Помогите пожалуйста решить

B1.
5*2ˣ - 9=4ˣ -5
-2²ˣ + 5*2ˣ -4=0
2²ˣ - 5*2ˣ +4=0

y=2ˣ
y²-5y+4=0
D=25-16=9
y₁=(5-3)/2=1 2ˣ=1 x=0
y₂=(5+3)/2=4 2ˣ=4 x=2
Ответ: 0; 2.

B2.
ОДЗ: x+1.5>0
x> -1.5

$\frac{log_{0.2}(x+1.5)}{log_{0.2} \frac{100}{4} }\ \textless \ 1 \\ \\ \frac{log_{0.2}(x+1.5)}{log_{0.2}25} \ \textless \ 1 \\ \\ \frac{log_{0.2}(x+1.5)}{log_{5^{-1}}5^2}\ \textless \ 1 \\ \\ \frac{log_{0.2}(x+1.5)}{-2log_{5}5}\ \textless \ 1 \\ \\ log_{0.2}(x+1.5)\ \textgreater \ -2 \\ x+1.5\ \textless \ (0.2)^{-2} \\ x+1.5\ \textless \ (5^{-1})^{-2} \\ x+1.5\ \textless \ 5^2 \\ x+1.5\ \textless \ 25 \\ x\ \textless \ 23.5$

{x>-1.5
{x<23.5<br>
x∈(-1.5; 23.5)

B3.
$\frac{(( \sqrt[3]{5}- \sqrt[3]{2})^2+4 \sqrt[3]{10})(( \sqrt[3]{5}- \sqrt[3]{2})^2+ \sqrt[3]{10}) }{ \sqrt[3]{5}+ \sqrt[3]{2} }= \\ \\ = \frac{( \sqrt[3]{5^2}-2 \sqrt[3]{10}+ \sqrt[3]{2^2}+4 \sqrt[3]{10})( \sqrt[3]{5^2}-2 \sqrt[3]{10}+ \sqrt[3]{2^2}+ \sqrt[3]{10} )}{ \sqrt[3]{5}+ \sqrt[3]{2} }= \\ \\ = \frac{( \sqrt[3]{5^2}+2 \sqrt[3]{10}+ \sqrt[3]{2^2})( \sqrt[3]{5^2}- \sqrt[3]{10}+ \sqrt[3]{2^2} )}{ \sqrt[3]{5}+ \sqrt[3]{2} }= \\ \\$
$= \frac{( \sqrt[3]{5}+ \sqrt[3]{2})^2( \sqrt[3]{5^2}- \sqrt[3]{10}+ \sqrt[3]{2^2} )}{ \sqrt[3]{5}+ \sqrt[3]{2}}= ( \sqrt[3]{5}+ \sqrt[3]{2})( \sqrt[3]{5^2}- \sqrt[3]{10}+ \sqrt[3]{2^2} )= \\ \\ =( \sqrt[3]{5} )^3+( \sqrt[3]{2} )^3=5+2=7$