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