Question

3*9^x - 5*6^x + 2*4^x≥0

Answer 1

$3* 9^{x} -5* 6^{x} +2* 4^{x} \geq 0 | : 4^{x} \neq 0$
$6* \frac{ 9^{x} }{ 4^{x} } -5* \frac{ 6^{x} }{ 4^{x} } +2* \frac{ 4^{x} }{ 4^{x} } \geq 0$
$5* ( \frac{9}{4} )^{x} -5* ( \frac{6}{4} )^{x} +2*1 \geq 0$
$6* (( \frac{3}{2} )^{2} ) ^{x} -5* ( \frac{3}{2} )^{x} +2 \geq 0$
$6*( ( \frac{3}{2} )^{x} )^{2} -5* ( \frac{3}{2} )^{x} +2 \geq 0$
- показательное квадратное неравенство, замена переменной:
$( \frac{3}{2} )^{x}=t, t\ \textgreater \ 0$
t²-5t+2≥0, метод интервалов:

1. t²-5t+2=0, t₁=4/6, t₂=1

2. ++++[4/6]--------[1]+++++++>t

3. t≤4/6, t≥1

обратная замена:
1.  $t_{1} \leq \frac{4}{6} , ( \frac{3}{2} )^{x} \leq \frac{4}{6} , ( \frac{3}{2} )^{x} \leq ( \frac{3}{2} )^{-1}$
- простейшее показательное неравенство, основание степени а=3/2, 3/2>1.  => знак неравенства не меняем
x≤-1
2.  $t_{1} \geq 1, ( \frac{3}{2} )^{x} \geq 1, ( \frac{3}{2} )^{x} \geq ( \frac{3}{2} )^{0}$
x≥0

ответ:  x∈(-∞; - 1]∪[0; ∞)

Answer 2

To solve the inequality $3 \cdot 9^x - 5 \cdot 6^x + 2 \cdot 4^x \geq 0$, we can use some algebraic techniques.

Let's start by factoring and simplifying the terms involving powers of numbers:

$3 \cdot 9^x = 3 \cdot (3^2)^x = 3 \cdot 3^{2x} = 3^{2x+1}$

$5 \cdot 6^x = 5 \cdot (2 \cdot 3)^x = 5 \cdot 2^x \cdot 3^x = 5 \cdot 2^x \cdot 3^{x}$

$2 \cdot 4^x = 2 \cdot (2^2)^x = 2 \cdot 2^{2x} = 2^{2x+1}$

Now, the inequality becomes:

$3^{2x+1} - 5 \cdot 2^x \cdot 3^{x} + 2^{2x+1} \geq 0$

Now, let's try to simplify the terms further. Notice that $3^{2x+1}$ and $2^{2x+1}$ have the same base, 3 and 2 respectively. We can combine them into a single term with a common base:

$3^{2x+1} + 2^{2x+1} = 3 \cdot 3^{2x} + 2 \cdot 2^{2x} = 3 \cdot (3^x)^2 + 2 \cdot (2^x)^2$

So, the inequality becomes:

$3 \cdot (3^x)^2 + 2 \cdot (2^x)^2 - 5 \cdot 2^x \cdot 3^{x} \geq 0$

Now, let's make a substitution to simplify things further. Let $y = 3^x$. Then $y^2 = (3^x)^2$ and $2^x = \frac{y}{2}$. The inequality becomes:

$3y^2 + 2\left(\frac{y}{2}\right)^2 - 5\left(\frac{y}{2}\right) \cdot y \geq 0$

Simplify:

$3y^2 + \frac{y^2}{2} - \frac{5}{2}y^2 \geq 0$

Combine the terms:

$\frac{7}{2}y^2 \geq 0$

Now, we need to find the range of values for $y$ (which is $3^x$) for which the inequality holds true. Since $y^2$ is always non-negative, the inequality will be true for any value of $y$ that is not equal to zero.

Therefore, the solution to the inequality is all real numbers $x$ except $x = \log_3{0}$ (which is undefined), or simply $x \neq -\infty$.

In interval notation, the solution is $(-∞, +∞)$ or $(-∞, -\log_3{0}) \cup (-\log_3{0}, +∞)$.