\(a=x^2;b=y^2;c=z^2\)

\(P=\left(a-b\right)\left(b-c\right)\left(c-a\right)=\left(x^2-y^2\right)\left(y^2-z^2\right)\left(z^2-x^2\right)\)

\(=\left(x-y\right)\left(x+y\right)\left(y-z\right)\left(y-z\right)\left(z-x\right)\left(z+x\right)\)

..............................

(x+2)²-(x-2)(x+2)=0

=> (x+2)[(x+2)-(x-2)]=0

=> (x+2)(x+2-x+2)=0

=>(x+2)+4=0

=> x+4=-4

=> x=-8

vậy ...

b) x⁴+4y4

= x⁴+4x²y²+4y⁴-4x²y²

= (x⁴+4x²y²+4y⁴)-4x²y²

=(x²+2y²)²-4x²y²

=(x²+2y²)²-(2xy)²

=(x²+2y²-2xy)(x²+2y²+2xy)

x⁴+4y⁴+4x²y²-4x²y²

=(x⁴+4x²y²+4y⁴)-4x²y²

=(x²+2y²)²-4x²y²

=(x²+2y²-2xy)(x²+2y²+2xy)

\(\dfrac{5.4^6.9^5+6^{10}.10^2}{8^4.3^{12}-6^{11}}=\dfrac{5.\left(2^2\right)^6.\left(3^2\right)^5+\left(2.3\right)^{10}.\left(5.2\right)^2}{\left(2^3\right)^4.3^{12}-\left(2.3\right)^{11}}\)

=\(\dfrac{5.2^{12}.3^{10}+2^{10}.3^{10}.5^2.2^2}{2^{12}.3^{12}-2^{11}.3^{11}}=\dfrac{5.2^{12}.3^{10}\left(1+5\right)}{2^{11}.3^{11}\left(2.3-1\right)}\)

=\(\dfrac{5.2.6}{1.3.5}=4\)

(3x+y)${}^{}$-2(x+y)(3x+y)+(x+y)${}^{}$

= [(3x-y)-(x+y)]²

=(3x-y-x-y)²

= (2x-2y)²

20 bao gạo 

tick mình nhé Luong Khanh Linh

(5x${}^{}$y${}^{}$+2x${}^{}$y${}^{}$-10x${}^{}$y${}^{}$):5x${}^{}$y${}^{}$

= x+ \(\dfrac{2}{5}\) -2y²

(5x${}^{}$y${}^{}$+2x${}^{}$y${}^{}$-10x${}^{}$y${}^{}$):5x${}^{}$y${}^{}$

-2y²