Áp dụng tc dstbn:

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=\dfrac{3x-4y+5z}{3\cdot2-3\cdot4+5\cdot4}=\dfrac{70}{14}=5\\ \Leftrightarrow\left\{{}\begin{matrix}x=10\\y=15\\z=20\end{matrix}\right.\)

? là phần đề thiếu bạn nhé, bạn xem lại đề.

Theo bài ra ta cs 

\(3.x=y;5.y=4.z\)và \(6x+7y+8z=456\)

Ta lại cs 

\(+,3x=y\Rightarrow\frac{x}{1}=\frac{y}{3}\Rightarrow\frac{x}{4}=\frac{y}{12}\)(1)

\(+,5y=4z\Rightarrow\frac{y}{4}=\frac{z}{5}\Rightarrow\frac{y}{12}=\frac{z}{15}\)(2)

Từ (1) ; (2) \(\Rightarrow\frac{x}{4}=\frac{y}{12}=\frac{z}{15}\)

ADTC dãy tỉ số bằng nhau ta cs 

\(\frac{x}{4}=\frac{y}{12}=\frac{z}{15}=\frac{6x+7y+8z}{6.4+7.12+8.15}=\frac{456}{228}=2\)

\(\Rightarrow\hept{\begin{cases}\frac{x}{4}=2\\\frac{y}{12}=2\\\frac{z}{15}=2\end{cases}\Rightarrow\hept{\begin{cases}x=2.4=8\\y=2.12=24\\z=2.15=30\end{cases}}}\)

đề ở câu đầu:

x-3/0,2=2x-1/-0,5

3x-1/2x+3=3x+2/2x-1

a: =>-0,5x+1,5=0,4x-0,2

=>-0,9x=-1,7

=>x=17/9

3x-1/2x+3=3x+2/2x-1

=>6x^2-3x-2x+1=6x^2+4x+9x+6

=>-5x+1=13x+6

=>-8x=5

=>x=-5/8

b: \(\Leftrightarrow\left(4x-1\right)\left(-x+7\right)=\left(4x+5\right)\left(-x-2\right)\)

=>\(-4x^2+28x+x-7=-4x^2-8x-5x-10\)

=>29x-7=-13x-10

=>42x=-3

=>x=-1/14

c: =>7x=5y và 2x-y=15

=>7x-5y=0 và 2x-y=15

=>x=25; y=35

a) Áp dụng t/c dtsbn:

 \(\dfrac{x}{7}=\dfrac{y}{13}=\dfrac{x+y}{7+13}=\dfrac{40}{20}=2\)

\(\Rightarrow\left\{{}\begin{matrix}x=2.7=14\\y=2.13=26\end{matrix}\right.\)

b) \(\dfrac{3}{x}=\dfrac{7}{y}\Rightarrow\dfrac{x}{3}=\dfrac{y}{7}\)

Và \(x+16=y\Rightarrow y-x=16\)

Áp dụng t/c dtsbn:

\(\dfrac{x}{3}=\dfrac{y}{7}=\dfrac{y-x}{7-3}=\dfrac{16}{4}=4\)

\(\Rightarrow\left\{{}\begin{matrix}x=4.3=12\\y=4.7=28\end{matrix}\right.\)

a) \(\dfrac{x}{7}=\dfrac{y}{13}\)

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

\(\dfrac{x}{7}=\dfrac{y}{13}=\dfrac{x+y}{7+13}=\dfrac{40}{20}=2\)

⇒\(\left\{{}\begin{matrix}x=2.7=14\\y=2.13=26\end{matrix}\right.\)

Đặt tính dồi tin

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a) Ta có: \(2x=3y\Rightarrow\frac{x}{3}=\frac{y}{2}\)

               \(3y=5z\Rightarrow\frac{y}{5}=\frac{z}{3}\)

\(\Rightarrow\frac{x}{15}=\frac{y}{10}=\frac{z}{6}=\frac{x-2y+3z}{15-2.10+3.6}=\frac{65}{13}=5\)

\(\Rightarrow x=5.15=75\)

      \(y=5.10=50\)

      \(z=5.6=30\)

b) Ta có: \(\frac{x}{5}=\frac{y}{3};\frac{y}{7}=\frac{z}{4}\Rightarrow\frac{x}{35}=\frac{y}{21}=\frac{z}{12}=\frac{x+y-z}{35+21-12}=\frac{132}{44}=3\)

\(\Rightarrow x=3.35=105\)

      \(y=3.21=63\)

      \(z=3.12=36\)

c) Gọi \(\frac{x}{4}=\frac{y}{7}=k\)

\(\Rightarrow x=4k;y=7k\)

\(\Rightarrow x.y=4k.7k=28k^2=112\)

\(\Rightarrow k^2=112:28=4\)

\(\Rightarrow k=\pm2\)

\(\Rightarrow x=\pm2.4=\pm8\)

     \(y=\pm2.7=\pm14\)

a,x.y=3=1x3=3x1=-1x(-3)=-3x(-1).

Vậy (x,y)=(1,3)=(3,1)=(-1,-3)=(-3,-1)

b,x.(y+1)=5=1x5=5x1=-1x(-5)=-5x(-1)

=>

Vậy (x,y)=(1,4)=(5,0)=(-1,-6)=(-1,-2).

c,(x-2)(y+3)=7=1x7=7x1=-1x(-7)=-7(-1)

=>

Vậy (x,y)=(3,4)=(9,-2)=(1,-10)=(-5,-4).