cho x-2y=6 tính:
A= x-y/x+6
B= 2x+6/3x-2y + 2y-6/4y-x
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x-2y=6
nên x=2y+6
\(P=\dfrac{2y+6-y}{2y+6+6}=\dfrac{y+6}{2y+12}=\dfrac{1}{2}\)
\(Q=\dfrac{2x+6}{3x-2y}+\dfrac{2y-6}{4y-x}\)
\(=\dfrac{2\left(2y+6\right)}{3\left(2y+6\right)-2y}+\dfrac{2y-6}{4y-\left(2y+6\right)}\)
\(=\dfrac{4y+12}{6y+18-2y}+\dfrac{2y-6}{4y-2y-6}\)
\(=\dfrac{4y+12}{4y+18}+\dfrac{2y-6}{2y-6}\)
\(=\dfrac{8y^2-24y+24y-48+8y^2+36y-24y-108}{\left(4y+18\right)\left(2y-6\right)}\)
\(=\dfrac{16y^2+12y-156}{4\left(2y+9\right)\cdot\left(y-3\right)}\)
\(=\dfrac{4y^2+3y-34}{\left(2y+9\right)\left(y-3\right)}\)
a) K= x-y/x+6
K=x-y/x+(x-2y)
K=x-y/x+x-2y
K=x-y/2x-2y
K=x-y/2(x-y)=> K=2
b) L= 2x+( x-2y)/3x-2y + 2y-(x-2y)/4y-x
L= 2x+x-2y/3x-2y + 2y-x+2y/4y-x
L=3x-2y/3x-2y + 4y-x/4y-x
L=1+1=2
a, \(x-2y+x^2-4y^2=\left(x-2y\right)+\left(x-2y\right)\left(x+2y\right)=\left(x-2y\right)\left(1+x+2y\right)\)
b, \(x^2-4x^2y^2+y^2+2xy=\left(x+y\right)^2-\left(2xy\right)^2\)
\(=\left(x+y-2xy\right)\left(x+y+2xy\right)\)
c, \(x^6-x^4+2x^3+2x^2=x^6+2x^3+1-x^4+2x^2-1\)
\(=\left(x^3+1\right)^2-\left(x^2-1\right)^2=\left(x^3-x^2+2\right)\left(x^3+x^2\right)\)
\(=x^2\left(x+1\right)\left(x^3-x^2+2\right)\)
d, \(x^3+3x^2+3x+1-8y^3=\left(x+1\right)^3-\left(2y\right)^3=\left(x+1-2y\right)\left(x+1+2y\right)\)
a) Ta có: \(x-2y+x^2-4y^2\)
\(=\left(x-2y\right)+\left(x-2y\right)\left(x+2y\right)\)
\(=\left(x-2y\right)\left(1+x+2y\right)\)
b: Ta có: \(x^2-4x^2y^2+y^2+2xy\)
\(=\left(x+y\right)^2-\left(2xy\right)^2\)
\(=\left(x+y-2xy\right)\left(x+y+2xy\right)\)