\(a,\left(a-b+c\right)-\left(a+c\right)=a-b+c-a-c=-b\)

\(b,\left(a+b\right)-\left(b-a\right)+c=a+b-b+a+c=2a+c\)

\(c,-\left(a+b-c\right)+\left(a-b-c\right)=-a-b+c+a-b-c=-2b\)

\(d,a\left(b+c\right)-a\left(b+d\right)=ab+ac-ab-ad=ac-ad=a\left(c-d\right)\)

\(e,a\left(b-c\right)+a\left(d+c\right)=ab-ac+ad+ac=ab+ad=a\left(b+d\right)\)

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

= a - b + c - a - c

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

= -b

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

= a + b - b + a + c

= 2a + (b - b) + c

= 2a + c

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

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

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

= -2b

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

= ab + ac - ab - ad

= (ab - ab) + (ac - ad)

= ac - ad

= a(c - d)

e) a(b - c) + a(d + c)

= a(b - c + d + c)

= a[b - (c - c) + d]

= d(b + d)

a ) 

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

 a - b + c - a -c = - b

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

0 + 0 - b = -b ( đpcm )

Áp dụng tính chất dãy tỉ số bằng nhau

\(\frac{x}{5}=\frac{y}{7}=\frac{z}{9}=\frac{x-y+z}{5-7+9}=\frac{315}{7}=45\)

  suy ra:   x/5 = 45   =>  x  =  225

               y/7 = 45  =>  y  =  315

               z/9 = 45  =>  z  =  405

Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=m\Rightarrow a=bm;c=dm\)

Ta có : \(\dfrac{a.b}{c.d}=\dfrac{b.m.b}{d.m.d}=\dfrac{b^2.m}{d^2.m}=\dfrac{b^2}{d^2}\)(1)

\(\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}=\dfrac{\left(bm+b\right)^2}{\left(dm+d\right)^2}=\dfrac{\left[b.\left(m+1\right)\right]^2}{\left[d.\left(m+1\right)\right]^2}=\dfrac{b^2.\left(m+1\right)^2}{d^2.\left(m+1\right)^2}=\dfrac{b^2}{d^2}\)(2)

Từ (1) và (2) suy ra :\(\dfrac{a.b}{c.d}=\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}\)

Vậy \(\dfrac{a.b}{c.d}=\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}\) khi \(\dfrac{a}{b}=\dfrac{c}{d}\)

Đc chưa bạn . Tick cho mk nha!

Xét tứ giác ABCD có

AB//CD

AD//BC

Do đó; ABCD là hình bình hành

=>AB=CD; AD=BC

1,

\(\frac{a}{b}=\frac{c}{d}\\ \Rightarrow\frac{a}{b}-1=\frac{c}{d}-1\\ \Rightarrow\frac{a}{b}-\frac{b}{b}=\frac{c}{d}-\frac{d}{d}\\ \Rightarrow\frac{a-b}{b}=\frac{c-d}{d}\\ \Rightarrow\frac{b}{a-b}=\frac{d}{c-d}\)

2,

Có: \(\frac{x}{y}=1,5=\frac{3}{2}\Rightarrow\frac{x}{3}=\frac{y}{2}\)

Đặt \(\frac{x}{3}=\frac{y}{2}=k\Rightarrow\left\{{}\begin{matrix}x=3k\\y=2k\end{matrix}\right.\)

Mà \(x\cdot y=24\)

\(\Rightarrow3k\cdot2k=24\\ \Rightarrow6k^2=24\\ \Rightarrow k^2=4\\ \Rightarrow\left[{}\begin{matrix}k=2\\k=-2\end{matrix}\right.\)

+ Với k = 2

\(\Rightarrow\left\{{}\begin{matrix}x=3\cdot2=6\\y=2\cdot2=4\end{matrix}\right.\)

+ Với k = -2

\(\Rightarrow\left\{{}\begin{matrix}x=3\cdot\left(-2\right)=-6\\y=2\cdot\left(-2\right)=-4\end{matrix}\right.\)

Vậy \(\left(x;y\right)\in\left\{\left(6;4\right);\left(-6;-4\right)\right\}\)

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

\(\Rightarrow\frac{a}{b}-1=\frac{c}{d}-1\)

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

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

\(\Rightarrow\frac{b}{a-b}=\frac{d}{c-d}\left(đpcm\right).\)

b) Ta có: \(\frac{x}{y}=1,5.\)

Đổi \(1,5=\frac{3}{2}\)

\(\Rightarrow\frac{x}{y}=\frac{3}{2}.\)

\(\Rightarrow\frac{x}{3}=\frac{y}{2}\) và \(x.y=24.\)

Đặt \(\frac{x}{3}=\frac{y}{2}=k\Rightarrow\left\{{}\begin{matrix}x=3k\\y=2k\end{matrix}\right.\)

Có: \(x.y=24\)

=> \(3k.2k=24\)

=> \(6.k^2=24\)

=> \(k^2=24:6\)

=> \(k^2=4\)

=> \(k=\pm2.\)

TH1: \(k=2.\)

\(\Rightarrow\left\{{}\begin{matrix}x=3.2=6\\y=2.2=4\end{matrix}\right.\)

TH2: \(k=-2.\)

\(\Rightarrow\left\{{}\begin{matrix}x=3.\left(-2\right)=-6\\y=2.\left(-2\right)=-4\end{matrix}\right.\)

Vậy \(\left(x;y\right)=\left(6;4\right),\left(-6;-4\right).\)

Chúc bạn học tốt!

Ta có: 2bd = c(b + d)
=> (a + c).d = bc + cd
=> ad + cd = bc + cd
=> ad = bc
=> \(\dfrac{a}{b}=\dfrac{c}{d}\) (đpcm)

À QUÊN, A',B',C' ĐỐI XỨNG QUA d

cho tỉ lệ thức ab = cd 

chứng minh rằng (2008a+2009c)(b+d)=(a+c)(2008+2009d)

Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow\hept{\begin{cases}a=bk\\c=dk\end{cases}}\)

Ta có :

\(\frac{2008a+2009c}{a+c}=\frac{2008bk+2009dk}{bk+dk}=\frac{k\left(2008b+2009d\right)}{k\left(b+d\right)}=\frac{2008b+2009d}{b+d}\)

\(\Rightarrow\frac{2008a+2009c}{a+c}=\frac{2008b+2009d}{b+d}\Rightarrow\left(2008a+2009c\right)\left(b+d\right)=\left(a+c\right)\left(2008b+2009d\right)\)

=> ĐPCM

Ta có : \(\frac{a}{b}=\frac{c}{d}=\frac{2008a}{2008b}=\frac{2009c}{2009d}\)

\(\Rightarrow\frac{a}{b}=\frac{c}{d}=\frac{2008a}{2008b}=\frac{2009c}{2009d}=\frac{a+c}{b+d}=\frac{2008a+2009c}{2008b+2009d}\)

\(\Rightarrow\left(2008a+2009c\right).\left(b+d\right)=\left(a+c\right).\left(2008b+2009d\right)\)