Question

i4,i5,i6,i7,i8,i9,i10.Va rooog !!

Answer 1

sper ca se intelege scrisul meu. si scuze ca nu l-am facut pe 10,nu stiam sa il fac

Answer 2

4.
$\sqrt[3]{125}+\sqrt{16}+\sqrt[3]{-27}=\\ \\=\sqrt[3]{5^{3}}+\sqrt{4^{2}}+\sqrt[3]{(-3)^{3}}=\\ \\=5+4+(-3)=\\ \\=9-3=\\ \\=6$

5.
$\log_{11}11+\log_{7}\frac{1}{7}=\\ \\=1+\log_{7}7^{-1}=\\ \\=1+(-1)=\\ \\=1-1=\\ \\=0$

6.
$\log_{3}81+\log_{5}25-\lg100000=\\ \\ =\log_{3}3^{4}+\log_{5}5^{2}-\lg10^{5}=\\ \\ =4+2-5=\\ \\ =6-5=\\ \\ =1$

7.
$a=-\sqrt[3]{27}=-\sqrt[3]{3^{3}}=-3\\ \\b=\log_{2}\frac{1}{16}=\log_{2}2^{-4}=-4\\ \\c=-2\\ \\ -4<-3<-2\\ \\ \rightarrow b<a<c\\ \\ ordonarea:b,a,c$

8.
$\log_{4}64+\sqrt[3]{1000}=\sqrt{16}+(\frac{1}{3})^{-2}\\ \\ \log_{4}4^{3}+\sqrt[3]{10^{3}}=4+3^{2}\\ \\3+10=4+9\\ \\13=13$

9.
$\log_{11}121=\log_{11}11^{2}=2\\ \\ \sqrt[3]{27}=\sqrt[3]{3^{3}}=3\\ \\ 2<3~\rightarrow \log_{11}121<\sqrt[3]{27}$

10.
$\log_{2}3=a\\ \\ \log_{2}6=\log_{2}(2 \cdot 3)=\log_{2}2+\log_{2}3=1+a$


Foarte usor...