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

⇒\(\dfrac{x}{-3}=\dfrac{y}{4}\) 

⇒\(\dfrac{2x}{-6}=\dfrac{3y}{12}\)

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

\(\dfrac{2x}{-6}=\dfrac{3y}{12}=\dfrac{3y-2x}{12-\left(-6\right)}=\dfrac{36}{18}=2\)

⇒\(\left\{{}\begin{matrix}x=2.-3=-6\\y=2.4=8\end{matrix}\right.\)

Giải:

a) \(\dfrac{-5}{8}=\dfrac{x}{16}\) 

\(\Rightarrow x=\dfrac{16.-5}{8}=-10\) 

\(\dfrac{3x}{9}=\dfrac{2}{6}\) 

\(\Rightarrow3x=\dfrac{2.9}{6}=3\) 

\(\Rightarrow x=1\)

b) \(\dfrac{x+3}{15}=\dfrac{1}{3}\)  

\(\Rightarrow x+3=\dfrac{1.15}{3}=5\) 

\(\Rightarrow x=2\)

\(\dfrac{6}{2x+1}=\dfrac{2}{7}\) 

\(\Rightarrow2x+1=\dfrac{6.7}{2}=21\) 

\(\Rightarrow x=10\)

c) \(\dfrac{4}{x-6}=\dfrac{y}{24}=\dfrac{-12}{18}\) 

\(\Rightarrow\dfrac{4}{x-6}=\dfrac{-12}{18}\) 

\(\Rightarrow x-6=\dfrac{18.4}{-12}=-6\) 

\(\Rightarrow x=0\) 

\(\Rightarrow\dfrac{y}{24}=\dfrac{-12}{18}\) 

\(\Rightarrow y=\dfrac{-12.24}{18}=-16\) 

 \(\dfrac{3-x}{-12}=\dfrac{16}{y+1}=\dfrac{192}{-72}\) 

\(\Rightarrow\dfrac{3-x}{-12}=\dfrac{192}{-72}\) 

\(\Rightarrow3-x=\dfrac{192.-12}{-72}=32\) 

\(\Rightarrow x=-29\) 

\(\Rightarrow\dfrac{16}{y+1}=\dfrac{192}{-72}\) 

\(\Rightarrow y+1=\dfrac{16.-72}{192}=-6\) 

d) \(\dfrac{-2}{3}< \dfrac{x}{5}< \dfrac{-1}{6}\) 

\(\Rightarrow\dfrac{-20}{30}< \dfrac{6x}{30}< \dfrac{-5}{30}\) 

\(\Rightarrow6x\in\left\{-18;-12;-6\right\}\) 

\(\Rightarrow x\in\left\{-3;-2;-1\right\}\) 

\(\dfrac{-1}{5}\le\dfrac{x}{8}\le\dfrac{1}{4}\) 

\(\Rightarrow\dfrac{-8}{40}\le\dfrac{5x}{40}\le\dfrac{10}{40}\) 

\(\Rightarrow5x\in\left\{-5;0;5;10\right\}\) 

\(\Rightarrow x\in\left\{-1;0;1;2\right\}\) 

e) \(\dfrac{x+46}{20}=x\dfrac{2}{5}\) 

\(\Rightarrow\dfrac{x+46}{20}=x+\dfrac{2}{5}\) 

\(\Rightarrow\dfrac{x+46}{20}=\dfrac{5x+2}{5}\) 

\(\Rightarrow5.\left(x+46\right)=20.\left(5x+2\right)\) 

\(\Rightarrow5x+230=100x+40\) 

\(\Rightarrow5x-100x=40-230\) 

\(\Rightarrow-95x=-190\) 

\(\Rightarrow x=-190:-95\) 

\(\Rightarrow x=2\) 

\(y\dfrac{5}{y}=\dfrac{86}{y}\) 

\(\Rightarrow y+\dfrac{5}{y}=\dfrac{86}{y}\) 

\(\Rightarrow\dfrac{y^2+5}{y}=\dfrac{86}{y}\) 

\(\Rightarrow y^2+5=86\) 

\(\Rightarrow y^2=86-5\) 

\(\Rightarrow y^2=81\) 

\(\Rightarrow\left[{}\begin{matrix}y=9\\y=-9\end{matrix}\right.\) 

Chúc bạn học tốt!

a: Sửa đề: \(\dfrac{12}{-6}=\dfrac{x}{5}=\dfrac{-y}{3}=\dfrac{2}{-z}=\dfrac{-t}{-9}\)

=>\(\dfrac{x}{5}=\dfrac{y}{-3}=\dfrac{-2}{z}=\dfrac{t}{9}=-2\)

=>\(x=-2\cdot5=-10;y=-2\cdot\left(-3\right)=6;z=\dfrac{-2}{-2}=1;t=9\cdot\left(-2\right)=-18\)

b: \(\dfrac{-24}{-6}=\dfrac{x}{3}=\dfrac{4}{y^2}=\dfrac{z^3}{-2}\)

=>\(\dfrac{x}{3}=\dfrac{4}{y^2}=\dfrac{z^3}{-2}=4\)

=>\(\left\{{}\begin{matrix}x=4\cdot3=12\\y^2=\dfrac{4}{4}=1\\z^3=-2\cdot4=-8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=12\\y\in\left\{1;-1\right\}\\z=-2\end{matrix}\right.\)

\(a,\dfrac{12}{5}=\dfrac{x}{1,5}\Rightarrow x=\dfrac{12\cdot1,5}{5}=3,6\\ b,\dfrac{x}{5}=\dfrac{3}{20}\Rightarrow x=\dfrac{5\cdot3}{20}=\dfrac{3}{4}\\ c,\dfrac{4}{x}=\dfrac{10}{9}\Rightarrow x=\dfrac{4\cdot9}{10}=\dfrac{18}{5}\\ d,\Rightarrow\dfrac{x}{15}=\dfrac{60}{x}\Rightarrow x^2=60\cdot15=900\Rightarrow\left[{}\begin{matrix}x=30\\x=-30\end{matrix}\right.\\ 2,\)

a, Áp dụng t/c dtsbn:

\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{6}=\dfrac{x+y-z}{3+5-6}=\dfrac{8}{2}=4\\ \Rightarrow\left\{{}\begin{matrix}x=12\\y=20\\z=24\end{matrix}\right.\)

b, Áp dụng t/c dtsbn:

\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{6}=\dfrac{x-y+z}{3-5+6}=\dfrac{-4}{4}=-1\\ \Rightarrow\left\{{}\begin{matrix}x=-3\\y=-5\\z=-6\end{matrix}\right.\)

c, Áp dụng t/c dtsbn:

\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{6}=\dfrac{2y}{10}=\dfrac{3z}{18}=\dfrac{x-2y+3z}{3-10+18}=\dfrac{-33}{11}=-3\\ \Rightarrow\left\{{}\begin{matrix}x=-9\\y=-15\\z=-18\end{matrix}\right.\)

d, Đặt \(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{6}=k\Rightarrow x=3k;y=5k;z=6k\)

\(x^2-4y^2+2z^2=-475\\ \Rightarrow9k^2-100k^2+72z^2=-475\\ \Rightarrow-19k^2=-475\\ \Rightarrow k^2=25\Rightarrow\left[{}\begin{matrix}k=5\\k=-5\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=15;y=25;z=30\\x=-15;y=-25;z=-30\end{matrix}\right.\)

a: 3x=7y

=>x/7=y/3=(x-y)/(7-3)=-16/4=-4

=>x=-28; y=-12

b: x/6=y/5

=>x/6=2y/10=(x+2y)/(6+10)=20/16=5/4

=>x=30/4=15/2; y=25/4

c: Áp dụng tính chất của DTSBN, ta được:

\(\dfrac{x}{2}=\dfrac{y}{-3}=\dfrac{z}{5}=\dfrac{2x+3y+5z}{2\cdot2+3\cdot\left(-3\right)+5\cdot5}=\dfrac{6}{20}=\dfrac{3}{10}\)

=>x=3/5; y=-9/10; z=3/2

d: x/2=y/3

=>x/8=y/12

y/4=z/5

=>y/12=z/15

=>x/8=y/12=z/15

Áp dụng tính chất của DTSBN, ta được:

\(\dfrac{x}{8}=\dfrac{y}{12}=\dfrac{z}{15}=\dfrac{x+y-z}{8+12-15}=\dfrac{10}{5}=2\)

=>x=16; y=24; z=30

Lời giải:
a. Đề có cả x,y. Bạn xem lại

b. 

PT $\Leftrightarrow 5x(x-3)-2(x-3)=0$

$\Leftrightarrow (x-3)(5x-2)=0$

$\Leftrightarrow x-3=0$ hoặc $5x-2=0$

$\Leftrightarrow x=3$ hoặc $x=\frac{2}{5}$

c.

PT $\Leftrightarrow (7x-2)(x-4)=0$

$\Leftrightarrow 7x-2=0$ hoặc $x-4=0$

$\Leftrightarrow x=\frac{2}{7}$ hoặc $x=4$

d. Đề thiếu.

Lời giải:
a. Thay $y=x+1$ vào điều kiện ban đầu có:

$3x+5(x+1)=13$
$8x+5=13$

$8x=8$

$x=1$

$y=x+1=2$
b. Thay $x=y+5$ vô điều kiện đầu thì:

$2(y+5)-3y=4$

$-y+10=4$

$-y=-6$

$y=6$

$x=6+5=11$

c. Thay $y=x-2$ vô điều kiện đầu thì:

$-x+5(x-2)=-6$

$4x-10=-6$

$4x=10+(-6)=4$

$x=1$

$y=x-2=1-2=-1$

a) Ta có: \(\left\{{}\begin{matrix}3x+5y=13\\x+1=y\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}3x+5y=13\\x-y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x+5y=13\\3x-3y=-3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}8y=16\\x+1=y\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=2\\x=y-1=2-1=1\end{matrix}\right.\)

b) Ta có: \(\left\{{}\begin{matrix}2x-3y=4\\x=y+5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x-3y=4\\x-y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x-3y=4\\2x-2y=10\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}-y=-6\\x=y+5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=6\\x=11\end{matrix}\right.\)

c) Ta có: \(\left\{{}\begin{matrix}-x+5y=-6\\y=x-2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}-x+5y=-6\\x-y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4y=-4\\y=x-2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=-1\\x=y+2=-1+2=1\end{matrix}\right.\)

A=5x-6y và y=-2x