\(a.\)

\(\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{x+y}{2+5}=\dfrac{-14}{7}=-2\)

\(\Rightarrow\left\{{}\begin{matrix}x=\left(-2\right)\cdot2=-4\\y=\left(-2\right)\cdot5=-10\end{matrix}\right.\)

\(b.\)

\(\dfrac{x}{7}=\dfrac{y}{5}=\dfrac{x-y}{7-5}=\dfrac{8}{2}=4\)

\(\Rightarrow\left\{{}\begin{matrix}x=4\cdot7=28\\y=5\cdot4=20\end{matrix}\right.\)

a, Áp dụng t/c dtsbn:

\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{x+y}{3+5}=\dfrac{16}{8}=2\\ \Rightarrow\left\{{}\begin{matrix}x=6\\y=10\end{matrix}\right.\\ b,x:2=y:\left(-5\right)\Rightarrow\dfrac{x}{2}=\dfrac{y}{-5}\)

Áp dụng t/c dtsbn:

\(\dfrac{x}{2}=\dfrac{y}{-5}=\dfrac{x-y}{2-\left(-5\right)}=\dfrac{-7}{7}=-1\\ \Rightarrow\left\{{}\begin{matrix}x=-2\\y=5\end{matrix}\right.\)

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

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

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

Bài 2: 

a: =>x=0 hoặc x=-3

b: =>x-2=0 hoặc 5-x=0

=>x=2 hoặc x=5

c: =>x-1=0

hay x=1

Bài 1: 

Ta có: \(3x=2y\)

nên \(\dfrac{x}{2}=\dfrac{y}{3}\)

mà x+y=-15

nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{x+y}{2+3}=\dfrac{-15}{5}=-3\)

Do đó:

\(\left\{{}\begin{matrix}\dfrac{x}{2}=-3\\\dfrac{y}{3}=-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-6\\y=-9\end{matrix}\right.\)

Vậy: (x,y)=(-6;-9)

Bài 2: 

a) Ta có: \(\dfrac{x}{4}=\dfrac{y}{3}=\dfrac{z}{5}\)

mà x+y-z=20

nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{x}{4}=\dfrac{y}{3}=\dfrac{z}{5}=\dfrac{x+y-z}{4+3-5}=\dfrac{20}{2}=10\)

Do đó:

\(\left\{{}\begin{matrix}\dfrac{x}{4}=10\\\dfrac{y}{3}=10\\\dfrac{z}{5}=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=40\\y=30\\z=50\end{matrix}\right.\)

Vậy: (x,y,z)=(40;30;50)

Bài 1: 

a) Ta có: \(\dfrac{17}{6}-x\left(x-\dfrac{7}{6}\right)=\dfrac{7}{4}\)

\(\Leftrightarrow\dfrac{17}{6}-x^2+\dfrac{7}{6}x-\dfrac{7}{4}=0\)

\(\Leftrightarrow-x^2+\dfrac{7}{6}x+\dfrac{13}{12}=0\)

\(\Leftrightarrow-12x^2+14x+13=0\)

\(\Delta=14^2-4\cdot\left(-12\right)\cdot13=196+624=820\)

Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{14-2\sqrt{205}}{-24}=\dfrac{-7+\sqrt{205}}{12}\\x_2=\dfrac{14+2\sqrt{2015}}{-24}=\dfrac{-7-\sqrt{205}}{12}\end{matrix}\right.\)

b) Ta có: \(\dfrac{3}{35}-\left(\dfrac{3}{5}-x\right)=\dfrac{2}{7}\)

\(\Leftrightarrow\dfrac{3}{5}-x=\dfrac{3}{35}-\dfrac{10}{35}=\dfrac{-7}{35}=\dfrac{-1}{5}\)

hay \(x=\dfrac{3}{5}-\dfrac{-1}{5}=\dfrac{3}{5}+\dfrac{1}{5}=\dfrac{4}{5}\)

ai giúp mik vs

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

\(\dfrac{x}{5}=\dfrac{y}{2}=\dfrac{z}{-3}=\dfrac{x.y.z}{5.2.-3}=\dfrac{240}{-30}=-8\)

\(\Rightarrow\dfrac{x}{5}=-8\Rightarrow x=-8.5=-40\)

\(\Rightarrow\dfrac{y}{2}=-8\Rightarrow y=-8.2=-16\)

\(\Rightarrow\dfrac{z}{-3}=-8\Rightarrow z=-8.-3=24\)

Vậy \(x=--40;y=-16\) và \(z=24\) 

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

\(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{2}=\dfrac{x^3-y^3+z^3}{3^3-4^3+2^3}=\dfrac{-29}{-29}=1\)

\(\Rightarrow\dfrac{x}{3}=1\Rightarrow x=3.1=3\)

\(\Rightarrow\dfrac{y}{4}=1\Rightarrow y=1.4=4\)

\(\Rightarrow\dfrac{z}{2}=1\Rightarrow z=1.2=2\)

Vậy \(x=3;y=4\) và \(z=2\) 

Áp dụng t/c dtsbn:

\(\dfrac{x}{5}=\dfrac{y}{7}=\dfrac{z}{11}=\dfrac{x+y+z}{5+7+11}=\dfrac{-46}{23}=-2\)

\(\Rightarrow\left\{{}\begin{matrix}x=\left(-2\right).5=-10\\y=\left(-2\right).7=-14\\z=\left(-2\right).11=-22\end{matrix}\right.\)

giúp m nhanh với sắp phải nộp bài rồi mấy bn