a, \(\dfrac{x}{-3}\)= \(\dfrac{7}{4}\) ⇒ x = \(\dfrac{7}{4}\)x (-3) ⇒ x = - \(\dfrac{21}{4}\)

b, \(\dfrac{x+9}{15-x}\) = \(\dfrac{2}{3}\)  ⇒ 3(x+9)  = 2( 15-x) ⇒ 3x + 27 = 30 - 2x

⇒ 3x + 2x = 30 - 27 ⇒

 5x = 3 ⇒ x = 3 : 5 ⇒ x = \(\dfrac{3}{5}\)

\(\begin{array}{l}a)\dfrac{x}{6} = \dfrac{{ - 3}}{4}\\x = \dfrac{{( - 3).6}}{4}\\x = \dfrac{{ - 9}}{2}\end{array}\)

Vậy \(x = \dfrac{{ - 9}}{2}\)

\(\begin{array}{l}b)\dfrac{5}{x} = \dfrac{{15}}{{ - 20}}\\x = \dfrac{{5.( - 20)}}{{15}}\\x = \dfrac{{ - 20}}{3}\end{array}\)

Vậy \(x = \dfrac{{ - 20}}{3}\)

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)

2.

a) \(\dfrac{x-1}{4}=\dfrac{9}{x-1}\)

\(\Leftrightarrow\left(x-1\right)^2=9.4\)

\(\Leftrightarrow\left(x-1\right)^2=36\)

\(\Leftrightarrow\left(x-1\right)^2=\left[{}\begin{matrix}6^2\\-6^2\end{matrix}\right.\)

\(\Leftrightarrow x-1=\left[{}\begin{matrix}6\\-6\end{matrix}\right.\)

\(\Leftrightarrow x=\left[{}\begin{matrix}7\\-5\end{matrix}\right.\)

1. a. \(\dfrac{3}{-1}=\dfrac{-15}{5}\); \(\dfrac{-1}{3}=\dfrac{5}{-15}\)

\(\dfrac{3}{-15}\)= \(\dfrac{5}{-1}\); \(\dfrac{-15}{3}=\dfrac{-1}{5}\)

b. \(\dfrac{4}{-3}=\dfrac{-12}{9};\dfrac{-3}{4}=\dfrac{9}{-12}\)

\(\dfrac{4}{-12}=\dfrac{9}{-3};\dfrac{-12}{4}=\dfrac{-3}{9}\)

c. \(\dfrac{3}{7}=\dfrac{c}{b};\dfrac{7}{3}=\dfrac{b}{c}\)

\(\dfrac{3}{c}=\dfrac{b}{7};\dfrac{c}{3}=\dfrac{7}{b}\)

d. \(\dfrac{a}{b}=\dfrac{y}{x};\dfrac{b}{a}=\dfrac{x}{y}\)

\(\dfrac{a}{y}=\dfrac{x}{b};\dfrac{y}{a}=\dfrac{b}{x}\)

chúc bạn học tốt

Mấy bài dễ tự làm nhé:D

1)

Đặt: \(\dfrac{a}{b}=\dfrac{c}{d}=k\Leftrightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)

\(\left\{{}\begin{matrix}\dfrac{a}{a+b}=\dfrac{bk}{bk+b}=\dfrac{bk}{b\left(k+1\right)}=\dfrac{k}{k+1}\\\dfrac{c}{c+d}=\dfrac{dk}{dk+d}=\dfrac{dk}{d\left(k+1\right)}=\dfrac{k}{k+1}\end{matrix}\right.\)

Ta có điều phải chứng minh

\(\left\{{}\begin{matrix}\dfrac{a}{a-b}=\dfrac{bk}{bk-b}=\dfrac{bk}{b\left(k-1\right)}=\dfrac{k}{k-1}\\\dfrac{c}{c-d}=\dfrac{dk}{dk-d}=\dfrac{dk}{d\left(k-1\right)}=\dfrac{k}{k-1}\end{matrix}\right.\)

Ta có điều phải chứng minh

Cái này chỉ cần làm quy tắc nhân chéo là ra rồi nhé :)

a) \(x=\dfrac{-2,6.42}{-12}\)=9,1

b) x = \(\dfrac{2,5.12}{1.5}\) = 20

c) Nhân chéo: 7.(x-1) = 6.(x+5)

<=> 7x - 7 = 6x +30

<=> 7x - 6x = 7 + 30 (chuyển vế)

-> x = 37

d) Nhân chéo: 25x² = 24.6 = 144

x² = \(\dfrac{144}{25}\)=5,76

-> x = \(\sqrt{5,76}\) = 2,4

e) Nhân chéo: (x-2)² = 4.9 = 36

Ta dễ thấy (x-2)² = 6²

-> x-2 = 6 -> x = 6+2 = 8

TICK NHÉ :)

\(D:\dfrac{x}{2}=\dfrac{y}{9}\)

Chọn D

j vậy bạn?????