Ôn tập: Phân thức đại số

Lời giải:

Xét mẫu số:

\(bc(y-z)^2+ac(x-z)^2+ab(x-y)^2=bc(y^2+z^2)+ac(x^2+z^2)+ab(x^2+y^2)-2(bcyz+acxz+abxy)\) (1)

Vì \(ax+by+cz=0\Rightarrow (ax+by+cz)^2=0\)

\(\Leftrightarrow a^2x^2+b^2y^2+c^2z^2+2(abxy+bcyz+acxz)=0\)

\(\Leftrightarrow -2(abxy+bcyz+acxz)=a^2x^2+b^2y^2+c^2z^2\)(2)

Từ \((1);(2)\Rightarrow \text{MS}=bc(y^2+z^2)+ac(x^2+z^2)+ab(x^2+y^2)+a^2x^2+b^2y^2+c^2z^2\)

\(=ax^2(a+b+c)+by^2(a+b+c)+cz^2(a+b+c)\)

\(=(a+b+c)(ax^2+by^2+cz^2)\)

Do đó:

\(P=\frac{ax^2+by^2+cz^2}{(a+b+c)(ax^2+by^2+cz^2)}=\frac{1}{a+b+c}=\frac{1}{2017}\)

a² -2a+b²+4b+4c²-4c+6=0

<=>(a²-2a+1)+(b²+4b+4)+(4c²-4c+1)=0

<=>(a-1)²+(b+2)²+(2c-1)²=0

\(\left[{}\begin{matrix}a-1=0\\b+2=0\\2c-1=0\end{matrix}\right.\left[{}\begin{matrix}a=1\\b=-2\\c=\dfrac{1}{2}\end{matrix}\right.\)

Ta có: \(a^2-2a+b^2+4b+4c^2-4c+6=\)

\(=\left(a^2-2a+1\right)+\left(b^2+4b+4\right)+\left(4c^2-4c+1\right)\)= 0

\(\Leftrightarrow\left(a-1\right)^2+\left(b+2\right)^2+\left(2c-1\right)^2=0\)

Mà \(\left(a-1\right)^2+\left(b+2\right)^2+\left(2c-1\right)^2\ge0\left(\forall a;b;c\right)\)

Dấu "=" xảy ra \(\Leftrightarrow\left(a-1\right)^2=0;\left(b+2\right)^2=0;\left(2c-1\right)^2=0\)

\(\Leftrightarrow a=1;b=-2;c=\dfrac{1}{2}\)

ĐKXĐ : \(\left\{{}\begin{matrix}x\ne-1\\x\ne1\end{matrix}\right.\)

Ta có :

\(\dfrac{8}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{2}{x^2+3}+\dfrac{1}{x+1}\)

\(=\dfrac{8}{\left(x^2+3\right)\left(x-1\right)\left(x+1\right)}+\dfrac{2\left(x-1\right)\left(x+1\right)}{\left(x^2+3\right)\left(x-1\right)\left(x+1\right)}+\dfrac{\left(x^2+3\right)\left(x-1\right)}{\left(x^2+3\right)\left(x-1\right)\left(x+1\right)}\)

\(=\dfrac{8+2\left(x-1\right)\left(x+1\right)+\left(x^2+3\right)\left(x-1\right)}{\left(x^2+3\right)\left(x-1\right)\left(x+1\right)}\)

\(=\dfrac{8+2x^2-2+x^3-x^2+3x-3}{\left(x^2+3\right)\left(x-1\right)\left(x+1\right)}\)

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

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

\(=\dfrac{1}{x-1}\)

\(\dfrac{8}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{2}{x^2+3}+\dfrac{1}{x+1}\)

ĐKXĐ : \(\left\{{}\begin{matrix}x^2+3\ne0\\x+1\ne0\\x-1\ne0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x^2\ne-3\left(vôlí\right)\\x\ne-1\\x\ne1\end{matrix}\right.\)

Ta có : \(\dfrac{8+2\left(x-1\right)\left(x+1\right)+\left(x^2+3\right)\left(x-1\right)}{\left(x^2+3\right)\left(x-1\right)\left(x+1\right)}\)

\(=\dfrac{8+2x^2-x+1+x^3-x^2+3x-3}{\left(x^2+3\right)\left(x-1\right)\left(x+1\right)}\)

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

Ta có:

\(xyz+xy+yz+zx+x+y+z=2011\)

\(\Leftrightarrow xy\left(z+1\right)+y\left(z+1\right)+x\left(z+1\right)+\) \(\left(z+1\right)=2012\)

\(\Leftrightarrow\left(z+1\right)\left(xy+y+x+1\right)=2012\)

\(\Leftrightarrow\left(z+1\right)\left[x\left(y+1\right)+\left(y+1\right)\right]=2012\)

\(\Leftrightarrow\left(x+1\right)\left(y+1\right)\left(z+1\right)=2012\)

Mà \(2012=1.2.2.503=503.4.1\)

\(\Leftrightarrow\left[{}\begin{matrix}x=502;y=1;z=1\\x=1005;y=1;z=0\\x=2011;y=0;z=0\end{matrix}\right.\)

Vậy...

\(\left(\dfrac{3x}{1-3x}+\dfrac{2x}{3x+1}\right):\dfrac{6x^2+10x}{1-6x+9x^2}\)\(\left(ĐKXĐ:x\ne\pm\dfrac{1}{3}\right)\)

\(=\left[\dfrac{-3x\left(3x+1\right)+2x\left(3x-1\right)}{\left(3x-1\right)\left(3x+1\right)}\right]:\dfrac{2x\left(3x+5\right)}{\left(3x-1\right)^2}\)

\(=\dfrac{-9x^2-3x+6x^2-2x}{\left(3x-1\right)\left(3x+1\right)}.\dfrac{\left(3x-1\right)^2}{2x\left(3x+5\right)}\)

\(=\dfrac{-3x^2-5x}{\left(3x-1\right)\left(3x+1\right)}.\dfrac{\left(3x-1\right)^2}{2x\left(3x+5\right)}\)

\(=\dfrac{-x\left(3x+5\right).\left(3x-1\right)^2}{\left(3x-1\right)\left(3x+1\right).2x\left(3x+5\right)}\)

\(=\dfrac{1-3x}{2\left(3x+1\right)}\)

\(=\dfrac{1-3x}{6x+2}\)

ĐKXĐ : \(x\ne0,x\ne\pm2\)

Câu a :

\(A=\left(\dfrac{1}{x-2}-\dfrac{2x}{4-x^2}+\dfrac{1}{x+2}\right).\left(\dfrac{2}{x}-1\right)\)

\(=\dfrac{x+2+2x+x-2}{\left(x-2\right)\left(x+2\right)}.\left(\dfrac{2}{x}-1\right)\)

\(=\dfrac{4x}{\left(x-2\right)\left(x+2\right)}\times\dfrac{2-x}{x}\)

\(=-\dfrac{4}{x+2}\)

Câu b :

Ta có : \(2x^2+x=0\Leftrightarrow x\left(2x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{1}{2}\end{matrix}\right.\)

Thay \(x=0\) vào A ta được \(-\dfrac{4}{0+2}=-2\)

Thay \(x=-\dfrac{1}{2}\) vào A ta được \(-\dfrac{4}{-\dfrac{1}{2}+2}=-\dfrac{8}{3}\)

Câu c :

Để \(A=\dfrac{1}{2}\) thì \(-\dfrac{4}{x+2}=\dfrac{1}{2}\)

\(\Leftrightarrow x+2=-8\Leftrightarrow x=-10\)

Câu d :

Để A nguyên dương thì \(-4⋮x+2\)

Xét :

\(Ư\left(-4\right)=-4;-2;-1;1;2;4\)

\(\left\{{}\begin{matrix}x+2=-4\\x+2=-2\\x+2=-1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-6\left(N\right)\\x=-4\left(N\right)\\x=-3\left(N\right)\end{matrix}\right.\)

\(\left\{{}\begin{matrix}x+2=1\\x+2=2\\x+2=4\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-1\left(N\right)\\x=0\left(L\right)\\x=2\left(L\right)\end{matrix}\right.\)

Vậy có 4 giá trị của x thì A nguyên : \(\left\{{}\begin{matrix}x=-6\\x=-4\\x=-3\\x=-1\end{matrix}\right.\)

Ta có : \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\)

\(\Rightarrow\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}=\dfrac{3}{xyz}\)

\(\Rightarrow A=\dfrac{yz}{x^2}+\dfrac{xz}{y^2}+\dfrac{xy}{z^2}=\dfrac{xyz}{x^3}+\dfrac{xyz}{y^3}+\dfrac{xyz}{z^3}\)

\(=xyz\left(\dfrac{1}{x^3}+\dfrac{1}{y^3}+\dfrac{1}{z^3}\right)=\dfrac{xyz.3}{xyz}=3\)