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Lizuuuuuuunya
@Lizuuuuuuunya
August 2022
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Помогите пожалуйста решить!
1)1/x(x+2)+2/(x+1)^2=2
2)1,5+1≤lx-1l
3)log5(2x-3)/log1/3log3 9 больше 0
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oganesbagoyan
Verified answer
1)
1/x(x+2)+2/(x+1)²=2 ; * * *ОДЗ x ∉{ -2 ;-1
;0 } * * *
1/(x
²+2x) +2/(x²+2x+1) =2; * * * замена t =x
²+2x * * *
1/t +1/(t+1) =2 ;
t+1 +t =2t(t+1)
⇔2t²=1 ⇒t =±1/√2.
а)x²+2x = -1/√2⇔x²+2x +1/√2 =0 (не имеет действительных корней)
.
б)x²+2x =1/√2 ⇔x²+2x -1/√2 =0 ⇒x= -1± √(1+1/√2).
-------
2)
1,5+1≤ lx-1l * * * ??? * * *
lx-1l ≥2,5 ⇔[ x-1≤ -2,5 ; x-1≥2,5 .⇔[ x≤ -1,5 ; x
≥ 3,5.
ответ: x∈ (-∞; -1,5] ∪ [3,5;∞).
-------
3)
log_5 (2x-3)/ log_1/3 log_3 9 >0 ;
log_5 (2x-3)/ log_1/3 2 >0 ; * * *т.к. 0<1/3<1 ,то log_1/3 2 <
0 * * *
log_5 (2x-3) < 0 * * * т.к. 5>1 ,т
о
⇔0< 2x-3 <1⇔ 3<2x< 4
⇔3/2 <x<2.
ответ: x∈ (1,5;2).
5 votes
Thanks 8
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Answers & Comments
Verified answer
1)1/x(x+2)+2/(x+1)²=2 ; * * *ОДЗ x ∉{ -2 ;-1;0 } * * *
1/(x²+2x) +2/(x²+2x+1) =2; * * * замена t =x²+2x * * *
1/t +1/(t+1) =2 ;
t+1 +t =2t(t+1) ⇔2t²=1 ⇒t =±1/√2.
а)x²+2x = -1/√2⇔x²+2x +1/√2 =0 (не имеет действительных корней).
б)x²+2x =1/√2 ⇔x²+2x -1/√2 =0 ⇒x= -1± √(1+1/√2).
-------
2)
1,5+1≤ lx-1l * * * ??? * * *
lx-1l ≥2,5 ⇔[ x-1≤ -2,5 ; x-1≥2,5 .⇔[ x≤ -1,5 ; x ≥ 3,5.
ответ: x∈ (-∞; -1,5] ∪ [3,5;∞).
-------
3)
log_5 (2x-3)/ log_1/3 log_3 9 >0 ;
log_5 (2x-3)/ log_1/3 2 >0 ; * * *т.к. 0<1/3<1 ,то log_1/3 2 < 0 * * *
log_5 (2x-3) < 0 * * * т.к. 5>1 ,то
⇔0< 2x-3 <1⇔ 3<2x< 4 ⇔3/2 <x<2.
ответ: x∈ (1,5;2).