Bài tập tự luyện :
Bài 1 : Chứng minh đẳng thức
a , \(\dfrac{\left(x+y\right)^2-\left(x-y\right)^2}{\left(y+z\right)^2-\left(y-z\right)^2}=\dfrac{x}{z}\) với y ,z \(\ne\) 0
b , \(\dfrac{\left(x+y\right)^2-\left(xy+1\right)^2}{\left(x+1\right)\left(y-1\right)}=\left(1-x\right)\left(y+1\right)\) với x \(\ne\) \(-1\) , y \(\ne1\)
Bài 2 : Chứng minh đẳng thức
a , \(\dfrac{16\left(x+y\right)^2-4\left(x-y\right)^2}{\left(x+3y\right)\left(3x+y\right)}=\dfrac{25\left(x+y\right)^2-9\left(x-y\right)^2}{\left(x+4y\right)\left(4x+y\right)}\) với x \(\ne-3y\) , y \(\ne-3x\) , x \(\ne-4y\) , y \(\ne-4x\)
b , \(\dfrac{\left(x-y\right)^2+\left(xy+1\right)^2}{\left(x^2+1\right)\left(y^2+1\right)}=\dfrac{\left(x^2+1\right)\left(y^2+1\right)}{\left(x+y\right)^2+\left(xy-1\right)^2}\)
c , \(\dfrac{\left(x^2+4x+3\right)\left(x^2+5x+4\right)\left(x^2+6x+5\right)}{\left(x+1\right)^3}\) = (x + 3 )(x+4 )(x+5 ) với x \(\ne-1\)
Bài 1:
a: \(\dfrac{\left(x+y\right)^2-\left(x-y\right)^2}{\left(y+z\right)^2-\left(y-z\right)^2}\)
\(=\dfrac{\left(x+y-x+y\right)\left(x+y+x-y\right)}{\left(y+z+y-z\right)\left(y+z-y+z\right)}\)
\(=\dfrac{2y\cdot2x}{2y\cdot2z}=\dfrac{xy}{yz}=\dfrac{x}{z}\)
b: \(\dfrac{\left(x+y\right)^2-\left(xy+1\right)^2}{\left(x+1\right)\left(y-1\right)}\)
\(=\dfrac{\left(x+y-xy-1\right)\left(x+y+xy+1\right)}{\left(x+1\right)\left(y-1\right)}\)
\(=\dfrac{\left[x\left(1-y\right)-\left(1-y\right)\right]\left(x+1\right)\left(y+1\right)}{\left(x+1\right)\left(y-1\right)}\)
\(=\dfrac{-\left(x-1\right)\left(y-1\right)\left(y+1\right)}{y-1}=-\left(x-1\right)\left(y+1\right)\)
