Помогите решить неравенства!!!! 1. log_1/2(2х+1)>-2 2. log_8(х^2+4х+3)≤1 3. log_1/2(4-х)+log_1/2(х-1)≥-1. Created by lapusic98. algebra-ru

1. 2x+1=2 2x+1=4 x=1,52x=1 x=0,52x+1=1 x=02x+1=1/2 x=1/4 вывод x∈(1,5;-∞)2.x²+4x+3=8x²+4x-5=0по виетеx1*x2=-5 x1=1x1+x2=-4 x2=-5 3.4-x+x-1=3=1 x=0=2 x=-1

2(х-1)≥-1...

lapusic98 @lapusic98

September 2021 1 6 Report

Помогите решить неравенства!!!!
1. log_1/2(2х+1)>-2
2. log_8(х^2+4х+3)≤1
3. log_1/2(4-х)+log_1/2(х-1)≥-1

Answers & Comments

greatnat 1. $log_{ \frac{1}{2} }(2x+1) > -2$
$\frac{1}{2} ^{-1} =2x+1$      $\frac{1}{2} ^{-2} =2x+1$
2x+1=2 2x+1=4 x=1,5
2x=1 x=0,5
$\frac{1}{2} ^{0} =2x+1$
2x+1=1 x=0
$\frac{1}{2} ^{1} =2x+1$
2x+1=1/2 x=1/4 вывод x∈(1,5;-∞)

2. $log_{8} ( x^{2} +4x+3) \leq 1$
$8^{1} = x^{2} +4x+3$
x²+4x+3=8
x²+4x-5=0
по виете
x1*x2=-5    x1=1
x1+x2=-4 x2=-5

3. $log_{ \frac{1}{2} } (4-x) + log_{ \frac{1}{2} } (x-1) \geq -1$
4-x+x-1=3
$\frac{1}{2} ^{x}$ =1    x=0
$\frac{1}{2} ^{x}$ =2 x=-1

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