Proszę o rozwiązanie od podpunktu K w zadaniu 2 Załącznik
Proszę to na jutro !!!; }
2k)
2x-3\5=1+x\3
2x-3\5=x
2x-3=5x
-3x=3\(-3)
x=-1
l)
3-y\2+1=y-1\7+2
2+3-y\2=y-1+14\7
7(2+3-y)=2(y-1+14)
14+21-7y=2y-2+28
-9y=9\(-9)
y=-1
m)
(a-2)²=(a-2)(a+2)-1
a²-4a+4=a²+4-1
-4a=-9\(-4)
a=-9\4
a=-3\2
n)
(x-2)²=0
x²-4x+4=0 Δ=b²-4ac=(-4)²-4*1*4=0 x₀=-b\2a=4\2=2
o)
(2x-3)(4-x)=-2(x-5)²
8x-2x²-12+3x=-2(x²-10x+25)
-2x²+11x-12=-2x²+20x-50
-6x²-9x+38=0\(-1)
6x²+9x-38
p)
(x+3\2)²=1\4(x-2)²+3
x²+6x+9\4=1\4x²-x+1+3
x²+6x+9=4(1\4x²-x+4)
x²+6x+9=x²-4x+16
10x=7\10
x=7\10
s)
3 2\3x+4 1\2=-1\6(x-4)
11\3x+9\2=-1\6x+4\6
11\3x+1\6x=4\6-9\2
23\6x=-23\6\ 23\6
t)
b-(7-b)=2(b+1)-9
b-7+b=2b+2-9
0=0
U
u)
2x(x-3)=(x-4)(2x+1)
2x²-6x=2x²+x-8x-4
x=-4
w)
-(5-4c)²=(4c-2)(-4c-2)
-(25-40c+16c²)=-16c²-8c+8c+4
-25+40c-16c²=-16c²+4
40c=29\40
c=29\40
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2k)
2x-3\5=1+x\3
2x-3\5=x
2x-3=5x
-3x=3\(-3)
x=-1
l)
3-y\2+1=y-1\7+2
2+3-y\2=y-1+14\7
7(2+3-y)=2(y-1+14)
14+21-7y=2y-2+28
-9y=9\(-9)
y=-1
m)
(a-2)²=(a-2)(a+2)-1
a²-4a+4=a²+4-1
-4a=-9\(-4)
a=-9\4
a=-3\2
n)
(x-2)²=0
x²-4x+4=0 Δ=b²-4ac=(-4)²-4*1*4=0 x₀=-b\2a=4\2=2
o)
(2x-3)(4-x)=-2(x-5)²
8x-2x²-12+3x=-2(x²-10x+25)
-2x²+11x-12=-2x²+20x-50
-6x²-9x+38=0\(-1)
6x²+9x-38
p)
(x+3\2)²=1\4(x-2)²+3
x²+6x+9\4=1\4x²-x+1+3
x²+6x+9=4(1\4x²-x+4)
x²+6x+9=x²-4x+16
10x=7\10
x=7\10
s)
3 2\3x+4 1\2=-1\6(x-4)
11\3x+9\2=-1\6x+4\6
11\3x+1\6x=4\6-9\2
23\6x=-23\6\ 23\6
x=-1
t)
b-(7-b)=2(b+1)-9
b-7+b=2b+2-9
0=0
U
u)
2x(x-3)=(x-4)(2x+1)
2x²-6x=2x²+x-8x-4
x=-4
w)
-(5-4c)²=(4c-2)(-4c-2)
-(25-40c+16c²)=-16c²-8c+8c+4
-25+40c-16c²=-16c²+4
40c=29\40
c=29\40