(2,4*x-36):(1+5/7)=-14

(2,4*x-36): 12/7 =-14

2,4*x-36= -14*12/7

2,4*x-36= -4

2,4*x= -4+36

2,4*x= 32

x=32:2,4

x= 40/3 vậy x= 40/3

Tick nha

\(a,36.19+36.81\)

\(=36.\left(19+81\right)\)

\(=36.100\)

\(=3600\)

\(b,8.12.125.5\)

\(=\left(125.8\right)\left(12.5\right)\)

\(=1000.60\)

\(=60000\)

\(c,\)chịu 

c,7 x 9 + 14 x 27 + 21 x 36 / 21 x 27 + 42 x 81 + 63 x 108

7 x 9 + 14 x 27 + 21 x 36 / 7 x 3 x 9 x 3 + 14 x 3 x 27 x 3 + 21 x 3 x 36 x 3

1 / 3 x 3 + 3 x 3 + 3 x 3

1 / 9 + 9 + 9

1 / 27

mấy bọn nguuuuuuuuuuuuuuuuuuuuuuuu

\(\frac{12}{5}\): x = \(\frac{8}{3}\)

x = \(\frac{8}{3}\): \(\frac{12}{5}\)

x   = \(\frac{10}{9}\)

lộn đề

Sai đe a bn ❌❌❌

Sửa đề: (x-15)/17

=>\(\left(\dfrac{x-15}{17}-5\right)+\left(\dfrac{x-36}{16}-4\right)+\left(\dfrac{x-58}{14}-3\right)+\left(\dfrac{x-76}{12}-2\right)=0\)=>x-100=0

=>x=100

Ta có: \(\frac{x-90}{10}+\frac{x-76}{12}+\frac{x-58}{14}+\frac{x-36}{16}+\frac{x-15}{17}=15\)

\(\Leftrightarrow\frac{x-90}{10}+\frac{x-76}{12}+\frac{x-58}{14}+\frac{x-36}{16}+\frac{x-15}{17}-15=0\)

\(\Leftrightarrow\frac{x-90}{10}-1+\frac{x-76}{12}-2+\frac{x-58}{14}-3+\frac{x-36}{16}-4+\frac{x-15}{17}-5=0\)

\(\Leftrightarrow\frac{x-100}{10}+\frac{x-100}{12}+\frac{x-100}{14}+\frac{x-100}{16}+\frac{x-100}{17}=0\)

\(\Leftrightarrow\left(x-100\right)\left(\frac{1}{10}+\frac{1}{12}+\frac{1}{14}+\frac{1}{16}+\frac{1}{17}\right)=0\)

mà \(\frac{1}{10}+\frac{1}{12}+\frac{1}{14}+\frac{1}{16}+\frac{1}{17}\ne0\)

nên x-100=0

hay x=100

Vậy: x=100

\(\frac{x-90}{10}+\frac{x-76}{12}+\frac{x-58}{14}+\frac{x-36}{16}+\frac{x-15}{17}=0\)

\(\Rightarrow\left(\frac{x-90}{10}-1\right)+\left(\frac{x-76}{12}-2\right)+\left(\frac{x-58}{14}-3\right)+\left(\frac{x-36}{16}-4\right)+\left(\frac{x-15}{17}-5\right)=0\)

\(\Rightarrow\frac{x-100}{10}+\frac{x-100}{12}+\frac{x-100}{14}+\frac{x-100}{16}+\frac{x-100}{17}=0\)

\(\Rightarrow\left(x-100\right).\left(\frac{1}{10}+\frac{1}{12}+\frac{1}{14}+\frac{1}{16}+\frac{1}{17}\right)=0\)

\(\Rightarrow x-100=0\left(Vì\frac{1}{10}+\frac{1}{12}+\frac{1}{14}+\frac{1}{16}+\frac{1}{17}\ne0\right)\)

\(\Rightarrow x=100\)

may gioi vai

a, \(\frac{5}{x+7}+\frac{8}{2x+14}=\frac{3}{2}\) Đkxđ : \(x\ne-7\)

⇔ \(\frac{5}{x+7}+\frac{8}{2\left(x+7\right)}=\frac{3}{2}\)

⇔ \(\frac{10}{2\left(x+7\right)}+\frac{8}{2\left(x+7\right)}=\frac{3\left(x+7\right)}{2\left(x+7\right)}\)

⇒ \(10+8=3\left(x+7\right)\)

⇔ \(10+8=3x+21\)

⇔ \(-3x=21-10-8\)

⇔ \(-3x=3\)

⇔ \(x=-1\) ( tm )

Ptr có tập nhiệm : S \(=\left\{-1\right\}\)

b, \(\frac{x+3}{x-3}-\frac{1}{x}=\frac{3}{x\left(x-3\right)}\) Đkxđ : \(x\ne3;x\ne0\)

⇔ \(\frac{x\left(x+3\right)}{x\left(x-3\right)}-\frac{1\left(x-3\right)}{x\left(x-3\right)}=\frac{3}{x\left(x-3\right)}\)

⇒ \(x\left(x-3\right)-1\left(x-3\right)=3\)

⇔ \(x^2-3x-x+3=3\)

⇔ \(x^2-4x=0\)

⇔ \(x\left(x-4\right)=0\)

⇔ \(\left\{{}\begin{matrix}x=0\\x-4=0\end{matrix}\right.\)

⇔\(\left\{{}\begin{matrix}x=0\left(ktm\right)\\x=4\left(tm\right)\end{matrix}\right.\)

Ptr có tập nhiệm : S \(=\left\{4\right\}\)