\(ĐKXĐ:\hept{\begin{cases}x\ne\pm2\\x\ne0\end{cases}}\)

a) \(P=\left(\frac{x^2}{x^3-4x}+\frac{6}{6-3x}+\frac{1}{x+2}\right):\left(x-2+\frac{10-x^2}{x+2}\right)\)

\(\Leftrightarrow P=\left(\frac{x^2}{x\left(x-2\right)\left(x+2\right)}-\frac{6}{3\left(x-2\right)}+\frac{1}{x+2}\right):\frac{x^2-4+10-x^2}{x-2}\)

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

\(\Leftrightarrow P=\frac{x^2-2x^2-4x+x^2-2x}{x\left(x-2\right)\left(x+2\right)}\cdot\frac{x-2}{6}\)

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

\(\Leftrightarrow P=\frac{-1}{x+2}\)

b) Khi \(\left|x\right|=\frac{3}{4}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{4}\\x=-\frac{3}{4}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}P=-\frac{1}{\frac{3}{4}+2}=-\frac{4}{11}\\P=-\frac{1}{-\frac{3}{4}+2}=-\frac{4}{5}\end{cases}}\)

c) Để P = 7

\(\Leftrightarrow-\frac{1}{x+2}=7\)

\(\Leftrightarrow7\left(x+2\right)=-1\)

\(\Leftrightarrow7x+14=-1\)

\(\Leftrightarrow7x=-15\)

\(\Leftrightarrow x=-\frac{15}{7}\)

Vậy để \(P=7\Leftrightarrow x=-\frac{15}{7}\)

d) Để \(P\inℤ\)

\(\Leftrightarrow1⋮x+2\)

\(\Leftrightarrow x+2\inƯ\left(1\right)=\left\{\pm1\right\}\)

\(\Leftrightarrow x\in\left\{-3;-1\right\}\)

Vậy để  \(P\inℤ\Leftrightarrow x\in\left\{-3;-1\right\}\)

A là đa thức bậc 4 nên A là bình phương của 1 đa thức bậc 2
Gọi đa thức bậc 2 đó là:\(cx^2+dx+e\)

\(A=\left(cx^2+dx+e\right)^2\)\(=c^2x^4+d^2x^2+e^2+2cdx^3+2cex^2+2dex\)
Đồng nhất hệ số:\(c^2=1;2cd=-6;d^2+ce=a;2de=b;e^2=1\)

Nếu \(c=1\) thì \(d=-3;e=\pm1\) 

   +,Với \(e=1\) thì \(a=10;b=-6\)

   +,Với \(e=-1\) thì \(a=8;b=6\)

Nếu \(c=-1\) tương tự

thiếu điều kiện gì ko e

m=3xyz+xy^2+x^2y+xz^2+x^2z+y^2z+yz^2

=(xy^2+x^2y+xyz)+(yz^2+y^2z+xyz)+(......)

=xy(x+y+z)+yz(x+y+z)+xz(x+y+z)=(xy+yz+xz)(x+y+z)

Trả lời :

Ta có :

\(x^2+2xy+7x+7y+y^2+10\)

\(=\left(x^2+2xy+y^2\right)+\left(7x+7y\right)+10\)

\(=\left(x+y\right)^2+7\left(x+y\right)+10\)

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

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

Hok tốt

a) \(x^2+2xy+7x+7y+y^2+10\)

\(=\left(x^2+2xy+y^2\right)+\left(7x+7y\right)+10\)

\(=\left(x+y\right)^2+7\left(x+y\right)+10\)

\(=\left(x+y\right)^2+2\left(x+y\right)+5\left(x+y\right)+10\)

\(=\left(x+y+2\right)\left(x+y+5\right).\)

b) \(x^2y+xy^2+x+y=2010\)

\(\Leftrightarrow xy\left(x+y\right)+\left(x+y\right)=2010\)

\(\Leftrightarrow11\left(x+y\right)+1\left(x+y\right)=2010\)

\(\Leftrightarrow12\left(x+y\right)=2010\)

\(\Leftrightarrow x+y=\frac{335}{2}\)

\(\Leftrightarrow\left(x+y\right)^2=\frac{112225}{4}\)

\(\Leftrightarrow x^2+2xy+y^2=\frac{112225}{4}\)

\(\Leftrightarrow x^2+y^2+22=\frac{112225}{4}\)

\(\Leftrightarrow x^2+y^2=\frac{112137}{4}.\)

Vậy \(x^2+y^2=\frac{112137}{4}.\)