Zadania w załączniku ;P Prosze o rozwiazanie.
Zad1
a. x²>2x²-x
-x²+x>0
x(-x+1)=0
x=0 v -x+1 =0
x=1
x∈(-∞,0)(1,∞)
b.3(x-3)²-27>0
3(x²-6x+9)-27>0
3x²-18x+27-27>0 |:3
x²-6x>0
x²-6x=0
x(x-6)=0
x=0 v x-6=0
x=6
x∈(
c.-4(x+2)²>4
-4(x²+4x+4)-4>0
-4x²-16x-16-4>0 |:4
-x²-4x -5>0
Δ=(-4)²-4*(-1)*(-5)=16-20 <0
czyli x∈R
Zad2.
3x²+6x<45
3x²+6x-45<0 |:3
x²+2x -15<0
Δ=2²-4*1*(-15)=4+60=64
=8
x
x∈(-5,3)
Całkowite liczby nalezace do przedziału to -4,-3,-2,-1,0,1,2
Zad3.
Gdy chceby zeby x∈R, to Δ musi byc ujemna(czyli <0)
a.mx²-9>0
Δ=0²-4*m*(-9)<0
36m<0
m<0
b.x²+mx+1>0
Δ=m²-4*1*1<0
m²-4<0
m²<4
m∈(-2,2)
" Life is not a problem to be solved but a reality to be experienced! "
© Copyright 2013 - 2025 KUDO.TIPS - All rights reserved.
Zad1
a. x²>2x²-x
-x²+x>0
x(-x+1)=0
x=0 v -x+1 =0
x=1
x∈(-∞,0)
(1,∞)
b.3(x-3)²-27>0
3(x²-6x+9)-27>0
3x²-18x+27-27>0 |:3
x²-6x>0
x²-6x=0
x(x-6)=0
x=0 v x-6=0
x=6
x∈(![(-\infty,0)\cup(6,\infty) (-\infty,0)\cup(6,\infty)](https://tex.z-dn.net/?f=%28-%5Cinfty%2C0%29%5Ccup%286%2C%5Cinfty%29)
c.-4(x+2)²>4
-4(x²+4x+4)-4>0
-4x²-16x-16-4>0 |:4
-x²-4x -5>0
Δ=(-4)²-4*(-1)*(-5)=16-20 <0
czyli x∈R
Zad2.
3x²+6x<45
3x²+6x-45<0 |:3
x²+2x -15<0
Δ=2²-4*1*(-15)=4+60=64
x![x_1=\frac{-2-8}{2\cdot1}=\frac{-10}{2}=-5\ \\ x_2=\frac{-2+8}{2\cdot1}=\frac{6}{2}=3 x_1=\frac{-2-8}{2\cdot1}=\frac{-10}{2}=-5\ \\ x_2=\frac{-2+8}{2\cdot1}=\frac{6}{2}=3](https://tex.z-dn.net/?f=x_1%3D%5Cfrac%7B-2-8%7D%7B2%5Ccdot1%7D%3D%5Cfrac%7B-10%7D%7B2%7D%3D-5%5C+%5C%5C+x_2%3D%5Cfrac%7B-2%2B8%7D%7B2%5Ccdot1%7D%3D%5Cfrac%7B6%7D%7B2%7D%3D3)
x∈(-5,3)
Całkowite liczby nalezace do przedziału to -4,-3,-2,-1,0,1,2
Zad3.
Gdy chceby zeby x∈R, to Δ musi byc ujemna(czyli <0)
a.mx²-9>0
Δ=0²-4*m*(-9)<0
36m<0
m<0
b.x²+mx+1>0
Δ=m²-4*1*1<0
m²-4<0
m²<4
m∈(-2,2)