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

TH1: \(a+b+c+d=0\)

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

TH2: \(a+b+c+d\ne0\)

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

Ta có:

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

⇔ \(\dfrac{2a+b+c+d}{a}-1=\dfrac{a+2b+c+d}{b}-1=\dfrac{a+b+2c+d}{c}-1\)

    \(=\dfrac{a+b+c+2d}{d}-1\)

⇔ \(\dfrac{a+b+c+d}{a}=\dfrac{a+b+c+d}{b}=\dfrac{a+b+c+d}{c}=\dfrac{a+b+c+d}{d}\)

Nếu a+b+c+d=0

⇒a+b=−(c+d);c+b=−(a+d);c+d=−(a+b);a+d=−(c+b)

Thay vào M, ta có:

\(M=\dfrac{a+b}{-\left(a+b\right)}=\dfrac{b+c}{-\left(b+c\right)}=\dfrac{c+d}{-\left(c+d\right)}=\dfrac{a+d}{-\left(a+d\right)}=-1\)

Nếu a+b+c+d ≠0

⇒ \(a=b=c=d\)

Thay vào M, ta có

\(M=\dfrac{a+b}{a+b}=\dfrac{b+c}{b+c}=\dfrac{c+d}{c+d}=\dfrac{d+a}{d+a}=1\)

Cắt cu 77

Đặt a/2019=b/2021=c/2023=k

=>a=2019k; b=2021k; c=2023k

(a-c)^2/4=(2023k-2019k)^2/4=(4k)^2/4=4k^2

(a-b)(b-c)=(2019k-2021k)(2021k-2023k)=4k^2

=>(a-c)^2/4=(a-b)(b-c)

Câu 1: Đặt \(\frac{a}{b}=\frac{c}{d}=k\)

=>a=bk; c=dk

\(\frac{a+2b}{a-2b}=\frac{bk+2b}{bk-2b}=\frac{b\left(k+2\right)}{b\left(k-2\right)}=\frac{k+2}{k-2}\)

\(\frac{c+2d}{c-2d}=\frac{dk+2d}{dk-2d}=\frac{d\left(k+2\right)}{d\left(k-2\right)}=\frac{k+2}{k-2}\)

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

BÀi 2:

a: 2x=3y

=>\(\frac{x}{3}=\frac{y}{2}\)

=>\(\frac{x}{21}=\frac{y}{14}\left(1\right)\)

5y=7z

=>\(\frac{y}{7}=\frac{z}{5}\)

=>\(\frac{y}{14}=\frac{z}{10}\left(2\right)\)

Từ (1),(2) suy ra \(\frac{x}{21}=\frac{y}{14}=\frac{z}{10}\)

mà 3x+5y-7z=30

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

\(\frac{x}{21}=\frac{y}{14}=\frac{z}{10}=\frac{3x+5y-7z}{3\cdot21+5\cdot14-7\cdot10}=\frac{30}{63}=\frac{10}{21}\)

=>x=10; \(y=14\cdot\frac{10}{21}=10\cdot\frac23=\frac{20}{3}\) ; \(z=\frac{10}{21}\cdot10=\frac{100}{21}\)

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

\(\frac{x-1}{2}=\frac{y+3}{4}=\frac{z-5}{6}=\frac{-3x-4y+5z+3-12-25}{-3\cdot2-4\cdot4+5\cdot6}=\frac{16}{8}=2\)

=>x-1=4; y+3=8; z-5=12

=>x=5; y=5; z=17

c: \(\frac12x=\frac23y=\frac34z\)

=>\(12\cdot\frac12x=12\cdot\frac23y=12\cdot\frac34z\)

=>6x=8y=9z

=>\(\frac{6x}{72}=\frac{8y}{72}=\frac{9z}{72}\)

=>\(\frac{x}{12}=\frac{y}{9}=\frac{z}{8}\)

mà x-y=15

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

\(\frac{x}{12}=\frac{y}{9}=\frac{z}{8}=\frac{x-y}{12-9}=\frac{15}{3}=5\)

=>\(\begin{cases}x=5\cdot12=60\\ y=5\cdot9=45\\ z=5\cdot8=40\end{cases}\)

Bạn viết sai đề rồi

\(\dfrac{a}{b+c}=\dfrac{b}{a+c}=\dfrac{c}{a+b}\)
\(\Rightarrow\dfrac{a}{b+c}+1=\dfrac{b}{a+c}+1=\dfrac{c}{a+b}+1\)

\(\Rightarrow\dfrac{a+b+c}{b+c}=\dfrac{a+b+c}{a+c}=\dfrac{a+b+c}{a+b}\)

\(\Rightarrow b+c=a+c=b+a\)

\(\Rightarrow a=b=c\)

\(\dfrac{a}{b+c}=\dfrac{b}{a+c}=\dfrac{c}{a+b}=\dfrac{a}{a+a}=\dfrac{1}{2}\)

Cậu chưa xét a+b+c=0 

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

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

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

Nếu \(a+b+c+d\ne0\Rightarrow a=b=c=d\)

\(\Rightarrow M=1+1+1+1=4\)

Nếu a + b + c + d = 0 => a + b = -(c + d) ; (b + c) = -(a + d) ; c + d = -(a+b) ; d + a = -(b + c)

\(\Rightarrow M=\left(-1\right)+\left(-1\right)+\left(-1\right)+\left(-1\right)=-4\)

Vậy M = 4 hoặc M = -4