\(\frac{a}{b}=\frac{c}{d}\Rightarrow ad=bc\)

a)\(\frac{a-b}{a+b}=\frac{c-d}{c+d}\)

\(\Leftrightarrow\left(a-b\right)\left(c+d\right)=\left(c-d\right)\left(a+b\right)\)

\(\Leftrightarrow ac-bc+ad-bd=ac-ad+bc-bd\)

\(\text{Thay }ad=bc\text{ vào}\Rightarrow ac-ad+ad-bd=ac-ad+ad-bd\)

\(\text{Đây là đẳng thức đúng }\Rightarrow\frac{a-b}{a+b}=\frac{c-d}{c+d}\text{ là đúng }\)

b)\(\text{Tương tự*}\)

a) \(\frac{a}{b}=\frac{c}{d}\Leftrightarrow\frac{a}{b}+1=\frac{c}{d}+1\Leftrightarrow\frac{a+b}{b}=\frac{c+d}{d}\Leftrightarrow\frac{b}{a+b}=\frac{d}{c+d}\)

\(\Leftrightarrow\frac{-2b}{a+b}+1=\frac{-2d}{c+d}+1\Leftrightarrow\frac{a-b}{a+b}=\frac{c-d}{c+d}\)

b) \(\frac{a}{b}=\frac{c}{d}\Leftrightarrow\frac{4a}{b}-5=\frac{4c}{d}-5\Leftrightarrow\frac{4a-5b}{b}=\frac{4c-5d}{d}\Leftrightarrow\frac{b}{4a-5b}=\frac{d}{4c-5d}\)

\(\Leftrightarrow\frac{11b}{4a-5b}+1=\frac{11d}{4c-5d}+1\Leftrightarrow\frac{4a+6b}{4a-5b}=\frac{4c+6d}{4c-5d}\Leftrightarrow\frac{2a+3b}{4a-5b}=\frac{2c+3d}{4c-5d}\)

\(\Leftrightarrow\frac{2a+3b}{2c+3d}=\frac{4a-5b}{4c-5d}\)

Do \(b^2=ac;c^2=bd\)

\(\Rightarrow\frac{a}{b}=\frac{b}{c};\frac{b}{c}=\frac{c}{d}\)

\(\Rightarrow\frac{a}{b}=\frac{b}{c}=\frac{c}{d}\)

\(\Rightarrow\frac{a^3}{b^3}=\frac{b^3}{c^3}=\frac{c^3}{d^3}=\frac{a}{b}\cdot\frac{b}{c}\cdot\frac{c}{d}=\frac{a}{d}\)

Áp dụng tính chất dãy tỉ số bằng nhau,ta được:

\(\frac{a^3}{b^3}=\frac{b^3}{c^3}=\frac{c^3}{d^3}=\frac{a}{d}=\frac{a^3+b^3+c^3}{b^3+c^3+d^3}\left(đpcm\right)\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{a+b+c}{b+c+d}\\ \Rightarrow\left(\dfrac{a+b+c}{b+c+d}\right)^3=\dfrac{a+b+c}{b+c+d}\cdot\dfrac{a+b+c}{b+c+d}\cdot\dfrac{a+b+c}{b+c+d}\\ =\dfrac{a}{b}\cdot\dfrac{c}{d}\cdot\dfrac{b}{c}=\dfrac{a}{d}\) => Điều pải chứng minh

Ta có: \(\frac{a}{b}=\frac{c}{d}=\frac{a+c}{b+d}\)

mà \(\frac{a}{b}=\frac{c}{d}=\frac{2c}{2d}=\frac{a+2c}{b+2d}\)

\(\Rightarrow\frac{a+c}{b+d}=\frac{a+2c}{b+2d}\)\(\Rightarrow\left(a+c\right)\left(b+2d\right)=\left(b+d\right)\left(a+2c\right)\)( đpcm )

3,-(a+b-c)+(a-b-c)

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

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

=0-b-b+0

=-b-b

=-2b(đpcm)

4,a(b+c)-a(b+d)

=ab+ac-ab+ad

=(ab-ab)+ac+ad

=0+ac+ad

=ac+ad

=a(c+d)(đpcm)

5,a(b-c)+a(d+c)

=ab-ac+ad+ac

=(-ac+ac)+ab+ad

=0+ab+ad

=ab+ad

=a(b+d)(đpcm)

k cho mình vs

1. ( a - b + c ) - ( a + c ) = - b

Ta có : VT = ( a - b + c ) - ( a + c )

                = a - b + c - a - c

                = - b = VP

=> ( a - b + c ) - ( a + c ) = - b  ( đpcm )

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

Ta có : VT = ( a + b ) - ( b - a ) + c

                = a + b - b + a + c

                = 2a + c = VP  

=> ( a + b ) - ( b - a ) + c = 2a + c ( đpcm )

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

Ta có : VT = - ( a + b - c ) + ( a - b - c )

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

                = - 2b = VP 

=> - ( a + b - c ) + ( a - b - c ) = - 2b ( đpcm )

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

Ta có : VT = a ( b + c ) - a ( b + d )

                = ab + ac - ab  - ad 

                = ac - ad

                = a ( c - d ) = VP 

=> a( b + c ) - a( b + d ) = a( c - d )  ( đpcm )

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

Ta có : VT = a( b - c ) + a ( d + c )

                = a ( b - c + d + c )

                = a( b + d ) = VP

=> a ( b - c ) + a ( d + c ) = a ( b + d ) ( đpcm )

VT là vế trái 

VP là vế phải .

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

=a+b-b+a+c

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

=2a+0+c

=2a+c(đpcm)