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Kiwi25
a fost răspuns

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

I4i5i6i7i8i9i10Va Rooog class=

Răspuns :

sper ca se intelege scrisul meu. si scuze ca nu l-am facut pe 10,nu stiam sa il fac
Vezi imaginea Slavitescudora
4.
[tex]\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[/tex]

5.
[tex]\log_{11}11+\log_{7}\frac{1}{7}=\\
\\=1+\log_{7}7^{-1}=\\
\\=1+(-1)=\\
\\=1-1=\\
\\=0[/tex]

6.
[tex]\log_{3}81+\log_{5}25-\lg100000=\\
\\ =\log_{3}3^{4}+\log_{5}5^{2}-\lg10^{5}=\\
\\ =4+2-5=\\
\\ =6-5=\\
\\ =1[/tex]

7.
[tex]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[/tex]

8.
[tex]\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[/tex]

9.
[tex]\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}[/tex]

10.
[tex]\log_{2}3=a\\
\\ \log_{2}6=\log_{2}(2 \cdot 3)=\log_{2}2+\log_{2}3=1+a[/tex]


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