Question

1) Tìm giá trị biểu thức : 
x + 3y + 2z biết z + y = 5, y + z =-8
2) Cho b²=ac, c²=bd (a,b,c,d khác 0). Chứng minh rằng:
( 12a + 3b - 5c/12b + 3c - 5d)³ = a/d
3) Cho a,b,c,d là các sô nguyên dương thỏa mãn a²+b²+c²+d²=2024²⁰²³. Chứng minh rằng a+b+c+d là hợp số

Answer 1

2:\(b^2=ac\)

=>\(\dfrac{b}{a}=\dfrac{c}{b}\)

\(c^2=bd\)

=>\(\dfrac{c}{b}=\dfrac{d}{c}\)

Do đó: \(\dfrac{b}{a}=\dfrac{c}{b}=\dfrac{d}{c}\)

=>\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=k\)

=>\(\left\{{}\begin{matrix}c=dk\\b=ck=dk^2\\a=bk=dk^2\cdot k=dk^3\end{matrix}\right.\)

\(\left(\dfrac{12a+3b-5c}{12b+3c-5d}\right)^3=\left(\dfrac{12\cdot dk^3+3\cdot dk^2-5\cdot dk}{12\cdot dk^2+3\cdot dk-5d}\right)^3\)

\(=\left(\dfrac{k\left(12dk^2+3dk-5d\right)}{12dk^2+3dk-5d}\right)^3=k^3\)
\(\dfrac{a}{d}=\dfrac{dk^3}{d}=k^3\)

Do đó: \(\dfrac{a}{d}=\left(\dfrac{12a+3b-5c}{12b+3c-5d}\right)^3\)