Question

Cho k÷x=a÷c;k÷y=b÷d và c+d=k. Chứng minh ax+by=k mũ 2

Answer 1

Kc=ax,kd=by

—>ax+by=kc+kd

ax=by=k+(c+d)

ax+by=k+k

—>ax+by=k^2

Answer 2

Kc=ax,kd=by

—>ax+by=kc+kd

ax=by=k+(c+d)

ax+by=k+k

—>ax+by=k^2

Answer 3

v

Kc=ax,kd=by

—>ax+by=kc+kd

ax=by=k+(c+d)

ax+by=k+k

—>ax+by=k^2

Kc=ax,kd=by

—>ax+by=kc+kd

ax=by=k+(c+d)

ax+by=k+k

—>ax+by=k^2

Kc=ax,kd=by

—>ax+by=kc+kd

ax=by=k+(c+d)

ax+by=k+k

—>ax+by=k^2

Kc=ax,kd=by

—>ax+by=kc+kd

ax=by=k+(c+d)

ax+by=k+k

—>ax+by=k^2

Kc=ax,kd=by

—>ax+by=kc+kd

ax=by=k+(c+d)

ax+by=k+k

—>ax+by=k^2

Kc=ax,kd=by

—>ax+by=kc+kd

ax=by=k+(c+d)

ax+by=k+k

—>ax+by=k^2

Kc=ax,kd=by

—>ax+by=kc+kd

ax=by=k+(c+d)

ax+by=k+k

—>ax+by=k^2

Kc=ax,kd=by

—>ax+by=kc+kd

ax=by=k+(c+d)

ax+by=k+k

—>ax+by=k^2

Kc=ax,kd=by

—>ax+by=kc+kd

ax=by=k+(c+d)

ax+by=k+k

—>ax+by=k^2

Kc=ax,kd=by

—>ax+by=kc+kd

ax=by=k+(c+d)

ax+by=k+k

—>ax+by=k^2

Kc=ax,kd=by

—>ax+by=kc+kd

ax=by=k+(c+d)

ax+by=k+k

—>ax+by=k^2

Kc=ax,kd=by

—>ax+by=kc+kd

ax=by=k+(c+d)

ax+by=k+k

—>ax+by=k^2

Kc=ax,kd=by

—>ax+by=kc+kd

ax=by=k+(c+d)

ax+by=k+k

—>ax+by=k^2

Kc=ax,kd=by

—>ax+by=kc+kd

ax=by=k+(c+d)

ax+by=k+k

—>ax+by=k^2

Kc=ax,kd=by

—>ax+by=kc+kd

ax=by=k+(c+d)

ax+by=k+k

—>ax+by=k^2

Answer 4

Kc=ax,kd=by

—>ax+by=kc+kd

ax=by=k+(c+d)

ax+by=k+k

—>ax+by=k^2