" Life is not a problem to be solved but a reality to be experienced! "
© Copyright 2013 - 2024 KUDO.TIPS - All rights reserved.
S=
T=
S+T=
zad2
(x+5)²-(x-1)²-12(x+2)=x²+10x+25-(x²-2x+1)-12x-24=x²+10x+25-x²+2x-1-12x-24=0
dla x=1+√3
(1+√3+5)²-(1+√3-1)²-12(1+√3+2)=(6+√3)²-(√3)²-12(3+√3)=36+12√3+3-3-36-12√3=0
zad3
a)4x=√2x²
-√2x²+4x=0
-x(√2x-4)=0
x=0 lub √2x-4=0
√2x=4
x=4/√2
x=4√2/2
x=2√2
b)2x²=1
x²=1/2
x=√1/√2
x=-√2/2 x=√2/2
c)-3x²+4x+2=0
Δ=b²-4ac
Δ=16-4*(-3)*2
Δ=16+24
Δ=40 √Δ=√40
√Δ=2√10
x1=(-b-√Δ)/2a
x1=(-4-2√10)/-6
x1=-2(2+√10)-6
x1=-(2+√10)/3
x2=(-b+√Δ)/2a
x2=(4+2√10)/-6
x2=2(2+√10)/-6
x2=-(2+√10)/3
d)x^4+6x³+8x²=0
x²(x²+6x+8)=0
x²=0 x²+6x+8=0
x=0 Δ=36-32
Δ=4 √Δ=2
x1=(-6-2)/2
x1=-4
x2=(-6+2) /2
x2=-2
e)6/(x+1)=4 mnożymy *(x+1)
x+1≠0
x≠-1 D:x∈R-{-1}
6=4(x+1)
6=4x+4
6-4=4x
2=4x dzielimy przez 4
x=2/4
x=1/2