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\(a,Đặt\dfrac{x}{y}=\dfrac{2}{3}\Leftrightarrow\dfrac{x}{2}=\dfrac{y}{3}=k\Leftrightarrow\left\{{}\begin{matrix}x=2k\\y=3k\end{matrix}\right.\\ A=\dfrac{2x-3y}{x-5y}=\dfrac{2\cdot2k-3\cdot3k}{2k-5\cdot3k}\\ =\dfrac{4k-9k}{2k-15k} \\ =\dfrac{5k}{13k}\\ =\dfrac{5}{13}\)
\(b,Thayx-y=7vàoB,tacó:\\ B=\dfrac{2x+7}{3x-y}+\dfrac{2y-7}{3y-x}\\ =\dfrac{2x+x-y}{3x-y}+\dfrac{2y-x+y}{3y-x}\\ =\dfrac{3x-y}{3x-y}+\dfrac{3y-x}{3y-x}\\ =1+1\\ =2\)
\(c,Đặt\dfrac{x}{3}=\dfrac{y}{5}=k\Leftrightarrow\left\{{}\begin{matrix}x=3k\\y=5k\end{matrix}\right.\\ C=\dfrac{5x^2+3y^2}{10x^2-3y^2}\\ =\dfrac{5\left(3k\right)^2+3\left(5k\right)^2}{10\left(3k\right)^2-3\left(5k\right)^2}\\ =\dfrac{45k^2+75k^2}{90k^2-75k^2}\\ =\dfrac{120k^2}{15k^2}\\ =8\)
\(d,\dfrac{a}{b}=\dfrac{5}{7}\Leftrightarrow\dfrac{a}{5}=\dfrac{b}{7}=k\Leftrightarrow\left\{{}\begin{matrix}a=5k\\b=7k\end{matrix}\right.\\ D=\dfrac{5a-b}{3a-2b}\\ =\dfrac{5\cdot5k-7k}{3\cdot5k-2\cdot7k}\\ =\dfrac{25k-7k}{15k-14k}\\ =\dfrac{18k}{k}=18\)
\(e,Thayx-y=5vàoE,tacó:\\ E=\dfrac{3x-5}{2x+y}-\dfrac{4y+5}{x+3y}\\ =\dfrac{3x-x+y}{2x+y}-\dfrac{4y+x-y}{x+3y}\\ =\dfrac{2x+y}{2x+y}-\dfrac{3y+x}{x+3y}\\ =1-1=0\)
Đặt \(\dfrac{x}{-4}=\dfrac{y}{-7}=\dfrac{z}{3}=k\left(k\ne0\right)\)
=>\(\left\{{}\begin{matrix}x=-4k\\y=-7k\\z=3k\end{matrix}\right.\)
Ta có P =\(\dfrac{-2\cdot\left(-4k\right)+\left(-7k\right)+5\cdot3k}{2\cdot\left(-4k\right)-3\left(-7k\right)-6\left(3k\right)}\)=\(\dfrac{8k+\left(-7k\right)+15k}{-8k+21k-18k}\)=
\(\dfrac{k\cdot\left(8+\left(-7\right)+15\right)}{k.\left(-8+21-18\right)}=\dfrac{-16}{5}\)
Vậy P= \(\dfrac{-16}{5}\)
Theo đề ta có:
\(\dfrac{x}{-4}=\dfrac{y}{-7}=\dfrac{z}{3}\)
Đặt k cho biểu thức trên
=>\(\dfrac{x}{-4}=\dfrac{y}{-7}=\dfrac{z}{3}\) =k
=> \(\left[{}\begin{matrix}\dfrac{x}{-4}=k\\\dfrac{y}{-7}=k\\\dfrac{z}{3}=k\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\left(-4\right).k\\y=\left(-7\right).k\\z=3.k\end{matrix}\right.\)
Thay \(\left[{}\begin{matrix}x=\left(-4\right).k\\y=\left(-7\right).k\\z=3.k\end{matrix}\right.\) vào biểu thức \(P=\dfrac{-2x+y+5z}{2x-3y-6z}\)
Ta được:
\(P=\dfrac{-2.\left(-4.k\right)+\left(-7.k\right)+5\left(3.k\right)}{2\left(-4.k\right)-3\left(-7.k\right)-6\left(3.k\right)}\)
=> \(P=\dfrac{8.k+\left(-7.k\right)+15.k}{-8.k+21.k-18.k}\)
=> \(P=\dfrac{k.\left(8+-7+15\right)}{k.\left(-8+21-18\right)}\)
=> P= \(-\dfrac{16}{5}\)
Vậy:....................
\(x-y=7\Rightarrow x=7+y\)
B=\(\dfrac{3x-7}{2x+y}-\dfrac{3y+7}{2x+y}\)
=\(\dfrac{3\left(7+y\right)-7}{2\left(7+y\right)+y}-\dfrac{3y+7}{2y+7+y}\)
=\(\dfrac{21+3y-7}{14+2y+y}-\dfrac{3y+7}{3y+7}\)
=\(\dfrac{14+3y}{14+3y}-\dfrac{3y+7}{3y+7}\)
=1-1=2
Vậy B=2
mk nhầm
B=1-1=0
Vậy B=0