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\(P=\frac{x}{xy+x+1}+\frac{y}{yz+y+1}+\frac{z}{xz+z+1}\)
\(=\frac{xz}{xyz+xz+z}+\frac{xyz}{xyz^2+xyz+xz}+\frac{z}{xz+z+1}\)(do \(xyz=1\))
\(=\frac{xz}{xz+z+1}+\frac{1}{z+1+xz}+\frac{z}{xz+z+1}\)(do \(xyz=1\))
\(=\frac{xz+z+1}{xz+z+1}=1\)
Bn đăng bài lên xong nói mình làm được r thế đăng lên làm gì vậy bạn?
\(a,P=\dfrac{x^2+6x+9}{x^2+3x}\\ =\dfrac{x^2+2\cdot3\cdot x+3^2}{x\left(x+3\right)}\\ =\dfrac{\left(x+3\right)^2}{x\left(x+3\right)}\\ =\dfrac{x+3}{x}\\ Q=\dfrac{x^2+3x}{x^2-9}\\ =\dfrac{x\left(x+3\right)}{\left(x+3\right)\left(x-3\right)}\\ =\dfrac{x}{x-3}\\ b,P\cdot Q=\dfrac{x+3}{x}\cdot\dfrac{x}{x-3}\\ =\dfrac{\left(x+3\right)\cdot x}{x\cdot\left(x-3\right)}\\ =\dfrac{x+3}{x-3}\\ P:Q=\dfrac{x+3}{x}:\dfrac{x}{x-3}\\ =\dfrac{x+3}{x}\cdot\dfrac{x-3}{x}\\ =\dfrac{x^2-9}{x^2}\)
a) P= \(\dfrac{x^2+6x+9}{x^2+3x}=\dfrac{\left(x+3\right)^2}{x\left(x+3\right)}=\dfrac{x+3}{x}\)
Q= \(\dfrac{x^2+3x}{x^2-9}=\dfrac{x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{x}{x-3}\)
b)\(P.Q=\dfrac{x+3}{x}.\dfrac{x}{x-3}=1\)
\(P:Q=\dfrac{x+3}{x}:\dfrac{x}{x-3}=\dfrac{x+3}{x}.\dfrac{x-3}{x}=\dfrac{x^2-9}{x^2}\)
a,\(A=\frac{6x+12}{\left(x+2\right)\left(2x-6\right)}=\frac{6\left(x+2\right)}{2\left(x+2\right)\left(x-3\right)}=\frac{3}{x-3}\)
b, Giá trị của x để phân thức có giá trị bằng (-2) :
\(\frac{3}{x-3}=-2\Rightarrow x=1,5\)
\(\left(\frac{x}{x-1}-\frac{x+1}{x}\right):\left(\frac{x}{x+1}-\frac{x-1}{x}\right)\)
\(=\left(\frac{x^2-\left(x-1\right)\left(x+1\right)}{\left(x-1\right).x}\right):\left(\frac{x^2-\left(x-1\right)\left(x+1\right)}{x\left(x+1\right)}\right)\)
\(=\frac{x^2-\left(x-1\right)\left(x+1\right)}{x\left(x-1\right)}.\frac{x\left(x+1\right)}{x^2-\left(x-1\right)\left(x+1\right)}\)
\(=\frac{x+1}{x-1}\)
Lời giải:
$P+Q=x+\frac{1}{x}+x-\frac{1}{x}=2x$
$P-Q=x+\frac{1}{x}-x+\frac{1}{x}=\frac{2}{x}$
$PQ=(x+\frac{1}{x})(x-\frac{1}{x})=x^2-\frac{1}{x^2}$
$P:Q=(x+\frac{1}{x}): (x-\frac{1}{x})=\frac{x^2+1}{x}: \frac{x^2-1}{x}=\frac{x^2+1}{x^2-1}$