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a fost răspuns

efectuati:
e) (x+√2)(x²-√2+2)-(x+√3)(x²-√3+3)
f) (x-√2)(x²+√2+2)-(x-√5)(x²+√5x+5)

Răspuns :

Nico234679ghd
e) (x+√2)(x²-√2+2)-(x+√3)(x²-√3+3) = x³-√2x+2x +√2x² -2 +2√2 -x³ +√3x -3x -√3 --x² +3-3√3 = -√2x -x+√2x² +1+2√2 +√3x -√3x² -3√3
f) (x-√2)(x²+√2+2)-(x-√5)(x²+√5x+5)= x³ +√2x +2x- √2x² -2√2 -x³ -√5x -5x+ √5x² +5+5√5 = √2x -3x -√2 x² +3-2√2 -√5 x +√5x² +5√5

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