Question

[Nguyễn Huyền Trâm]

Cho \(\dfrac{a}{b} = \dfrac{c}{d}\) . Chứng minh :

a, \((a+c).((b-d)=(a-c).(b-d)\)

b, \((a+c).b=(b+d).a\)

c, \(a.(b-d)=b(a-c)\)

d, \((b+d).c=(a+c).d\)

e, \((b-d).c=(a-c).d\)

f, \((a+b).(c-d)=(a-b).(c+d)\)

g, \((2a+3c).(2b-3d)=(2a-3c).(2b+3d)\)

h, \((4a+3b).(4c-3d)=(4a-3b).((4c+3d)\)

i, \((2a+3b).(4c-5d)=(4a-5b).(2c+3d)\)

k, \((4a+5b).(7c-11d)=(7a-11b).(4c+5d)\)

Answer 1

[Akai Haruma]

Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk,c=dk\). Khi đó ta có:

a)

\((a+c)(b-d)=(bk+dk)(b-d)=k(b+d)(b-d)\)

\((a-c)(b+d)=(bk-dk)(b+d)=k(b-d)(b+d)=k(b+d)(b-d)\)

\(\Rightarrow (a+c)(b-d)=(a-c)(b+d)\) (đpcm)

b)

\((a+c)b=(bk+dk)b=k(b+d).b=bk(b+d)\)

\((b+d).a=(b+d).bk=bk(b+d)\)

\(\Rightarrow (a+c)b=(b+d)a\)

c)

\(a(b-d)=bk(b-d)\)

\(b(a-c)=b(bk-dk)=bk(b-d)\)

\(\Rightarrow a(b-d)=b(a-c)\)

d)

\((b+d).c=(b+d).dk=dk(b+d)\)

\((a+c)d=(bk+dk)d=k(b+d)d=dk(b+d)\)

\(\Rightarrow (b+d)c=(a+c)d\)

Answer 2

[Akai Haruma]

e)

\((b-d).c=(b-d).dk=dk(b-d)\)

\((a-c)d=(bk-dk)d=k(b-d)d=dk(b-d)\)

\(\Rightarrow (b-d)c=(a-c)d\)

f)

\((a+b)(c-d)=(bk+b)(dk-d)=b(k+1)d(k-1)=bd(k-1)(k+1)\)

\((a-b)(c+d)=(bk-b)(dk+d)=b(k-1)d(k+1)=bd(k-1)(k+1)\)

\(\Rightarrow (a+b)(c-d)=(a-b)(c+d)\)

g)

\((2a+3c)(2b-3d)=(2bk+3dk)(2b-3d)=k(2b+3d)(2b-3d)\)

\((2a-3c)(2b+3d)=(2bk-3dk)(2b+3d)=k(2b-3d)(2b+3d)\)

\(\Rightarrow (2a+3c)(2b-3d)=(2a-3c)(2b+3d)\)

h)

\((4a+3b)(4c-3d)=(4bk+3b)(4dk-3d)=b(4k+3)d(4k-3)=bd(4k+3)(4k-3)\)

\((4a-3b)(4c+3d)=(4bk-3b)(4dk+3d)=b(4k-3)d(4k+3)=bd(4k+3)(4k-3)\)

\(\Rightarrow (4a+3b)(4c-3d)=(4a-3b)(4c+3d)\)

i,k: Hoàn toàn tương tự.