Question

Vă rog să mă ajutați cu rezolvarea! ​

Answer 1

$Formule:~A_n^k=\dfrac{n!}{(n-k)!} ;~~~~C_n^k=\dfrac{n!}{k!(n-k)!} ;~~~~P_n=n!=1\cdot2\cdot3\cdot...\cdot n\\\\1)~2\cdot A_4^2+C_3^1=P_4+3\\\\A_4^2=\dfrac{4!}{(4-2)!}=\dfrac{\not1\cdot\not2\cdot3\cdot4}{\not1\cdot\not2} =12\\\\C_3^1=\dfrac{3!}{1!(3-1)!}=\dfrac{\not1\cdot\not2\cdot3}{\not1\cdot\not2} =3\\\\P_4=4!=1\cdot2\cdot3\cdot4=24\\\\2\cdot A_4^2+C_3^1=2\cdot12+3=24+3=27\\\\P_4+3=24+3=27\\\\2\cdot A_4^2+C_3^1=P_4+3\implies 27=27$

$2)~C_4^0+C_4^1+C_4^2+C_4^3+C_4^4=2^4\\\\C_4^0=\dfrac{4!}{0!(4-0)!}=\dfrac{\not1\cdot\not2\cdot\not3\cdot\not4}{\not1\cdot\not2\cdot\not3\cdot\not4} =1\\\\C_4^1=\dfrac{4!}{1!(4-1)!}=\dfrac{\not1\cdot\not2\cdot\not3\cdot4}{\not1\cdot\not2\cdot\not3}= 4\\\\C_4^2=\dfrac{4!}{2!(4-2)!}=\dfrac{\not1\cdot\not2\cdot3\cdot4}{\not1\cdot\not2\cdot1\cdot2} =\dfrac{12}{2}=6$

$C_4^3=\dfrac{4!}{3!(4-3)!}=\dfrac{\not1\cdot\not2\cdot\not3\cdot4}{\not1\cdot\not2\cdot\not3}=4\\\\C_4^4=\dfrac{4!}{4!(4-4)!}=\dfrac{\not1\cdot\not2\cdot\not3\cdot\not4}{\not1\cdot\not2\cdot\not3\cdot\not4}=1\\\\C_4^0+C_4^1+C_4^2+C_4^3+C_4^4=1+4+6+4+1=16=2^4$

$3)~P_4-C_6^2-9=4!-\dfrac{6!}{2!(6-2)!}-9= 1\cdot2\cdot3\cdot4-\dfrac{\not1\cdot\not2\cdot\not3\cdot\not4\cdot5\cdot6}{\not1\cdot\not2\cdot1\cdot2\cdot\not3\cdot\not4} -9=\\\\=24-\dfrac{30}{2} -9=24-15-9=0$

$4)~\dfrac{log_532-log_52}{log_54} =\dfrac{log_52^5-log_52}{log_52^2}=\dfrac{5log_52-log_52}{2log_52}=\dfrac{4log_52}{2log_52} =\dfrac{4}{2}=2$

$5)~C_4^2-C_6^1+A_4^2-P_3=\dfrac{4!}{2!(4-2)!}-\dfrac{6!}{1!(6-1)!}+\dfrac{4!}{(4-2)!} -3!=\\\\=\dfrac{\not1\cdot\not2\cdot3\cdot4}{\not1\cdot\not2\cdot1\cdot2}-\dfrac{\not1\cdot\not2\cdot\not3\cdot\not4\cdot\not5\cdot6}{\not1\cdot\not2\cdot\not3\cdot\not4\cdot\not5}+\dfrac{\not1\cdot\not2\cdot3\cdot4}{\not1\cdot\not2}-1\cdot2\cdot3=\\\\=\dfrac{12}{2}-6+12-6=6-6+12-6=6$

$6)~(P_3+P_2):(A_4^2-4)=(3!+2!):\bigg(\dfrac{4!}{(4-2)!}-4\bigg)=\\\\=(1\cdot2\cdot3+1\cdot2):\bigg(\dfrac{\not1\cdot\not2\cdot3\cdot4}{\not1\cdot\not2}-4\bigg)= (6+2):(12-4)=8:8=1$

$7)~lg25+lg2-lg5=lg(25\cdot2)-lg5=lg50-lg5=lg\dfrac{50}{5}=lg10=1 \\\\Formule:~log_ax+log_ay=log_a(x\cdot y);~~~~log_ax-log_ay=log_a\bigg(\frac{x}{y} \bigg)$

$8)~1+2+2^2+2^3+2^4+2^5=1+2+4+8+16+32=63$

$9)~\Big(3-\sqrt{3}\Big)^2+\Big(3+\sqrt{3}\Big)^2=\\\\=3^2-2\cdot3\cdot\sqrt{3} +\Big(\sqrt{3}\Big)^2 +3^2+2\cdot3\cdot\sqrt{3}+\Big(\sqrt{3}\Big)^ 2= \\\\=9-6\sqrt{3}+3+9+6\sqrt{3}+3=24\in\mathbb{N}\\\\Formule:~(a+b)^2=a^2+2ab+b^2;~~~(a-b)^2=a^2-2ab+b^2$

$10)~log_545+log_51-log_59=log_5(45\cdot1)-log_59=log_545-log_59=\\\\=log_5\bigg(\dfrac{45}{9}\bigg) =log_55=1$

$11)~C_6^5-P_3+A_6^3=\dfrac{6!}{5!(6-5)!} -3!+\dfrac{6!}{(6-3)!}=\\\\=\dfrac{\not1\cdot\not2\cdot\not3\cdot\not4\cdot\not5\cdot6}{\not1\cdot\not2\cdot\not3\cdot\not4\cdot\not5}-1\cdot2\cdot3+\dfrac{\not1\cdot\not2\cdot\not3\cdot4\cdot5\cdot6}{\not1\cdot\not2\cdot\not3} = 6-6+120=120$