Question

Chứng minh đẳng thức sau:

a) \(\dfrac{x^2-y^2}{x^2-y^2+xz-yz}=\dfrac{x+y}{x+y+z}\)

b) \(\dfrac{x^2+y^2-z^2+2xy}{x^2+z^2-y^2-2xz}=\dfrac{x+y+z}{x-z-y}\)

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

d) \(\dfrac{4x^3-8x^2+3x-6}{12x^3+4x^2+9x+3}=\dfrac{x-2}{3x+1}\)

m.n jup mk vs mai nộp bài

Answer 1

a) \(\dfrac{x^2-y^2}{x^2-y^2+xz-yz}=\dfrac{\left(x-y\right)\left(x+y\right)}{\left(x+y\right)\left(x-y\right)+z\left(x-y\right)}\)

\(=\dfrac{\left(x-y\right)\left(x+y\right)}{\left(x-y\right)\left(x+y+z\right)}=\dfrac{x+y}{x+y+z}\)

b) \(\dfrac{x^2+y^2-z^2+2xy}{x^2+z^2-y^2-2xz}=\dfrac{\left(x+y\right)^2-z^2}{\left(x-z\right)^2-y^2}=\dfrac{\left(x+y-z\right)\left(x+y+z\right)}{\left(x-y-z\right)\left(x-z+y\right)}\)\(=\dfrac{x+y+z}{x-y-z}\)

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

d) \(\dfrac{4x^2\left(x-2\right)+3\left(x-2\right)}{4x^2\left(3x+1\right)+3\left(3x+1\right)}=\dfrac{\left(x-2\right)\left(4x^2+3\right)}{\left(3x+1\right)\left(4x^2+3\right)}=\dfrac{x-2}{3x+1}\)

Answer 2

a) áp dụng hằng đẳng thức