Question

[hoa anh dao]

1/ Tìm \(a1;a2;a3;.......;a9\)biết :

        \(\frac{a1-1}{9}=\frac{a2-2}{8}=\frac{a3-3}{7}=.......=\frac{a9-9}{1}\)và  \(a1+a2+a3+.........+a9=90\)

2/ Tìm x,y,z biết :

a)        \(x:y:z=3:4:5\)và  \(2x^2+2y^2-3z^2=-100\)

b)         \(\frac{x-1}{2}=\frac{y+3}{4}=\frac{z-5}{6}\)và  \(5z-3x-4y=50\)

Answer 1

[ST]

Bài 1:

Áp dụng TCDTSBN có:

\(\frac{a1-1}{9}=\frac{a2-2}{8}=...=\frac{a9-9}{1}=\frac{a1-1+a2-2+...+a9-9}{9+8+...+1}=\frac{\left(a1+...+a9\right)-\left(1+2+...+9\right)}{45}=\frac{90-45}{45}=1\)

\(\Rightarrow\frac{a1-1}{9}=1\Rightarrow a1=10\)

\(\frac{a2-2}{8}=1\Rightarrow a2=10\)

.....

\(\frac{a9-9}{1}=1\Rightarrow a9=10\)

Vậy a1=a2=...=a9=10

2,

a, \(\frac{x}{3}=\frac{y}{4}=\frac{z}{5}\Rightarrow\frac{x^2}{9}=\frac{y^2}{16}=\frac{z^2}{25}\Rightarrow\frac{2x^2}{18}=\frac{2y^2}{32}=\frac{3z^2}{75}=\frac{2x^2+2y^2-3z^2}{18+32-75}=\frac{-100}{-25}=4\)

=> x=6, y=8, z=10

b, \(\frac{x-1}{2}=\frac{y+3}{4}=\frac{z-5}{6}\Rightarrow\frac{3x-3}{6}=\frac{4y+12}{16}=\frac{5z-25}{30}=\frac{5z-25-3x+3-4y-12}{30-6-16}=\frac{\left(5x-3x-4y\right)-\left(25-3+12\right)}{8}=\frac{50-34}{8}=2\)

=> x-1/2 = 2 => x=5

y+3/4=2=>y=5

z-5/6=2=>z=17

Answer 2

[zzz_Thúy_Loan_zzz]

Bài 1 : Giải

a1−19=a2−28=a3−37=...=a9−91a1−19=a2−28=a3−37=...=a9−91
Theo tính chất dãy tỉ số bằng nhau →a1−19=a2−28=a3−37=...=a9−91=a1−1+a2−2+a3−3+a4−4+...+a9−99+8+7+...+3+2+1=(a1+a2+a3+...+a9)−4545=90−4545=1→a1−19=a2−28=a3−37=...=a9−91=a1−1+a2−2+a3−3+a4−4+...+a9−99+8+7+...+3+2+1=(a1+a2+a3+...+a9)−4545=90−4545=1
a1−1=9→a1=10a2−2=8→a2=10a3−3=7→a3=10...a9−9=1→a9=10a1−1=9→a1=10a2−2=8→a2=10a3−3=7→a3=10...a9−9=1→a9=10
Vậy a1=a2=a3=...=a9=10