(y + 1) + (y + 2) + (y + 3 ) + (y + 4) = 50
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\(y+y.\frac{1}{3}.\frac{9}{2}+y.\frac{7}{2}=25\)
\(y+y.6+y.\frac{7}{2}=25\)
\(y.\left(1+6+\frac{7}{2}\right)=25\)
\(y.\frac{21}{2}=25\)
\(y=25:\frac{21}{2}\)
\(y=25.\frac{2}{21}\)
\(y=\frac{50}{21}\)
\(y.5+y.3+y+y=50\)
\(y.\left(5+3+1+1\right)=50\)
\(y.10=50\)
\(y=5\)
(y + 1) + (y + 2) + (y + 3) + (y + 4) + (y + 5) = 50
y + 1 + y + 2 + y + 3 + y + 4 + y + 5 = 50
(y + y + y + y + y ) + (1 + 2 + 3 + 4 + 5) =50
y x (1 + 1 + 1 + 1 + 1) + 15 = 50
y x 5 + 15 = 50
y x 5 = 50 - 15
y x 5 = 35
y = 35 : 5
y = 7
Ta có : 3(x-1)= 3(z-3)
->x=z-2 (1)
2(y-2) = 3(z-3)
->y=(3z-5)/2 (2)
Thay (1),(2) vào 2x+3y-z=50 ta suy ra:
z=123/11
->x=101/11;y=157/11
Lời giải:
$3(x-1)=2(y-2); 4(y-2)=3(z-3)$
$\Rightarrow \frac{x-1}{2}=\frac{y-2}{3}; \frac{y-2}{3}=\frac{z-3}{4}$
$\Rightarrow \frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$
Áp dụng TCDTSBN:
$\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$
$=\frac{2x-2}{4}=\frac{3y-6}{9}=\frac{z-3}{4}$
$=\frac{2x-2+3y-6-(z-3)}{4+9-4}$
$=\frac{2x+3y-z-5}{9}=\frac{50-5}{9}=5$
$\Rightarrow x-1=10; y-2=15; z-3=20$
$\Rightarrow x=11; y=17; z=23$
y x 4 + ( 1 + 2 + 3 + 4 ) = 50
y x 4 + 10 = 50
y x 4 = 50 - 10
y x 4 = 40
y = 40 : 4
y = 10