1.Oblicz wartość wyrażenia:
a) (x-y) (x+y) + (x+2y) (x-2y) dla
b)
2.Usuń niewymierność z mianownika:
a)
Zadanie 1:
(x-y)(x+y)+(x+2y)(x-2y) = x²-y²+x²-4y² = 2x²-5y²
dla:
x = √2
y = 1-√3
2*√2*√2-5*(1-√3)² = 2*2-5(1-2√3+3) = 4-5(4-2√3) = 4-20+10√3 = 10√3-16 = 2(5√3-8)
(√5+x)²-(√5-x)² = 5+2√5*x+x²-(5-2√5*x+x²) = 5+2√5*x+x²-5+2√5*x-x² = 4√5*x
x = √5-1
4*√5(√5-1) = 4*5-4√5 = 20-4√5
Zadanie 2:
1/2√5 *√5/√5 = √5/(2*5) = √5/10
(√3-1)/(4-2√3) *(4+2√3)/(4+2√3) = (2√3+2)/(16-12) = (2√3+2)/4 = (√3+1)/2
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Zadanie 1:
a)
(x-y)(x+y)+(x+2y)(x-2y) = x²-y²+x²-4y² = 2x²-5y²
dla:
x = √2
y = 1-√3
2*√2*√2-5*(1-√3)² = 2*2-5(1-2√3+3) = 4-5(4-2√3) = 4-20+10√3 = 10√3-16 = 2(5√3-8)
b)
(√5+x)²-(√5-x)² = 5+2√5*x+x²-(5-2√5*x+x²) = 5+2√5*x+x²-5+2√5*x-x² = 4√5*x
dla:
x = √5-1
4*√5(√5-1) = 4*5-4√5 = 20-4√5
Zadanie 2:
a)
1/2√5 *√5/√5 = √5/(2*5) = √5/10
b)
(√3-1)/(4-2√3) *(4+2√3)/(4+2√3) = (2√3+2)/(16-12) = (2√3+2)/4 = (√3+1)/2