b) \(\frac{3x-2y}{4}=\frac{2z-4x}{3}=\frac{4y-3z}{2}\)

\(\Leftrightarrow\frac{12x-8y}{4^2}=\frac{6z-12x}{3^2}=\frac{8y-6z}{2^2}=\frac{12x-8y+6z-12x+8y-6z}{4^2+3^2+2^2}=0\)(tính chất dãy tỉ số bằng nhau)

=> \(\hept{\begin{cases}12x=8y\\6z=12x\\8y=6z\end{cases}\Leftrightarrow\hept{\begin{cases}3x=2y\\2z=4x\\4y=3z\end{cases}}\Rightarrow\hept{\begin{cases}\frac{x}{2}=\frac{y}{3}\\\frac{x}{2}=\frac{z}{4}\\\frac{y}{3}=\frac{z}{4}\end{cases}}}\Rightarrow\frac{x}{2}=\frac{y}{3}=\frac{z}{4}\left(\text{đpcm}\right)\)

Mày làm ngu vl

a) Ta có: f(2)-f(-1)=(m-1).2-[(m-1).(-1)]=7

<=> 2m-2+m-1=7 <=> 3m=10 => m=10/3

b) m=5 => f(x)=4x

 => f(3-2x)=4(3-2x)=20 <=> 3-2x=5 => 2x=-2 => x=-1

a) Ta có: f(2)-f(-1)=(m-1).2-[(m-1).(-1)]=7

<=> 2m-2+m-1=7 <=> 3m=10 => m=10/3

b) m=5 => f(x)=4x

 => f(3-2x)=4(3-2x)=20 <=> 3-2x=5 => 2x=-2 => x=-1

Ta có: 2^m+ 2019= 2020

  =>  2^m = 2020 - 2019 = 1

  => 2^m=2⁰  => m=0

Ta lại có:\(|n-2020|\)+ x = 2020

        =>\(|n-2020|\)= 2020 - x

        =>\(\orbr{\begin{cases}n-2020=2020-x\\n-2020=-2020+x\end{cases}}\)

        =>\(\orbr{\begin{cases}n=2020+2020-x=4040-x\\n=2020-2020+x=x\end{cases}}\)

                        Vậy m=0; n=4040 - x hoặc n = x

        Chúc bn học tốt!

Trần Hồ Hoàng Vũ

Chỗ n - 2020 = -2020 - x mới đúng chứ đâu phải -2020 + x

Ta có : \(x\left(x-y\right)-y\left(x-y\right)=\frac{3}{10}-\frac{-3}{50}\)

\(\left(x-y\right)^2=\frac{9}{25}\)

\(\Rightarrow\orbr{\begin{cases}x-y=\frac{3}{5}\\x-y=\frac{-3}{5}\end{cases}}\) \(\Rightarrow\orbr{\begin{cases}x=\frac{3}{10}:\frac{3}{5}=\frac{1}{2}\\x=\frac{3}{10}:\frac{-3}{5}=\frac{-1}{2}\end{cases}}\) \(\Rightarrow\orbr{\begin{cases}y=\frac{-3}{50}:\frac{3}{5}=\frac{-1}{10}\\y=\frac{-3}{50}:\frac{-3}{5}=\frac{1}{10}\end{cases}}\)

Vậy \(x=\frac{1}{2};y=\frac{-1}{10}\)hoặc  \(x=\frac{-1}{2};y=\frac{1}{10}\)

\(x=y=6\) nha cậu thi hk tot nha

X = 12           Y = 4

d: x+y=5

nên x=5-y

Ta có: xy=6

=>y(5-y)=6

=>y²-5y+6=0

=>(y-2)(y-3)=0

=>y=2 hoặc y=3

=>x=3 hoặc x=2

a: \(\Leftrightarrow\left(x-3;y+4\right)\in\left\{\left(1;-7\right);\left(-1;7\right);\left(-7;1\right);\left(7;-1\right)\right\}\)

hay \(\left(x,y\right)\in\left\{\left(4;-11\right);\left(2;3\right);\left(-4;-3\right);\left(10;-5\right)\right\}\)

chịu