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mariannataimaz
@mariannataimaz
July 2022
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irinaorlova2
2.
2х+3\х²-2х -х-3\х²+2х=0
2х+3\х(х-2) -х-3\х(х+2)=0
(2х+3)(х+2)-(х-3)(х-2)=0
2х²+5х+6-х²+5х-6=0
х²+10х=0
х(х+10)=0
х=0 х+10=0
х=-10
3.
(х-2)(х+2)\х-3<0
x-2<0 x+2<0 x-3<0
x<2 x<-2 x<3
x∈]-∞ -2[
x²-10x+25 \x²-4x-12≥0 D=-4²-4*1*(-12)=16+48=64=8²
(x-5)(x-5) \ (X-2)(X+6)≥0 X=4-8\2=-2 X=4+8\2=6
X-5≥0 X-2>0 X+6>0
X≥5 X>2 X>6
X∈]6 +∞[
4.
(1\n²-n + 1\n²+n):n+3\ n²-1= (1(n+1)\n(n-1)(n+2) + (n-1)\n(n+1)(n-1)) : n+3\n(n-1)= 2n\n(n-1)(n+1) : n+3\n(n-1)=2n\n(n-1)(n+1)* n(n-1)\n+3= = 2n\(n+1)(п+3) = 2п\п²+4п+3
п=-1 2*(-1)\(-1)²+4(-1)+3=-2\1-4+3=0
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Answers & Comments
2х+3\х²-2х -х-3\х²+2х=0
2х+3\х(х-2) -х-3\х(х+2)=0
(2х+3)(х+2)-(х-3)(х-2)=0
2х²+5х+6-х²+5х-6=0
х²+10х=0
х(х+10)=0
х=0 х+10=0
х=-10
3.
(х-2)(х+2)\х-3<0
x-2<0 x+2<0 x-3<0
x<2 x<-2 x<3
x∈]-∞ -2[
x²-10x+25 \x²-4x-12≥0 D=-4²-4*1*(-12)=16+48=64=8²
(x-5)(x-5) \ (X-2)(X+6)≥0 X=4-8\2=-2 X=4+8\2=6
X-5≥0 X-2>0 X+6>0
X≥5 X>2 X>6
X∈]6 +∞[
4.
(1\n²-n + 1\n²+n):n+3\ n²-1= (1(n+1)\n(n-1)(n+2) + (n-1)\n(n+1)(n-1)) : n+3\n(n-1)= 2n\n(n-1)(n+1) : n+3\n(n-1)=2n\n(n-1)(n+1)* n(n-1)\n+3= = 2n\(n+1)(п+3) = 2п\п²+4п+3
п=-1 2*(-1)\(-1)²+4(-1)+3=-2\1-4+3=0