Question

VA ROG AM NEVOIE PÂNĂ MÂINE!!!
1. {a,b,c} direct proportionale {2,3,4}
a+b+c= 180
2. {a,b} invers proportionale {3,7}
a+b=100
3. {x,y,z} invers proportionale {1,5; 2,4; 3}
4x-6y+z=6
4. {a,b,c,d} direct proportionale {0,2; 0,5; 0,(3);
0,(6)}
3a-c+d-b=13​

Answer 1

1. {a,b,c} d.p. {2,3,4}

$\frac{a}{2} = \frac{b}{3} = \frac{c}{3} = k$

k-coeficient de proportionalitate

$\frac{a}{2} = k = > a = 2k$

$\frac{b}{3} = k = > b = 3k$

$\frac{c}{4} = k = > c = 4k$

a+b+c= 180

2k+3k+4k=180

9k=180

k=20

a=2k=>a=2×20=>a=40

b=3k=>b=3×20=>b=60

c=4k=>c=4×20=>c=80

2. {a,b} i.p. {3,7}

3a=7b=k

$3a = k = > a = \frac{k}{3}$

$7b = k = > b = \frac{k}{7}$

a+b=100

$\frac{k}{3} + \frac{k}{7} = 100 \\ aduci \: la \: acelasi \: numitor - > 21 \\ \frac{7k}{21} + \frac{3k}{21} = 100 \\ \frac{10k}{21} = 100 \\ 10k = 21 \times 100 \\ 10k = 2100 \\ k = 2100 \div 10 \\ k = 210$

$a = \frac{k}{3} = > a = \frac{210}{3} = 70$

$b = \frac{k}{7} = > b = \frac{210}{7} = 30$

3. {x,y,z} i.p.{1,5; 2,4; 3}

$1.5 = \frac{15}{10} = \frac{3}{2} \\ 2.4 = \frac{24}{10} = \frac{12}{5}$

$\frac{3}{2} \times x = \frac{12}{5} \times y = 3z = k \\ \frac{3x}{2} = \frac{12y}{5} = 3z = k$

$\frac{3x}{2} = k = > 3x = 2k = > x = \frac{2k}{3} \\ \frac{12y}{5} = k = > 12y = 5k = > y = \frac{5k}{12} \\ 3z = k = > z = \frac{k}{3}$

4x-6y+z=6

$4 \times \frac{2k}{3} - 6 \times \frac{5k}{12} + \frac{k}{3} = 6 \\ \frac{8k}{3} - \frac{30k}{12} + \frac{k}{3} = 6 \\ \frac{8k}{3} - \frac{5k}{2} + \frac{k}{3} = 6 \\ \frac{9k}{3} - \frac{5k}{2} = 6 \\ aduci \: la \: acelasi \: numtor \\ \frac{18k}{6} - \frac{15k}{6} = 6 \\ \frac{3k}{6} = 6 = > 3k = 6 \times 6 = > \\ 3k = 36 = > k = 12$

$x = \frac{2k}{3} = \frac{2 \times 12}{3} = \frac{24}{3} = 8$

$y = \frac{5k}{12} = \frac{5 \times 12}{12} = 5$

$z = \frac{k}{3} = \frac{12}{3} = 4$