Question

Urgent!!! Va rog!!!!!!

Answer 1

subiectul 1
1) √6>√2
2) (-10/12) : (-5/6)=(-10/12) x (-6/5)=1
3)√(-5-7)²=√(-12)²=√144=12
4) 3x-1-2x+5=x+4
5) √2 * sin 45⁰=√2 *√2 /2=1
6) triunghiul dreptunghic isoscel are un unghi ascutit de (180⁰-90⁰)/2=90⁰/2=45⁰

subiectul 2
1) (√3+1)(√3-1)+6=3-1+6=8
2)(a+b)/2=(3+√2+3-√2)/2=6/2=3
3) ipotenuza=√(5²+12²)=√(25+144)=√169=13 cm.
P=5+12+13=30 cm  (raspunc corect A
4)diagonala=√(8²+6²)=√(64+36)=√100=10 cm.
5) x-3=-10
x=-7
raspuns corect A .  -7

subiectul 3
1.(-3/8) : (6/8) +(11/7) *(-7/22)=
=(-3/8) x (8/6) -1/2=
=-1/2 -1/2=-1
2) 3*(x-2)=9
3x-6=9
3x=15
x=5
3) consideram <A=90⁰
AB=√(25²+15²)=√(625+225)=√400=20 cm.
notam CD=x
BD=25-x
AD²=15²-x²
AD²=20²-(25-x²)=400-625-x²+50x=-225-x²+50x
15²-x²=-225-x²+50x
225=-225+50x
50x=500
CD=x=10 cm
AD=√(15²-x²)=√(225-100)=√125=5√3 cm

Answer 2

$\displaystyle Subiectul ~1 \\ \\ 1). \sqrt{2} = \sqrt{2} ^2=2~;~ \sqrt{6} = \sqrt{6} ^2=6 \\ 2\ \textless \ 6 \Rightarrow \sqrt{2} \ \textless \ \sqrt{6} \\ \\ 2). \left(\begin{}{- \frac{10}{12}}&&\\&&\\&&\end{.}\right):\left(\begin{}{- \frac{5}{6} }&&\\&&\\&&\end{.}\right)= \frac{2}{2} =1 \\ \\ \\ 3). \sqrt{(-5-7)^2} =|-5-7|=|-12|=12 \\ \\ 4).3x-1-2x+5=x+4 \\ \\ 5). \sqrt{2} \cdot sin~45^o= \sqrt{2} \cdot \frac{1}{ \sqrt{2} }= \frac{ \sqrt{2} }{ \sqrt{2} }=1$

$\displaystyle 6).Triunghiul~dreptunghic~isoscel~are~un~unghi~ascutit~cu \\ masura~de~45^o. \\ \\ Subiectul~2 \\ \\ 1).( \sqrt{3} +1)( \sqrt{3} -1)+6=3-1+6=2+6=8 \\ \\ 2).a=3+ \sqrt{2} \\ b=3- \sqrt{2} \\ m_a= \frac{a+b}{2} = \frac{3+ \sqrt{2} +3- \sqrt{2} }{2} = \frac{3+3}{2} = \frac{6}{2} =3 \\ \\ 3).c_1=5~cm~;~c_2=12~cm \\ ip^2=c_1^2+c_2^2 \Rightarrow ip^2=5^2+12^2 \Rightarrow ip^2=25+144 \Rightarrow ip^2=169 \Leftrightarrow \\ \Leftrightarrow ip= \sqrt{169} \Rightarrow ip=13~cm$
$P=5+12+13=17+13=30~cm$

$4).L=8~cm~;~l=6~cm \\ AC^2=AB^2+BC^2 \Rightarrow AC^2=8^2+6^2 \Rightarrow AC^2=64+36 \Leftrightarrow \\ \Leftrightarrow AC^2=100 \Rightarrow AC= \sqrt{100} \Rightarrow AC=10 ~cm \\ \\ 5).x-3=-10 \\ x=-10+3 \\ x=-7$

$\displaystyle Subiectul ~3 \\ \\ 1). \left(\begin{}{- \frac{3}{8} }&&\\&&\\&&\end{.}\right):\left(\begin{}{ \frac{6}{8} }&&\\&&\\&&\end{.}\right)+\left(\begin{}{ \frac{11}{7} }&&\\&&\\&&\end{.}\right) \cdot \left(\begin{}{- \frac{7}{22} }&&\\&&\\&&\end{.}\right)=\left(\begin{}{- \frac{3}{8} }&&\\&&\\&&\end{.}\right) \cdot \left(\begin{}{ \frac{8}{6} }&&\\&&\\&&\end{.}\right)+\left(\begin{}{- \frac{77}{154} }&&\\&&\\&&\end{.}\right)=\\ \\ =- \frac{24}{48} - \frac{77}{154} =$ $\displaystyle =- \frac{1}{2} - \frac{1}{2}= - \frac{2}{2} =-1$
$\displaystyle 2).3(x-2)=-9 \\ 3x-6=-9 \\ 3x=-9+6 \\ 3x=-3 \\ x=- \frac{3}{3} \\ x=-1$