Question

[Trần Dương Quang Hiếu]

Tìm x;y;z biết:

\(\frac{9x-13}{16}=\frac{7y-11}{19}=\frac{z-5}{3}\)và 5x-7y+3z=9

Answer 1

[Đỗ Lê Tú Linh]

\(\frac{9x-13}{16}=\frac{7y-11}{19}=\frac{z-5}{3}\)

\(\frac{\left(9x-13\right)\cdot\frac{5}{9}}{16\cdot\frac{5}{9}}=\frac{7y-11}{19}=\frac{3\left(z-5\right)}{3\cdot3}\)

\(\frac{5x-\frac{65}{9}}{\frac{80}{9}}=\frac{7y-11}{19}=\frac{3z-15}{9}\)

Áp dụng tc dãy tỉ số bằng nhau, ta được:

\(\frac{5x-\frac{65}{9}}{\frac{80}{9}}=\frac{7y-11}{19}=\frac{3z-15}{9}=\frac{5x-\frac{65}{9}-7y+11+3z-15}{\frac{80}{9}-19+9}=\frac{\left(5x-7y+3z\right)-\left(\frac{65}{9}-11-15\right)}{\frac{80}{9}-\frac{171}{9}+\frac{81}{9}}\)

\(=\frac{9-\left(\frac{65}{9}-\frac{99}{9}-\frac{135}{9}\right)}{-\frac{10}{9}}=\frac{9+\frac{169}{9}}{-\frac{10}{9}}=-\frac{88}{9}:-\frac{10}{9}=\frac{44}{5}\)

=>x=[(44/5*16)+13]/9=769/45

=>y=[(44/5*19)+11]/7=891/35

=>z=(44/5*3)+5=157/5

hình như sai hay sao, số lớn quá