Question

Cineva care imi poate face tot testul?

Answer 1

Partea I

1. $m=1^n+2^n+3^n+4^n+5^n, n \in \mathbb{N} \\ b.~n=3 \rightarrow m=1^3+2^3+...+5^3 \\ \\ m=\frac{5^2(5+1)^2}{4}= \\ \\ m= \frac{25*36}{4}=25*9 \\ \\ \bold{m=225} \\ \\ a.~Cel~mai~mic~n~pentru~care~m~este~divizibil~cu~5~este~0 \\ 1^0+2^0+3^0+4^0+5^0=1+1+1+1+1=5$

2.$a=\frac12+\frac13+\frac14+\frac15+\frac16 \\ \\ a=(\frac11+\frac12+\frac13+\frac14+\frac15+\frac16)-\frac11 \\ \\ a=\frac{2+1}{2*1}+\frac{4+3}{4*3}+\frac{6+5}{6*5}-1 \\ \\ a=^{30)}\frac{3}{2}+^{5)}\frac{7}{12}+^{2)}\frac{11}{30}-1 \\ \\ a=\frac{90}{60}+\frac{35}{60}+\frac{22}{60}-1 \\ \\a=\frac{147}{60}-1 \\ \\ a=2,45-1 \\ \\ \bold{a=1,45}$

3.
a)Notezi numarul cu x
1/5·x=x/4-10
x/5=x/4-10
x/4-x/5=10
Amplifici prima fractie cu 5 si a doua cu 4
5x/20-4x/20=10
x/20=10
x=200

b)12/100·1025=12·1025/100=3·1025/25=3·41=123

4.
a/b=0,6=6/10=2/5⇒2b=5a
a) b/a=52

b) (5a+b)/(10a+2b)=
(2b+b)/(4b+2b)=3b/6b=1/2

5. a)
A={x∈Z| |2x+5|=7}
2x+5∈{7,-7} |-5
2x∈{2,-12} |:2
x∈{1,-6}

B={x∈N| |3x+5|<11}
I. 3x+5<11| -5
3x<6 |:3
x<2
II. -(3x+5)<11
-3x-5<11 |+5
-3x<16 |:(-3)
x>-16/3

⇒x∈(-16/3,2) 
b) este in ataşare

6 b) este in atasare, nu stiu sa fac a)

7. a)
$\left \{ {{\frac1x}+\frac2y=2 \atop {2x+y=4}} \right \Leftrightarrow \left \{ {{\frac1x}+\frac2y=2 \atop {y=4-2x}} \right \Leftrightarrow \left \{ {{\frac1x}+\frac2y=2 \atop {y=2(2-x)}} \right \Leftrightarrow \left \{ {{\frac1x}+\frac{2}{2(2-x)} \atop {{y=2(2-x)}}} \right \\ \\ \Leftrightarrow \left \{ {{\frac1x}+\frac{1}{2-x}=2 \atop {{y=2(2-x)}}} \right \Leftrightarrow \left \{ {{\frac{2-x}{x(2-x)}}+\frac{x}{x(2-x)}=2 \atop {{y=2(2-x)}} \right \Leftrightarrow$$\left \{ {{\frac{2-x+x}{x(2-x)}}=2 \atop {{y=2(2-x)}} \right \Leftrightarrow \left \{ {{\frac{2}{x(2-x)}}=2 \atop {{y=2(2-x)}} \right \Leftrightarrow \left \{ {{x(2-x)=1} \atop {{y=2(2-x)}} \right \Leftrightarrow \left \{ {{x(2-x)=1} \atop {{y=2(2-x)}} \right \left \{ {x=1} \atop {{y=2(2-1)}} \right \\ \\ \Leftrightarrow \left \{ {{x=1} \atop {{y=2}} \right \rightarrow \bold{(x,y)=(1,2)}$

b)
|x+2|+|y-3|=0
|x+2|=0
x+2=0⇒x=-2
|y-3|=0
y-3=0⇒y=3

Partea a IIa

1. a)
$a+b=7 \\ \\ \frac1a+\frac1b=\frac{7}{12} \\ \\ \frac{b}{ab}+\frac{a}{ab}=\frac{7}{12} \\ \\ \frac{7}{ab}=\frac{7}{12} \\ \\ ab=12 \\ a+b=7 \rightarrow a=3~si~b=4$

b)
$4*\frac{x}{100}=3 \\ \\ 4x=300 \\ \\ \bold{x=75\%}$

2. b)
a²+b²+c²≥ab+ac+bc
a²+b²+c²-ab-ac-bc≥0 |·2
a²+a²+b²+b²+c²+c²-2ab-2bc-2ca≥0
a²+b²-2ab+b²+c²-2bc+c²+a²-2ca≥0(a²+b²-2ab)+(b²+c²-2bc)+(c²+a²-2ca)≥0(a-b)²+(b-c)²+(c-a)²≥0 (Adevarat) pentru ca:
(a-b)²>0
(b-c)²>0
(c-a)²>0