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

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 \(x = \dfrac{{ - 20}}{3}\)

a) \(\dfrac{x}{27}=\dfrac{-2}{3,6}\)

=> x. 3,6 = 27. (-2)

=> x.3,6 = -54

x = (-54) : 3,6

x = -15

b) -0,52 : x = -9,36 : 16,38

- 0,52 : x = \(\dfrac{-4}{7}\)

x = \(\dfrac{-4}{7}\) . ( -0,52)

x = \(\dfrac{52}{175}\)

Lời giải:

a)

b)

-0,52 : x = -9,36: 16,38 => -0,52 . 16,38 = x. (-9,36) => x = -8,5176: ( -9,36) => x = 0,91

c)

a)\(\dfrac{x}{27}=\dfrac{-2}{3,6}\)

=>x.3,6=27.(-2)

=>x . 3,6=-54

=>x=-54:3.6

=>x=-54:\(\dfrac{18}{5}\)

=>x=-54.\(\dfrac{5}{18}\)

=>x=-15

b)-0,52:x=-9,36:16,38

-\(\dfrac{13}{25}:x=-\dfrac{234}{25}:\dfrac{819}{50}\)

\(-\dfrac{13}{25}:x=-\dfrac{4}{7}\)

\(x=-\dfrac{4}{7}:\left(-\dfrac{13}{25}\right)\)

x=\(\dfrac{52}{175}\)

c)\(\dfrac{4\dfrac{1}{4}}{2\dfrac{7}{8}}=\dfrac{x}{1,61}\)

\(\dfrac{\dfrac{17}{4}}{\dfrac{23}{8}}=\dfrac{x}{1,61}\)

\(\dfrac{17}{4}:\dfrac{23}{8}=\dfrac{x}{1,61}\)

\(\dfrac{17}{4}.\dfrac{8}{23}=\dfrac{x}{1,61}\)

\(\dfrac{34}{23}=\dfrac{x}{1,61}\)

=>34.1,61=x.23

rồi bạn tự tính tự làm nốt nhé

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 2: 

b) Ta có: \(\dfrac{y}{3}=\dfrac{z}{7}\)

nên \(\dfrac{y}{12}=\dfrac{z}{28}\)

mà \(\dfrac{x}{11}=\dfrac{y}{12}\)

nên \(\dfrac{x}{11}=\dfrac{y}{12}=\dfrac{z}{28}\)

hay \(\dfrac{2x}{22}=\dfrac{y}{12}=\dfrac{z}{28}\)

mà 2x-y+z=152

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

\(\dfrac{2x}{22}=\dfrac{y}{12}=\dfrac{z}{28}=\dfrac{2x-y+z}{22-12+28}=\dfrac{152}{38}=4\)

Do đó:

\(\left\{{}\begin{matrix}\dfrac{x}{11}=4\\\dfrac{y}{12}=4\\\dfrac{z}{28}=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=44\\y=48\\z=112\end{matrix}\right.\)

Vậy: (x,y,z)=(44;48;112)

bạn ghi rõ đề câu b ra đi ạ

a)\(\dfrac{x}{27}=-\dfrac{2}{3,6}\)

\(\Rightarrow\)\(3,6x=27.(-2)\)

\(\Leftrightarrow\)\(3,6x=54\)

\(x=-15\)

\(a,E=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)^2}:\dfrac{x-1+\sqrt{x}+2-x}{\sqrt{x}\left(\sqrt{x}-1\right)}\left(x>0;x\ne1\right)\\ E=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)^2}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}+1}=\dfrac{x}{\sqrt{x}-1}\\ b,E>1\Leftrightarrow\dfrac{x-\sqrt{x}+1}{\sqrt{x}-1}>0\\ \Leftrightarrow\sqrt{x}-1>0\left[x-\sqrt{x}+1=\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\right]\\ \Leftrightarrow x>1\left(tm\right)\)

\(c,E=\dfrac{x}{\sqrt{x}-1}=\dfrac{x-1+1}{\sqrt{x}-1}=\sqrt{x}+1+\dfrac{1}{\sqrt{x}-1}\\ E=\sqrt{x}-1+\dfrac{1}{\sqrt{x}-1}+2\ge2\sqrt{\dfrac{\sqrt{x}-1}{\sqrt{x}-1}}+2=2+2=4\\ E_{min}=4\Leftrightarrow\sqrt{x}-1=1\Leftrightarrow x=4\)

Ta có : \(\dfrac{5}{3} = \dfrac{x}{9} \Rightarrow 5.9 = 3x \Leftrightarrow 45 = 3x \Rightarrow x = 45:3\)

\( \Rightarrow \) x = 15

Vậy x = 15

\(a,\dfrac{x-1}{x+5}=\dfrac{6}{7}\\ \Leftrightarrow\left(x-1\right).7=6\left(x+5\right)\\ \Rightarrow7x-7=6x+30\\ \Rightarrow7x-6x=7+30\\ \Rightarrow x=37\)

Vậy \(x=37\)

\(b,\dfrac{x^2}{6}=\dfrac{24}{25}\\ \Leftrightarrow x^2.25=24.6\\ \Rightarrow x^2.5^2=144\\ \Rightarrow\left(5x\right)^2=144\\ \Rightarrow\left(5x\right)^2=\left(\pm12\right)^2\\ \Rightarrow\left\{{}\begin{matrix}5x=12\\5x=-12\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{12}{5}\\x=-\dfrac{12}{5}\end{matrix}\right.\)

Vậy \(x=\pm\dfrac{12}{5}\)