Решите неравенство подробно Ответ: (-бескон ; 2] ;[3 ;+бескон). Created by oihgsgjs. algebra-ru

Решите неравенство подробно Ответ: (-бескон ; 2] ;[3 ;+бескон)...

hlopushinairina 4^x-3·2^(x+2)+32≥0;⇒
4^x=2^(2x);2^(x+2)=2²·2^x=4·2^x;⇒
2^(2x)-12·2^x+32≥0;⇒
2^x=y;y>0;⇒
y²-12y+32≥0;
y₁,₂=6⁺₋√36-32=6⁺₋√4=6⁺₋2;
y₁=8;y₂=4;⇒
y≤4;y≥8;
2^x≤4;⇒x≤2;⇒(-∞;2];
2^x≥8;⇒x≥3;⇒[3;+∞)

2 votes Thanks 2

julyap $4 ^{x} -3*2 ^{x+2} +32 \geq 0 \\ 2 ^{2x} -3*2 ^{x} *4+32 \geq 0 \\ 2 ^{x} =t \\ t^{2} -3t*4+32 \geq 0 \\ t ^{2} -12t+32 \geq 0 \\ t ^{2} -12t+32=0 \\ D=144-128=16 \\ \sqrt{D} =4 \\ t _{1} = \frac{12+4}{2} = \frac{16}{2} =8 \\ t _{2} = \frac{12-4}{2} =4$

$2 ^{x} =8 \\ 2 ^{x} =2 ^{3} \\ x=3 \\ 2 ^{x} =4 \\ 2 ^{x} =2 ^{2} \\ x=2 \\ \\ (x-2)(x-3) \geq 0 \\ \\$
+ - +
--------------|--------------|----------------------> x
2 3

$x \in (- \infty ;2]\cup [3;+ \infty )$

0 votes Thanks 1

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