\(\frac{-7}{6}=\frac{x}{18}=\frac{-98}{y}=\frac{-14}{z}=\frac{t}{102}=\frac{u}{-18}\)

\(\text{+) }\frac{-7}{6}=\frac{x}{18}\)

\(\Rightarrow\left(-7\right).18=6.x\)

\(\Rightarrow-126=6.x\)

\(\Rightarrow x=-21\)

\(\text{+) }\frac{-7}{6}=\frac{-98}{y}\)

\(\Rightarrow\left(-7\right).y=6.\left(-98\right)\)

\(\Rightarrow y=\frac{6.\left(-98\right)}{-7}\)

\(\Rightarrow y=-84\)

\(\text{+) }\frac{-7}{6}=\frac{-14}{z}\)

\(\Rightarrow\left(-7\right).z=\left(-14\right).6\)

\(\Rightarrow\left(-7\right).z=-84\)

\(\Rightarrow z=12\)

\(\frac{-7}{6}=\frac{t}{102}\)

\(\Rightarrow\left(-7\right).102=t.6\)

\(\Rightarrow t=\frac{\left(-7\right).102}{6}=\frac{\left(-7\right).17.6}{6}\)

\(\Rightarrow t=-119\)

\(\frac{-7}{6}=\frac{u}{-18}\)

\(\Rightarrow\frac{-7}{6}=\frac{-u}{18}\)

\(\Rightarrow\left(-7\right).18=\left(-u\right).6\)

\(\Rightarrow-126=\left(-u\right).6\)

\(\Rightarrow-u=-21\)

\(\Rightarrow u=21\)

cảm ơn bạn nha

Tham khảo!

\(\dfrac{-7}{6}=\dfrac{x}{18}=\dfrac{-98}{y}=\dfrac{-14}{z}=\dfrac{t}{102}=\dfrac{u}{-78}\)

\(\dfrac{-7}{6}=\dfrac{x}{18}\Leftrightarrow6.x=\left(-7\right).18\Rightarrow x=\dfrac{\left(-7\right).18}{6}=-21\)

\(\dfrac{-7}{6}=\dfrac{-98}{y}\Leftrightarrow\left(-7\right).y=6.\left(-98\right)\Rightarrow y=\dfrac{6.\left(-98\right)}{-7}=84\)

\(\dfrac{-7}{6}=\dfrac{-14}{z}\Leftrightarrow\left(-7\right).z=6.\left(-14\right)\Rightarrow z=\dfrac{6.\left(-14\right)}{-7}=12\)

\(\dfrac{-7}{6}=\dfrac{t}{102}\Leftrightarrow6.t=\left(-7\right).102\Rightarrow t=\dfrac{\left(-7\right).102}{6}=-119\)

\(\dfrac{-7}{6}=\dfrac{u}{-78}\Leftrightarrow6.u=\left(-7\right).\left(-78\right)\Rightarrow u=\dfrac{\left(-7\right).\left(-78\right)}{6}=91\)

\(\text{Vậy }x=-21;y=84;y=84;z=12;t=-119;u=91\)

Bài 9:

Ta có: \(\dfrac{12}{-6}=\dfrac{x}{5}=\dfrac{-y}{3}=\dfrac{z}{-17}=\dfrac{-t}{-9}\)

\(\Leftrightarrow\dfrac{x}{5}=\dfrac{-y}{3}=\dfrac{-z}{17}=\dfrac{t}{9}=-2\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{5}=-2\\\dfrac{-y}{3}=-2\\\dfrac{-z}{17}=-2\\\dfrac{t}{9}=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-10\\-y=-6\\-z=-34\\t=-18\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-10\\y=6\\z=34\\t=-18\end{matrix}\right.\)

Vậy: (x,y,z,t)=(-10;6;34;-18)

Bài 11:

Ta có: \(\dfrac{-7}{6}=\dfrac{x}{18}=\dfrac{-98}{y}=\dfrac{-14}{z}=\dfrac{t}{102}=\dfrac{u}{-78}\)

\(\Leftrightarrow\dfrac{x}{18}=\dfrac{-98}{y}=\dfrac{-14}{z}=\dfrac{t}{102}=\dfrac{u}{-78}=\dfrac{-7}{6}\)

Ta có: \(\dfrac{x}{18}=\dfrac{-7}{6}\)

\(\Leftrightarrow x=\dfrac{18\cdot\left(-7\right)}{6}=-21\)

Ta có: \(\dfrac{-98}{y}=\dfrac{-7}{6}\)

\(\Leftrightarrow y=\dfrac{-98\cdot6}{-7}=84\)

Ta có: \(\dfrac{-14}{z}=\dfrac{-7}{6}\)

\(\Leftrightarrow z=\dfrac{-14\cdot6}{-7}=12\)

Ta có: \(\dfrac{u}{-78}=\dfrac{-7}{6}\)

\(\Leftrightarrow u=\dfrac{-78\cdot\left(-7\right)}{6}=\dfrac{78\cdot7}{6}=91\)

Ta có: \(\dfrac{t}{102}=\dfrac{-7}{6}\)

\(\Leftrightarrow t=\dfrac{-7\cdot102}{6}=-7\cdot17=-119\)

Vậy: (x,y,z,t,u)=(-21;84;12;-119;91)

Nguyễn Lê Phước Thịnh giải giùm mk bài 10 đc ko ạ

Bài 2:

\(a,\dfrac{2}{x}=\dfrac{x}{8}\\ \Rightarrow x.x=8.2\\ \Rightarrow x^2=16\\ \Rightarrow x=\pm4\)

\(b,\dfrac{2x-9}{240}=\dfrac{39}{80}\\ \Rightarrow80\left(2x-9\right)=240.39\\ \Rightarrow160x-720=9360\\ \Rightarrow160x=10080\\ \Rightarrow x=63\)

\(c,\dfrac{x-1}{9}=\dfrac{8}{3}\\ \Rightarrow3\left(x-1\right)=8.9\\ \Rightarrow3\left(x-1\right)=72\\ \Rightarrow x-1=24\\ \Rightarrow x=25\)

1)\(\dfrac{-3}{6}=\dfrac{x}{-2}=\dfrac{-18}{y}=\dfrac{-z}{24}\)

\(\Rightarrow x=-\dfrac{3}{6}\cdot\left(-2\right)=1\)

\(\Rightarrow y=-18:\dfrac{-3}{6}=36\)

\(\Rightarrow z=-\dfrac{3}{6}\cdot\left(-24\right)=12\)

câu cuối làm tương tự

1)\(\dfrac{-3}{6}=\dfrac{x}{-2}=\dfrac{-18}{y}=\dfrac{-z}{24}\)

\(\Rightarrow x=-\dfrac{3}{6}\cdot\left(-2\right)=1\)

\(\Rightarrow y=-18:\dfrac{-3}{6}=36\)

\(\Rightarrow z=-\dfrac{3}{6}\cdot\left(-24\right)=12\)

1)\(\dfrac{-3}{6}=\dfrac{x}{-2}=\dfrac{-18}{y}=\dfrac{-z}{24}\)

\(\Rightarrow x=-\dfrac{3}{6}\cdot\left(-2\right)=1\)

\(\Rightarrow y=-18:\dfrac{-3}{6}=36\)

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

3) \(\dfrac{x}{-3}=\dfrac{4}{y}\Rightarrow\left\{{}\begin{matrix}x=4\\y=-3\end{matrix}\right.\)

Thanks

Bài 1

a) (x + 3)(x + 2) = 0

x + 3 = 0 hoặc x + 2 = 0

*) x + 3 = 0

x = 0 - 3

x = -3 (nhận)

*) x + 2 = 0

x = 0 - 2

x = -2 (nhận)

Vậy x = -3; x = -2

b) (7 - x)³ = -8

(7 - x)³ = (-2)³

7 - x = -2

x = 7 + 2

x = 9 (nhận)

Vậy x = 9

Bài 3

20a + 10b = 2010

10b = 2010 - 20a

b = (2010 - 20a) : 10

*) a = 0

b = (2010 - 20.0) : 10 = 201

*) a = 1

b = (2010 - 10.1) : 10 = 200

*) a = 2

b = (2010 - 10.2) : 10 = 199

Vậy ta có ba cặp số nguyên (a; b) thỏa mãn:

(0; 201); (1; 200); (2; 199)