Question

Ajutați-mă va rog frumos din suflet urgent fisa aceasta îmi trebuie urgent

Answer 1

Răspuns:

Explicație pas cu pas:

1.

a) x-9=10  ⇒x=10+9  ⇒x=19

x+5=-6  ⇒x=-6-5  ⇒x=-11

x-3=-103 ⇒x=-103+3  ⇒x=-100

x+8=8  ⇒x=8-8  ⇒x=0

b) x·2=20  ⇒x=20:2  ⇒x=10

x·(-5)=-30  ⇒x=(-30):(-5)  ⇒x=6

x:6=1  ⇒x=6

x:(-3)=9  ⇒x=9·(-3)  ⇒x=-27

c) 2x-1=7  ⇒2x=7+1  ⇒2x=8  ⇒x=4

-3x+8=8  ⇒-3x=8-8  ⇒-3x=0  ⇒x=0

3x+5=x+9  ⇒3x-x=9-5  ⇒2x=4  ⇒x=2

4x+7=2x-2  ⇒4x-2x=-2-7  ⇒2x=-9  ⇒x=-9/2

2.

a) x²=4  ⇒x=±2

x²=1  ⇒±1

x²=36  ⇒x=±6

x²=-15  ⇒x=±i√15  

b) x²-9=0  ⇒x²=9  ⇒x=±3

x²-25=0  ⇒x²25  ⇒x=±5

9x²-4=0  ⇒9x²=4  ⇒x=±2/3

49x²=81  ⇒x=±9/7

c) x²+3x=0  ⇒x(x+3)=0  ⇒x=0; x=-3

x²-6x=0  ⇒x(x-6)=0  ⇒x=0; x=6

x²+x=0  ⇒x(x+1)=0  ⇒x0; x=-1

10x²-20x=0  ⇒10x(x-2)=0  ⇒x=0; x=2

3.      

a) x²+2x-3=0  a=1; b=2; c=-3  ⇒Δ=b²-4ac=4+4·1·3=4+12=16  

x₁,₂=(-b±√Δ)/2a  ⇒x₁=(-2+4)/2=1    x₂=(-2-4)/2=-3 x₁=1    x₂=-3

x²+6x+5=0  a=1; b=6; c=5  ⇒Δ=b²-4ac=36-20=16

x₁,₂=(-b±√Δ)/2a  ⇒x₁=(-6+4)/2=-1   x₂=(-6-4)/2=-5  x₁=-1   x₂=-5  

x²-2x-35=0  a=1; b=-2; c=-35  ⇒Δ=b²-4ac=4+140=144  

x₁,₂=(-b±√Δ)/2a  ⇒x₁=(2+12)/2=7  x₂=(2-12)/2=-5  x₁=7  x₂=-5  

3x²+5x-2=0  a=3; b=5; c=-2  ⇒Δ=b²-4ac=25+24=49

x₁,₂=(-b±√Δ)/2a  ⇒x₁=(-5+7)/6=2/6=1/3   x₂=(-5-7)6=-12/6=-2 x₁=1/3   x₂=-2

5x²-9x-2=0   a=5; b=-9; c=4  ⇒Δ=b²-4ac=81+40=121

x₁,₂=(-b±√Δ)/2a  ⇒x₁=(9+11)/10=2   x₂=(9-11)/10=-2/10=-1/5 x₁=2   x₂=-1/5

b) x²+4x+4=0  ⇒(x+2)²=0  ⇒x₁=x₂=-2

x²-16x+64=0  ⇒(x-8)²=0  ⇒x₁=x₂=8

9x²+6x+1=0 ⇒(3x+1)²=0  ⇒x₁=x₂=-1/3

4x²-4x+1=0  ⇒(2x-1)²=0  ⇒x₁=x₂=1/2

x²-2x+1=0  ⇒(x-1)²=0  ⇒x₁=x₂=1

c) 3x²+5x+4=0   a=3; b=5; c=4  ⇒Δ=b²-4ac=25-48=-23  Δ<0  ⇒Ecuatia nu are solutii reale

2x²-6x+6=0   a=2; b=-96 c=6  ⇒Δ=b²-4ac=36-48=-12  Δ<0  ⇒Ecuatia nu are solutii reale

10x²+2x+1=0   a=10; b=2; c=1  ⇒Δ=b²-4ac=4-40=-36  Δ<0  ⇒Ecuatia nu are solutii reale

x²+2x+3=0   a=1; b=2; c=3  ⇒Δ=b²-4ac=4-12=-8  Δ<0  ⇒Ecuatia nu are solutii reale

3x²+6x+5=0   a=3; b=6; c=5  ⇒Δ=b²-4ac=36-60=-24  Δ<0  ⇒Ecuatia nu are solutii reale

4. f:R→R; f(x)=2x+3

A(0;3); B(-1;1); C(-1;1); D(1;4); E(0;0)∈Gf?

f(0)=2·0+3=3  ⇒A(0;3)∈Gf

f(-1)=2·(-1)+3=1  ⇒ B(-1;-1)∉Gf

f(-1)=2·(-1)+3=1   ⇒C(-1;1))∈Gf

f(1)=2·1+3=5  ⇒D(1;4)∉Gf

f(0)=2·0+3=3  ⇒E(0;0)∉Gf