Question

zadania z matamatyki :
rozwiąż:
a) (x-1)do kwadratu - (x+4) do kwadratu = 2x+1
b) (x+5) do kwadratu +(x-3) do kwadratu = x+1
c) (x+1) do kwadratu +(x-3)(x+3-2=2(x-1)do kwadratu
d) (2-x)(2+x)+(x+3)do kwadratu = x-7
e) (x+3)do kwadratu - (4-x)(4+x)=2(x-1)do kwadratu +1
f) (x-5)do kwadratu - (x-3)do kwadratu = 4 (x-2)

Answer 1

f) (x-5)²-(x-3)²=4(x-2)

x²-10x²+25 - (x²-6x²+9)=4x-8

x²-10x²+25-x²+6x²-9=4x-8

4x²-4x=-8 - 25+9

4x²-4x -24 =0

 

Δ=b²-4ac

Δ=16 - 4 · 4 · (-24)

Δ= 16 + 384

Δ=400

√Δ=20

 

x1=-b -√Δ / 2a

x1=-4-20 / 8

x1=-16/8

x1=-2

 

x2=-b+√Δ / 2a

x2= -4 +20 / 8

x2= 16/8

x2=2

 

e) (x+3)²-(4-x)(4+x)=2(x-1)²

x²+6x+9-(4-x²)=2(x²-2x+2)

x²+6x+9-4+x²=2x²-4x+4

2x=-1 /÷2

x=-½

 

d)(2-x)(2+x)+(x+3)²=x-7

4-x² +x²+6x+9=x-7

5x= -20 / ÷5

x= -4

 

c)(x+1)²+(x-3)(x+3)-2=2(x-1)²

x²+2x+1 + x²-9-2=2(x²-2x+1)

2x²+2x - 10= 2x²-4x+2

6x= 12 /÷6

x=2

 

b)(x+5)²+(x-3)²=x+1

x²+10x+25 +x²-6x+9=x+1

2x²+3x+33= 0

Δ= 9- 4·2·33

Δ=9-264

Δ= -255

brak miejsc zerowych

 

a)(x-1)²-(x+4)²=2x+1

x²-2x+1-(x²+8x+16)=2x+1

x²-2x+1-x²-8x-16=2x+1

-12x=16 /÷(-12)

x= -16/12

x=-4/3