Răspuns:
Ai rezolvare pt ex 1 si 2
Explicație pas cu pas:
a=(2+√2-3+2√2)(2+√2+3-2√2)+11-8√2
a=(3√2-1)(5-√2)+11-8√2
a=15√2-6-5+√2+11-8√2
a=8√2
b=3∛3-4∛3+5∛3
b=4∛3
a²+b³=128+192=320
b. [tex]2^{3x-1} <4^{2x+3} \\2^{3x-1} <2^{2(2x+3)} \\[/tex]
3x-1<4x+6
-x<7
x>-7
x∈(-7,+∞)
2. a. x-1>0 x>1 x∈(1,+∞)
2x²-3x+1>0
Δ=9-8=1
x₁=2
x₂=1
x∈(-∞,1)∪(2,+∞)
b.
[tex]a=log_{2}( \frac{1024}{1023}\cdot \frac{1023}{1022} \cdot... \cdot \frac{2}{1})=log_{2} 1024=10[/tex]
[tex]b=log_{3}81+ log_{\frac{1}{2}} 0,25+lg1000+log_{\frac{1}{3}} 27}=4+2+3-3=6[/tex]
a+b=16
