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\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\Leftrightarrow\dfrac{x}{2}=\dfrac{2y}{6}=\dfrac{3z}{12}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{2}=\dfrac{2y}{6}=\dfrac{3z}{12}=\dfrac{x+2y-3z}{2+6-12}=\dfrac{-20}{-4}=5\)
\(\Rightarrow\left\{{}\begin{matrix}x=5.2=10\\y=5.3=15\\z=5.4=20\end{matrix}\right.\)
\(\dfrac{x}{y}=\dfrac{3}{4}\Leftrightarrow\dfrac{x}{3}=\dfrac{y}{4}\)
\(\Rightarrow\dfrac{-3x}{-9}=\dfrac{5y}{20}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{-3x}{-9}=\dfrac{5y}{20}=\dfrac{-3x+5y}{-9+20}=\dfrac{33}{11}=3\)
\(\Rightarrow\left\{{}\begin{matrix}x=3.3=9\\y=3.4=12\end{matrix}\right.\)
a: \(\dfrac{2x-y}{3x+2y}=\dfrac{5}{2}\)
\(\Leftrightarrow15x+10y=4x-2y\)
=>11x=-12y
=>\(\dfrac{x}{-12}=\dfrac{y}{11}\)
Đặt \(\dfrac{x}{-12}=\dfrac{y}{11}=k\)
=>x=-12k; y=11k
\(P=\dfrac{5x+4y}{25x-y}=\dfrac{5\cdot\left(-12k\right)+4\cdot11k}{25\cdot\left(-12k\right)-11k}=\dfrac{16}{311}\)
b: \(\dfrac{x-5y}{x-3y}=\dfrac{4}{3}\)
=>4x-12y=3x-15y
=>x=-3y
\(\Leftrightarrow\dfrac{x}{-3}=\dfrac{y}{1}=k\)
=>x=-3k; y=k
\(P=\dfrac{x^3+2y^3}{x^3-y^3}=\dfrac{-27k^3+2k^3}{-27k^3-k^3}=\dfrac{-25}{-28}=\dfrac{25}{28}\)
Bài 1:
1)
\(\dfrac{3x+2}{4}\) = \(\dfrac{5x-3}{3}\)
<=> 3(3x + 2) = 4(5x - 3)
<=> 9x + 6 = 20x - 12
<=> 6 +12 = 20x - 9x
<=> 11x = 18
<=> x = \(\dfrac{18}{11}\)
Vậy: x = \(\dfrac{18}{11}\)
2)
\(\dfrac{x-1}{3x+2}\)= \(\dfrac{1}{5}\)
<=> 5(x - 1) = 3x + 2
<=> 5x - 5 = 3x + 2
<=> 5x - 3x = 2 +5
<=> 2x = 7
<=> x = \(\dfrac{7}{2}\)
Vậy : x = \(\dfrac{7}{2}\)
Bài 1 :
1) Ta có :
\(\dfrac{3x+2}{4}=\dfrac{5x-3}{3}\\ \Leftrightarrow4\cdot\left(5x-3\right)=3\cdot\left(3x+2\right)\\ \Leftrightarrow20x-12=9x+6\\ \Leftrightarrow20x-18=9x\\ \Leftrightarrow20x-9x=18\\ \Leftrightarrow11x=18\\ \Leftrightarrow x=\dfrac{18}{11}\\ Vậy.,...\)
2) Ta có :
\(\dfrac{x-1}{3x+2}=\dfrac{1}{5}\Leftrightarrow5\cdot\left(x-1\right)=3x+2\\ \Leftrightarrow5x-5=3x+2\\ \Leftrightarrow5x-3x-5=2\\ \Leftrightarrow2x-5=2\\ \Leftrightarrow2x=7\\ \Leftrightarrow x=\dfrac{7}{2}\)
Vậy ....
Bài 2 ;
1) Áp dụng tính chất của dãy tỉ số bằng nhau ta có :
\(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{x+y}{3+4}=\dfrac{21}{7}=3\\ \Rightarrow\left\{{}\begin{matrix}x=3\cdot3=9\\y=3\cdot4=12\end{matrix}\right.\\ Vậy...\)
2) Ta có : \(3x=5y\Leftrightarrow\dfrac{x}{5}=\dfrac{y}{3}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có :
\(\dfrac{x}{5}=\dfrac{y}{3}=\dfrac{x-y}{5-3}=\dfrac{-16}{2}=-8\\ \Rightarrow\left\{{}\begin{matrix}x=-8\cdot5=-40\\y=-8\cdot3=-24\end{matrix}\right.\\ Vậy....\)
3) Ta có : \(4x=7y\Leftrightarrow\dfrac{x}{7}=\dfrac{y}{4}=\dfrac{x^2}{7^2}=\dfrac{y^2}{4^2}=\dfrac{x\cdot y}{7\cdot4}\\ \Leftrightarrow\dfrac{x}{7}=\dfrac{y}{4}=\dfrac{112}{28}=4\\ \Rightarrow\left\{{}\begin{matrix}x=4\cdot7=28\\y=4\cdot4=16\end{matrix}\right.\\ Vậy...\)
a: \(\dfrac{-0.2}{x}=\dfrac{x}{-0.8}\)
\(\Leftrightarrow x^2=\dfrac{1}{5}\cdot\dfrac{4}{5}=\dfrac{4}{25}\)
=>x=2/5 hoặc x=-2/5
c: \(\dfrac{x-1}{x-2}=\dfrac{-3}{4}\)
=>4(x-1)=-3(x-2)
=>4x-4=-3x+6
=>7x=10
hay x=10/7
d: \(\dfrac{2-x}{5-x}=\dfrac{x+3}{x+2}\)
\(\Leftrightarrow\dfrac{x+3}{x+2}=\dfrac{x-2}{x-5}\)
\(\Leftrightarrow\left(x+3\right)\left(x-5\right)=\left(x-2\right)\left(x+2\right)\)
\(\Leftrightarrow x^2-2x-15=x^2-4\)
=>-2x=11
hay x=-11/2
a: \(\Leftrightarrow x^3=-216\)
=>x=-6
b: \(\Leftrightarrow\dfrac{x}{2}=\dfrac{y}{\dfrac{5}{2}}=\dfrac{z}{\dfrac{7}{4}}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{2}=\dfrac{y}{\dfrac{5}{2}}=\dfrac{z}{\dfrac{7}{4}}=\dfrac{3x+5y+7z}{3\cdot2+5\cdot\dfrac{5}{2}+7\cdot\dfrac{7}{4}}=\dfrac{123}{\dfrac{123}{4}}=4\)
=>x=8; y=10; z=7
a) \(\dfrac{x^3}{8}=\dfrac{y^3}{64}=\dfrac{z^3}{216}\)
Từ \(\dfrac{x^3}{8}=\dfrac{y^3}{64}=\dfrac{z^3}{216}\Rightarrow\dfrac{x^3}{2^3}=\dfrac{y^3}{4^3}=\dfrac{z^3}{6^3}\)
\(\Leftrightarrow\dfrac{x^2}{2^2}=\dfrac{y^2}{4^2}=\dfrac{z^2}{6^2}\Leftrightarrow\dfrac{x^2}{4}=\dfrac{y^2}{16}=\dfrac{z^2}{36}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\dfrac{x^2}{4}=\dfrac{y^2}{16}=\dfrac{z^2}{36}=\dfrac{x^2+y^2+z^2}{4+16+36}=\dfrac{14}{56}=\dfrac{1}{4}\)
\(\Rightarrow\dfrac{x^2}{4}=\dfrac{1}{4}\Rightarrow x^2=\dfrac{1}{4}\cdot4\Rightarrow x^2=1\Rightarrow x=1\)
\(\dfrac{y^2}{16}=\dfrac{1}{4}\Rightarrow y^2=\dfrac{1}{4}\cdot16\Rightarrow y^2=4\Rightarrow y=2\)
\(\dfrac{z^2}{36}=\dfrac{1}{4}\Rightarrow z^2=\dfrac{1}{4}\cdot36\Rightarrow z^2=9\Rightarrow z^2=3\)
Xin lỗi mình chỉ làm được câu a)
e, Đặt \(\dfrac{x}{4}=\dfrac{y}{5}=k\left(k\in Z\right)\)
\(\Leftrightarrow x=4k,y=5k\) (1)
Theo bài ra ta có: xy = 80
Từ (1) \(\Rightarrow4k.5k=80\Rightarrow20.k^2=80\Rightarrow k^2=4\Rightarrow\left[{}\begin{matrix}k^2=2^2\\k^2=\left(-2\right)^2\end{matrix}\right.\left[{}\begin{matrix}k=2\\k=-2\end{matrix}\right.\)
+ Với k = 2 \(\Rightarrow\left\{{}\begin{matrix}x=8\\y=10\end{matrix}\right.\)
+ Với k = -2 \(\Rightarrow\left\{{}\begin{matrix}x=-8\\y=-10\end{matrix}\right.\)
Vậy \(\left(x,y\right)\in\left\{\left(8,10\right);\left(-8,-10\right)\right\}\)
a) \(\Rightarrow\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{-2}=\dfrac{5x}{15}=\dfrac{3z}{-6}=\dfrac{5x-y+3z}{15-5-6}=\dfrac{-16}{4}=-4\Rightarrow\left[{}\begin{matrix}\dfrac{x}{3}=-4\\\dfrac{y}{5}=-4\\\dfrac{z}{-2}=-4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-12\\y=-20\\z=8\end{matrix}\right.\)
\(a,Đặt\dfrac{x}{y}=\dfrac{2}{3}\Leftrightarrow\dfrac{x}{2}=\dfrac{y}{3}=k\Leftrightarrow\left\{{}\begin{matrix}x=2k\\y=3k\end{matrix}\right.\\ A=\dfrac{2x-3y}{x-5y}=\dfrac{2\cdot2k-3\cdot3k}{2k-5\cdot3k}\\ =\dfrac{4k-9k}{2k-15k} \\ =\dfrac{5k}{13k}\\ =\dfrac{5}{13}\)
\(b,Thayx-y=7vàoB,tacó:\\ B=\dfrac{2x+7}{3x-y}+\dfrac{2y-7}{3y-x}\\ =\dfrac{2x+x-y}{3x-y}+\dfrac{2y-x+y}{3y-x}\\ =\dfrac{3x-y}{3x-y}+\dfrac{3y-x}{3y-x}\\ =1+1\\ =2\)
\(c,Đặt\dfrac{x}{3}=\dfrac{y}{5}=k\Leftrightarrow\left\{{}\begin{matrix}x=3k\\y=5k\end{matrix}\right.\\ C=\dfrac{5x^2+3y^2}{10x^2-3y^2}\\ =\dfrac{5\left(3k\right)^2+3\left(5k\right)^2}{10\left(3k\right)^2-3\left(5k\right)^2}\\ =\dfrac{45k^2+75k^2}{90k^2-75k^2}\\ =\dfrac{120k^2}{15k^2}\\ =8\)
\(d,\dfrac{a}{b}=\dfrac{5}{7}\Leftrightarrow\dfrac{a}{5}=\dfrac{b}{7}=k\Leftrightarrow\left\{{}\begin{matrix}a=5k\\b=7k\end{matrix}\right.\\ D=\dfrac{5a-b}{3a-2b}\\ =\dfrac{5\cdot5k-7k}{3\cdot5k-2\cdot7k}\\ =\dfrac{25k-7k}{15k-14k}\\ =\dfrac{18k}{k}=18\)
\(e,Thayx-y=5vàoE,tacó:\\ E=\dfrac{3x-5}{2x+y}-\dfrac{4y+5}{x+3y}\\ =\dfrac{3x-x+y}{2x+y}-\dfrac{4y+x-y}{x+3y}\\ =\dfrac{2x+y}{2x+y}-\dfrac{3y+x}{x+3y}\\ =1-1=0\)
\(\dfrac{x}{y}=\dfrac{3}{4}\Rightarrow\dfrac{x}{3}=\dfrac{y}{4}\)
Áp dụng t/c dtsbn:
\(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{-3x}{-9}=\dfrac{5y}{20}=\dfrac{-3x+5y}{-9+20}=\dfrac{33}{11}=3\)
\(\Rightarrow\left\{{}\begin{matrix}x=3.3=9\\y=3.4=12\end{matrix}\right.\)
thanks bạn