Question

3 si 4
Dau coroana si 28 de puncte.
Materie clasa a 10 a.
Multumesc anticipat.​

Answer 1

 

3)

$\displaystyle\bf\\2^{2x+1}+2^{x+2}=160\\\\2^{2x}\cdot2^1+2^{x}\cdot2^2=160\\\\2\cdot2^{2x}+4\cdot2^x=160\\\\2\cdot\Big(2^x\Big)^2+4\cdot2^x=160\\\\Facem~substitutia\!:~~~\boxed{\bf2x=y}\\\\2y^2+4y-160=0~~\Big|:2\\\\y^2+2y-80=0\\\\y_{12}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{-2\pm\sqrt{4+320}}{2}=\\\\~~~~~~~=\frac{-2\pm\sqrt{324}}{2}=\frac{-2\pm18}{2}=-1\pm9\\\\y_1=-1+9=8\\y_2=-1-9=-10~~~Solutie~eliminata~deoarece~~2^x>0$

.

$\displaystyle\bf\\Rezulta~solutia:~~~\boxed{\bf y=8}\\\\Ne~intoarcem~la~substitutie.\\\\2^x=y\\\\2^x=8\\\\2^x=2^3\\\\\boxed{\bf x=3}$

.

$\displaystyle\bf\\4)\\\\Avem~22~de~elevi~din~care~12~sunt~fete.\\\\12~fete\\\\22-12=10~baieti\\\\Comitetul~format~din~3~fete~si~2~baieti.\\\\Calculam~numnarul~de~moduri~(n):\\\\n=C_{12}^3\times C_{10}^2=\\\\n=\frac{12\times11\times10}{1\times2\times3}\times\frac{10\times9}{2}=\\\\=(2\times11\times10)\times(5\times9)=220\times45=9900~moduri.$

 

Answer 2

$\it 2^{2x+1}+2^{x+2}=160|_{:2}\ \Rightarrow 2^{2x}+2^{x+1}=80 \Rightarrow 2^{2x}+2\cdot2^x- 80=0 \Rightarrow \\ \\ \Rightarrow 2^{2x}+10\cdot2^x-8\cdot2^x-80=0 \Rightarrow 2^x(2^x+10) -8(2^x+10)=0 \Rightarrow \\ \\ \Rightarrow (\underbrace{2^x+10}_{>0})(2^x-8)=0=0 \Rightarrow 2^x-8=0 \Rightarrow 2^x=8 \Rightarrow 2^x=2^3 \Rightarrow x=3$

$\it 4)\ C^3_{12}\ \cdot\ C^2_{10}= 220\cdot45=9900$