A=(a-b+c)-(b-c-d)+(c-d+a)

A=a-b+c-b+c+d+c-d+a

A=2a-2b-3c

B=( a + b - c ) + ( b + c - a ) - ( a - c )

B=a + b - c + b + c - a - a + c

B=2b + c - a

C = - ( 4a + 5b + c) - ( 5b + 3c )

C = -4a - 5b - c - 5b -3c

C= -4a - 10b - 4c

D= ( a - 3b + c) - ( 2a -b +c)

D= a - 3b +c - 2a + b -c

D= a - 2b

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.

-Biểu thức cần tính bằng:

\(2a-b-c+a-b+a-c+c-b\)

\(=4a-3b-c\)

-Chọn D.

Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)

=>\(a=bk;c=dk\)

1: \(\dfrac{2a+3c}{2b+3d}=\dfrac{2\cdot bk+3\cdot dk}{2b+3d}=\dfrac{k\left(2b+3d\right)}{2b+3d}=k\)

\(\dfrac{2a-3c}{2b-3d}=\dfrac{2bk-3dk}{2b-3d}=\dfrac{k\left(2b-3d\right)}{2b-3d}=k\)

Do đó: \(\dfrac{2a+3c}{2b+3d}=\dfrac{2a-3c}{2b-3d}\)

2: \(\dfrac{4a-3b}{4c-3d}=\dfrac{4\cdot bk-3b}{4\cdot dk-3d}=\dfrac{b\left(4k-3\right)}{d\left(4k-3\right)}=\dfrac{b}{d}\)

\(\dfrac{4a+3b}{4c+3d}=\dfrac{4bk+3b}{4dk+3d}=\dfrac{b\left(4k+3\right)}{d\left(4k+3\right)}=\dfrac{b}{d}\)

Do đó: \(\dfrac{4a-3b}{4c-3d}=\dfrac{4a+3b}{4c+3d}\)

3: \(\dfrac{3a+5b}{3a-5b}=\dfrac{3bk+5b}{3bk-5b}=\dfrac{b\left(3k+5\right)}{b\left(3k-5\right)}=\dfrac{3k+5}{3k-5}\)

\(\dfrac{3c+5d}{3c-5d}=\dfrac{3dk+5d}{3dk-5d}=\dfrac{d\left(3k+5\right)}{d\left(3k-5\right)}=\dfrac{3k+5}{3k-5}\)

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

4: \(\dfrac{3a-7b}{b}=\dfrac{3bk-7b}{b}=\dfrac{b\left(3k-7\right)}{b}=3k-7\)

\(\dfrac{3c-7d}{d}=\dfrac{3dk-7d}{d}=\dfrac{d\left(3k-7\right)}{d}=3k-7\)

Do đó: \(\dfrac{3a-7b}{b}=\dfrac{3c-7d}{d}\)