Question

Cine ma poate ajuta cu cateva raspunsuri acestui test , va implor , Multumesc !

Answer 1

$\displaystyle Subiectul ~1 \\ \\ 1).-3 \sqrt{7} \cdot (- \sqrt{7} )=3 \cdot 7=21 \\ \\ 2).ip^2=c_1^2+c_2^2 \Rightarrow 15^2=12^2+c_2^2 \Rightarrow 225=144+c_2^2 \Leftrightarrow \\ \Leftrightarrow c_2^2=225-144 \Rightarrow c_2^ 2=81 \Rightarrow c_2= \sqrt{81} \Rightarrow c_2=9 ~m \\ P=15+12+9=27+9=36~m \\ \\ 3). \sqrt{(-58+54)^2} : \sqrt{(103-105)^2} =|-58+54|:|103-105|= \\ =|-4|:|-2|=4:2=2 \\ \\ 4).A= \frac{d_1 \cdot d_2}{2} = \frac{2,4 \cdot 2,5}{2} = \frac{6}{2} =3~cm^2$

$\displaystyle 5). \left(\begin{}{- \frac{1}{6} }&&\\&&\\&&\end{.}\right)^{-27+25}= \left(\begin{}{- \frac{1}{6} }&&\\&&\\&&\end{.}\right) ^{-2}=36$

$\displaystyle 6).A= \frac{(B+b) \cdot h}{2} = \frac{(9+5) \cdot 7}{2} = \frac{14 \cdot 7}{2} = \frac{98}{2} =49~m^2$

$\displaystyle Subiectul~2 \\ \\ 1).a=9- \sqrt{72}=9-6 \sqrt{2} \\ b=3(3+2 \sqrt{2} ) =9+6 \sqrt{2} \\ m_a= \frac{a+b}{2} = \frac{9-6 \sqrt{2} +9+6 \sqrt{2} }{2} = \frac{9+9}{2} = \frac{18}{2} =9 \\ \\ m_g= \sqrt{a \cdot b} = \sqrt{(9-6 \sqrt{2} )(9+6 \sqrt{2} })= \sqrt{81-72} = \sqrt{9} =3$

$3).9x^3-4x^5=x^3(9-4x^2)=x^3(3-2x)(3+2x) \\ \\ 4).P=6 \cdot l=6 \cdot 12=72~cm$

$\displaystyle Subiectul~1 \\ \\ 1).4 \sqrt{2} =(4 \sqrt{2} )^2=32 \\ 2 \sqrt{4} =(2 \sqrt{4} )^2=16 \\ 3 \sqrt{3} =(3 \sqrt{3} )^2=27 \\ Dintre~numerele~4 \sqrt{2} ,~2 \sqrt{4} ,~3 \sqrt{3} ~mai~mare~este~numarul~4 \sqrt{2} . \\ \\ 2). \left(\begin{}{- \frac{2}{5} }&&\\&&\\&&\end{.}\right)^{25}: \left(\begin{}{- \frac{2}{5} }&&\\&&\\&&\end{.}\right) ^{27}=\left(\begin{}{- \frac{2}{5} }&&\\&&\\&&\end{.}\right)^{25-27}=\left(\begin{}{- \frac{2}{5} }&&\\&&\\&&\end{.}\right) ^{-2}= \frac{25}{4}$

$\displaystyle 3). \sqrt{(8-11)^2} - \sqrt{(-6+10)^2} =|8-11|-|-6+10|= \\ =|-3|-|4|=3-4=-1 \\ \\ 4).n= \sqrt{3} \\ n(5n-3)-3n(n-1) \\ \sqrt{3} (5 \sqrt{3} -3)-3 \sqrt{3} ( \sqrt{3} -1)=15-3 \sqrt{3} -9+3 \sqrt{3} =15-9=6 \\ \\ 6).l_m= \frac{B+b}{2} = \frac{27+19}{2} = \frac{46}{2} =23~m$