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freeforever
@freeforever
October 2021
1
16
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[tex] x^{2} [/tex]+|2x|[tex] \geq [/tex]63
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ТатМих
Verified answer
X≥0
x²+2x≥63
x²+2x-63≥0
D=4+252=256
√D=16
x1=(-2-16)/2=-18/2=-9
x2=(-2+16)/2=14/2=7
(x+9)(x-7)≥0
+ - +
////////////-9...................7/////////////////////////
x=[-∞;-9] ∨ x=[7;+∞]
x≥0
x=[7;+∞]
2.
x≤0
x²-2x-63≥0
D=4+252=256
√256=16
x1=(2-16)/2=-7
x2=(2+16)/2=9
(x+7)(x-9)≥0
+ - +
////////////-7.....................9////////////////////
x=[-∞;-7]∨x=[9;+∞]
x≤0
x=[-∞;-7]
Ответ:
x=[-∞;-7]∨x=[7;+∞]
.
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Verified answer
X≥0x²+2x≥63
x²+2x-63≥0
D=4+252=256
√D=16
x1=(-2-16)/2=-18/2=-9
x2=(-2+16)/2=14/2=7
(x+9)(x-7)≥0
+ - +
////////////-9...................7/////////////////////////
x=[-∞;-9] ∨ x=[7;+∞]
x≥0
x=[7;+∞]
2.
x≤0
x²-2x-63≥0
D=4+252=256
√256=16
x1=(2-16)/2=-7
x2=(2+16)/2=9
(x+7)(x-9)≥0
+ - +
////////////-7.....................9////////////////////
x=[-∞;-7]∨x=[9;+∞]
x≤0
x=[-∞;-7]
Ответ:x=[-∞;-7]∨x=[7;+∞].