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

E=-a-b+c+d-d-c+b+2a

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

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

E=(2a-a)+(-b+b)+(-d+d)+(-c+c)=a+0+0+0=a

thanks nhiều nha ĐỨC THỊNH

F=(a-2b-c-2d)-(3d-2c-3b+a)+15

F=a-2b-c-2d-3d+2c+3b-a+15

F=a+(-2b)+(-c)+(-2d)+(-3d)+2c+3b+(-a)+15

F=(-2b+3b)+(-c+2c)+[-2d+(-3d)]+(-a+a)+15

F=b+c+(-5d)+0+15=b+c+(-5d)+15

`#3107.101107`

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

Ta có:

\(\dfrac{3b}{a}=\dfrac{3d}{c}\Rightarrow3bc=3da\Rightarrow bc=da\)

Vậy, từ tỉ lệ thức \(\dfrac{a}{b}=\dfrac{c}{d}\) ta có thể suy ra tỉ lệ thức \(\dfrac{3b}{a}=\dfrac{3d}{c}\)

\(\Rightarrow B.\)

b.\(ĐK:x;y\in Z^+;x;y\ne0\)

\(\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{5}\)

\(\Leftrightarrow\dfrac{5}{x}+\dfrac{5}{y}=1\)

\(\Leftrightarrow\dfrac{5}{x}=1-\dfrac{5}{y}\)

\(\Leftrightarrow\dfrac{5}{x}=\dfrac{y-5}{y}\)

\(\Leftrightarrow\dfrac{x}{5}=\dfrac{y}{y-5}\)

\(\Leftrightarrow x=\dfrac{5y}{y-5}\)

\(\Leftrightarrow x=5+\dfrac{25}{y-5}\) ( bạn chia \(5y\) cho \(y-5\) ý )

Để x;y là số nguyên dương thì \(25⋮y-5\) hay \(y-5\in U\left(25\right)=\left\{\pm1;\pm5;\pm25\right\}\)

TH1: 

\(y-5=1\) 

\(\Leftrightarrow\left\{{}\begin{matrix}y=6\\x=30\end{matrix}\right.\) ( tm )   ( bạn thế y=6 vào \(x=5+\dfrac{25}{y+5}\) nhé )

Xét tương tự, ta ra được nghiệm nguyên dương của phương trình:

\(\left\{{}\begin{matrix}x=30\\y=6\end{matrix}\right.\)  \(\left\{{}\begin{matrix}x=10\\y=10\end{matrix}\right.\)  \(\left\{{}\begin{matrix}x=6\\y=30\end{matrix}\right.\)

Bài 1: Đặt \(\dfrac{a}{c}=\dfrac{b}{d}=k\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=ck\\b=dk\end{matrix}\right.\)

\(\dfrac{a}{a+c}=\dfrac{ck}{ck+c}=\dfrac{ck}{c\left(k+1\right)}=\dfrac{k}{k+1}\)

\(\dfrac{b}{b+d}=\dfrac{dk}{dk+d}=\dfrac{k}{k+1}\)

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

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

   = a+b-2c+a-b-c-2a+b+c

   = b-2c

B=-(2a-b+c) + (b-2c-3a) -(-5a-3c+b)

  = -2a+b-c+b-2c-3a+5a+3c-b

  = b-c

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

  = a-b-2c-2b-3c+a+2a-3b

  = -6b-5c

D=(5a-3b+c) +( 2a-3b+5) -( b-c+a)

   = 5a-3b+c+2a-3b+5-b+c-a

   = 6a-7b+2c

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

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

\(B=-\left(2a-b+c\right)+\left(b-2c-3a\right)-\left(-5a-3c+b\right)\)

\(=-2a+b-c+b-2c-3a+5a+3c-b=b\)

\(C=\left(3a-b-2c\right)-\left(2b+3c-a\right)+\left(2a-3b\right)\)

\(=3a-b-2c-2b-3c+a+2a-3b=6a-6b-5c\)

\(D=\left(5a-3b+c\right)+\left(2a-3b+5\right)-\left(b-c+a\right)\)

\(=5a-3b+c+2a-3b+5-b+c-a=6a-7b+2c\)

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

Khi đó, ta có : \(\frac{3bk+2b}{2bk+3b}=\frac{\left(3k+2\right)b}{\left(2k+3\right)b}=\frac{3k+2}{2k+3}\)(1)

       \(\frac{3dk+2d}{2dk+3d}=\frac{\left(3k+2\right).d}{\left(2k+3\right).d}=\frac{3k+2}{2k+3}\)(2)

Từ (1) và (2), suy ra :  \(\frac{3a+2b}{2a+3b}=\frac{3c+2d}{2c+3d}\)

Đặt a/b=c/d=k

=>a=bk; c=dk

a: \(\dfrac{3a+2b}{a}=\dfrac{3bk+2b}{bk}=\dfrac{3k+2}{k}\)

\(\dfrac{3c+2d}{c}=\dfrac{3dk+2d}{dk}=\dfrac{3k+2}{k}\)

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

b: \(\dfrac{2a-3b}{b}=\dfrac{2bk-3b}{b}=2k-3\)

\(\dfrac{2c-3d}{d}=\dfrac{2dk-3d}{d}=2k-3\)

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

c: \(\dfrac{a}{a-2b}=\dfrac{bk}{bk-2b}=\dfrac{k}{k-2}\)

\(\dfrac{c}{c-2d}=\dfrac{dk}{dk-2d}=\dfrac{k}{k-2}\)

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