\(\dfrac{1}{2}\left(6x-2y\right)\left(3x+y\right)=\dfrac{1}{2}.2\left(3x-y\right)\left(3x+y\right)=9x^2-y^2\)

\(\left(\dfrac{2}{3}z-\dfrac{2}{5}x\right)\left(\dfrac{1}{3}z+\dfrac{1}{5}x\right).\dfrac{1}{2}=\left(\dfrac{1}{3}z-\dfrac{1}{5}x\right)\left(\dfrac{1}{3}z+\dfrac{1}{5}z\right).2.\dfrac{1}{2}=\dfrac{1}{9}z^2-\dfrac{1}{25}x^2\)

\(\left(5y-3x\right).\dfrac{1}{4}\left(12x+20y\right)=\left(5y-3x\right)\left(5y+3x\right).4.\dfrac{1}{4}=25y^2-9x^2\)

\(\left(\dfrac{3}{4}y-\dfrac{1}{2}x\right)\left(x+\dfrac{3}{2}y\right)=\left(\dfrac{3}{2}y-x\right)\left(\dfrac{3}{2}y+x\right)=\dfrac{9}{4}y^2-x^2\)

\(\left(a+b+c\right)\left(a+b+c\right)=\left(a+b+c\right)^2=a^2+b^2+c^2+2ab+2bc+2ac\)

\(\left(x-y+z\right)\left(x+y-z\right)=x^2-\left(y-z\right)^2=x^2-y^2-z^2+2yz\)

a: \(\dfrac{1}{2}\left(6x-2y\right)\left(3x+y\right)=\left(3x-y\right)\cdot\left(3x+y\right)=9x^2-y^2\)

b: \(\left(\dfrac{2}{3}z-\dfrac{2}{5}x\right)\left(\dfrac{1}{3}z+\dfrac{1}{5}x\right)\cdot\dfrac{1}{2}\)

\(=\left(\dfrac{1}{3}z-\dfrac{1}{5}x\right)\left(\dfrac{1}{3}z+\dfrac{1}{5}x\right)\)

\(=\dfrac{1}{9}z^2-\dfrac{1}{25}x^2\)

c: \(\left(5y-3x\right)\cdot\dfrac{1}{4}\cdot\left(12x+20y\right)\)

\(=\left(5y-3x\right)\left(5y+3x\right)\)

\(=25y^2-9x^2\)

d: \(\left(\dfrac{3}{4}y-\dfrac{1}{2}x\right)\left(\dfrac{3}{2}y+x\right)\cdot2\)

\(=\left(\dfrac{3}{2}y-x\right)\left(\dfrac{3}{2}y+x\right)\)

\(=\dfrac{9}{4}y^2-x^2\)

e: \(\left(a+b+c\right)\left(a+b-c\right)\)

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

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

\(a,4x=5y\:\Rightarrow\frac{x}{5}=\frac{y}{4}\Rightarrow\frac{x}{15}=\frac{y}{12}\)

\(4y=6z\Rightarrow\frac{y}{6}=\frac{z}{4}\Rightarrow\frac{y}{12}=\frac{z}{8}\)

\(\Rightarrow\frac{x}{15}=\frac{y}{12}=\frac{z}{8}\)

\(\Rightarrow\frac{x}{15}=\frac{2y}{24}=\frac{3z}{24}\)

\(\Rightarrow\frac{x-2y+3z}{15-24+24}=\frac{x}{15}=\frac{y}{12}=\frac{z}{8}\)

\(\Rightarrow\frac{5}{15}=\frac{x}{15}=\frac{y}{12}=\frac{z}{8}\)

\(\Rightarrow\frac{1}{3}=\frac{x}{15}=\frac{y}{12}=\frac{z}{8}\)

\(\Rightarrow\hept{\begin{cases}x=\frac{1}{3}\cdot15=5\\y=\frac{1}{3}\cdot12=4\\z=\frac{1}{3}\cdot8=\frac{8}{3}\end{cases}}\)

mọi người giúp mk câu b, c, d còn lại nha

Áp dụng t/c dãy tỉ số bằng nhau:

a.

\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{2x}{6}=\dfrac{4y}{20}=\dfrac{2x+4y}{6+20}=\dfrac{28}{26}=\dfrac{14}{13}\)

\(\Rightarrow\left\{{}\begin{matrix}x=3.\dfrac{14}{13}=\dfrac{52}{13}\\y=5.\dfrac{14}{13}=\dfrac{70}{13}\end{matrix}\right.\)

(Em có nhầm đề 26 thành 28 ko nhỉ, số xấu quá)

b.

\(4x=5y\Rightarrow\dfrac{x}{5}=\dfrac{y}{4}=\dfrac{3x}{15}=\dfrac{-2y}{-8}=\dfrac{3x-2y}{15-8}=\dfrac{35}{7}=5\)

\(\Rightarrow\left\{{}\begin{matrix}x=5.5=25\\y=4.2=20\end{matrix}\right.\)

c.

\(\dfrac{x}{-3}=\dfrac{y}{-7}=\dfrac{2x}{-6}=\dfrac{4y}{-28}=\dfrac{2x+4y}{-6-28}=\dfrac{68}{-34}=-2\)

\(\Rightarrow\left\{{}\begin{matrix}x=-3.\left(-2\right)=6\\y=-7.\left(-2\right)=14\end{matrix}\right.\)

d.

\(\dfrac{x}{2}=\dfrac{y}{-3}=\dfrac{z}{4}=\dfrac{4x}{8}=\dfrac{-3y}{9}=\dfrac{-2z}{-8}=\dfrac{4x-3y-2z}{8+9-8}=\dfrac{16}{9}\)

\(\Rightarrow\left\{{}\begin{matrix}x=2.\dfrac{16}{9}=\dfrac{32}{9}\\y=-3.\dfrac{16}{9}=-\dfrac{48}{9}\\z=4.\dfrac{16}{9}=\dfrac{64}{9}\end{matrix}\right.\)

ta có \(\frac{x}{y}=\frac{7}{20}\Rightarrow\frac{x}{7}=\frac{y}{20}\Rightarrow\frac{x}{14}=\frac{y}{40}\Rightarrow\frac{2x}{28}=\frac{5y}{200}\left(1\right)\)

       \(\frac{y}{z}=\frac{5}{8}\Rightarrow\frac{y}{5}=\frac{z}{8}\Rightarrow\frac{y}{40}=\frac{z}{64}\Rightarrow\frac{5y}{200}=\frac{2z}{128}\left(2\right)\)

          \(\left(1\right)\&\left(2\right)\Rightarrow\frac{2x+5y-2z}{28+200-128}=\frac{100}{100}=1\)

                       \(\frac{2x}{28}=1\Rightarrow x=\frac{28.1}{2}=14\)

                          \(\frac{5y}{200}=1\Rightarrow y=\frac{200.1}{5}=40\)

                          \(\frac{2z}{128}=1\Rightarrow z=\frac{128.1}{2}=64\)

\(\frac{x}{y}=\frac{7}{20};\frac{y}{z}=\frac{5}{8}\Rightarrow\frac{x}{7}=\frac{y}{20};\frac{y}{5}=\frac{z}{8}\Rightarrow\frac{x}{35}=\frac{y}{100};\frac{y}{100}=\frac{z}{160}\Rightarrow\frac{x}{35}=\frac{y}{100}=\frac{z}{160}\)

áp dụng tính chất của dãy tỉ số bằng nhau ta có:

\(\frac{x}{35}=\frac{y}{100}=\frac{z}{160}=\frac{2x+5y-2z}{2.35+5.100-2.160}=\frac{100}{250}\)= số lẽ sai đề     

ngu

2x−3y/5=5y−2z/3=3z−5x/2=10x-15y/25=15y-6z/9=6z-10x/4=...+..+..../25+9+4=0/31=0

=> 2x=3y;  5y=2z ;  3z=5x => x/3=y/2; y/2=z/5

=> x/3=y/2 =z/5 = 12x/36=5y/10=3z/15= (12x+5y-3z)/31

      x/3 = 3y/6=2z/10 = (x-3y+2z)/7

=>  (12x+5y-3z)/ (x-3y+2z)=31/7