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.

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

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

\(2xy-x^2-y^2+16\)

\(=-\left(x^2-2xy+y^2-16\right)\)

\(=-\left(x-y-4\right)\left(x-y+4\right)\)

\(\left(4x+1\right)\left(12x-1\right)\left(3x-2\right)\left(x+1\right)-4\) (Sửa đề)

\(=[\left(4x+1\right)\left(3x+2\right)][\left(12x-1\right)\left(x+1\right)]-4\)

\(=\left(12x^2+11x+2\right)\left(12x^2+11x-1\right)-4\)

Đặt \(12x^2+11x-1=n\)

\(=\left(n+3\right)n-4\)

\(=n^2+3n-4\)

\(=n^2-n+4n-4\)

\(=n\left(n-1\right)+4\left(n-1\right)\)

\(=\left(n-1\right)\left(n+4\right)\)

\(=\left(12x^2+11x-1-1\right)\left(12x^2+11x-1+4\right)\)

\(=\left(12x^2+11x-2\right)\left(12x^2+11x+3\right)\)

\(\left(3x+4\right)\left(x+1\right)\left(6x+7\right)^2=6\)

\(\Leftrightarrow\left(3x^2+7x+4\right)\left(36x^2+84x+49\right)=6\)(1)

Đặt \(\left(3x^2+7x+4\right)=n\)lúc đó (1):

\(\left(12n+1\right)n=6\)

\(\Rightarrow\hept{\begin{cases}n=0,75\\n=\frac{2}{3}\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x=\frac{-2}{3}\\x=\frac{-5}{3}\end{cases}}\)

Có điều kiện không bạn.