 2xy

Quy đồng: \(\frac{2x-3y}{2xy}\) giữ nguyên

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

b) \(\frac{2}{x^2-4x}=\frac{2}{x\left(x-4\right)};\frac{x}{x^2-16}=\frac{x}{\left(x-4\right)\left(x+4\right)}\)

MTC: x (x-4)(x+4)

Quy đồng : \(\frac{2}{x\left(x-4\right)}=\frac{2\left(x+4\right)}{x\left(x-4\right)\left(x+4\right)}=\frac{2x+8}{x\left(x-4\right)\left(x+4\right)}\)

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

Học tốt nhé ^3^

\(\left(x-1\right)^2-\left(x-2\right)^2-\left(x-3\right)^2+\left(x-4\right)^2\)

\(=\left(x-1-x+2\right)\left(x-1+x-1\right)-\left(x-3+x-4\right)\left(x-3-x+4\right)\)

\(=2x-2-2x+7\)

\(=5\)

tại sao lại ra dòng thứ 2 vậy bn