x-3/-9 = -4/x-3

(x-3)^2=-9.-4=36

=>TH1 x-3= 6

   TH2  x-3 = -6

=> TH1: X= 9

     TH2: X=-3

=>(x-3)^2=36

=>x-3=6 hoặc x-3=-6

=>x=-3; x=9

`a)5/8x+2/5=1/5`

`=>5/8x=1/5-2/5`

`=>5/8x=-1/5`

`=>x=-1/5:5/8=-8/25`

`b)5/7:x+11/7=18/7`

`=>5/7:x=18/7-11/7`

`=>5/7:x=1`

`=>x=5/7`

`c)(-1,2).(-3/24)+(0,4-1 4/15):1 2/3`

`=(-6/5).(-1/8)+(2/5-19/15):5/3`

`=3/20+(-13/15)*3/5`

`=3/20-13/25=-37/100`

a)5/8.x+2/5=1/5

5/8.x=1/5 - 2/5

5/8.x=-1/5

x=(-1/5):5/8

x=(-1/5).8/5

x=-8/25.      Vậy x=-8/25

b)5/7:x +11/7=18/7

5/7:x=1

x=5/7:1

x=5/7.      Vậy x=5/7

a) Ta có: \(\dfrac{5}{8}x+\dfrac{2}{5}=\dfrac{1}{5}\)

\(\Leftrightarrow x\cdot\dfrac{5}{8}=\dfrac{-1}{5}\)

\(\Leftrightarrow x=\dfrac{-1}{5}:\dfrac{5}{8}=\dfrac{-1}{5}\cdot\dfrac{8}{5}=\dfrac{-8}{25}\)

b) Ta có: \(\dfrac{5}{7}:x+\dfrac{11}{7}=\dfrac{18}{7}\)

\(\Leftrightarrow\dfrac{5}{7}:x=1\)

hay \(x=\dfrac{5}{7}\)

x,y∈Z⇒x+1,xy-1∈Z và x+1,xy-1 thuộc Ư(3)

Ta có bảng:

Vậy \(\left(x,y\right)\in\left\{\left(2;1\right)\left(-2;1\right);\left(-4;0\right)\right\}\)

Vì x,y là chữ số nên \(x,y\in\left\{0;1;2;3;4;5;6;7;8;9\right\}\)

Ta có \(\left(x+1\right)\left(xy-1\right)=3=3.1=1.3\)

Với \(\left\{{}\begin{matrix}x+1=3\\xy-1=1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)

Với \(\left\{{}\begin{matrix}x+1=1\\xy-1=3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0\\-1=3\end{matrix}\right.\left(loại\right)\)

Vậy \(\left(x;y\right)=\left(2;1\right)\)

a:ta có: \(2x^2\ge0\)

\(\Leftrightarrow2x^2+1>0\forall x\)

vậy: H(x) vô nghiệm

ko cần trả lời đâu

\(a.\dfrac{12}{3}=\dfrac{20}{5}=4\\ b.\dfrac{9}{-3}=\dfrac{-15}{5}=-3\)

a, Xét \(\dfrac{x}{3}=4\Rightarrow x=12;\dfrac{20}{y}=4\Rightarrow y=\dfrac{20}{4}=5\)

b, \(\dfrac{9}{-x}=-3\Rightarrow-x=-3\Leftrightarrow x=3\)

\(\dfrac{y}{5}=-3\Rightarrow y=-15\)

a: \(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{3}{4}=\dfrac{1}{5}\\x-\dfrac{3}{4}=-\dfrac{1}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{19}{20}\\x=\dfrac{11}{20}\end{matrix}\right.\)

\(a,\left|x-\dfrac{3}{4}\right|=\dfrac{1}{5}\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{3}{4}=\dfrac{1}{5}\\x-\dfrac{3}{4}=\dfrac{-1}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{19}{20}\\x=\dfrac{11}{20}\end{matrix}\right.\)

\(b,\dfrac{-1}{3}+\left|x\right|=0,5\)

\(\Leftrightarrow\left|x\right|=\dfrac{5}{6}\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{6}\\x=\dfrac{-5}{6}\end{matrix}\right.\)

mọi người ơi giúp em vs ạ , e đang rất cần 

\(1+2+...+n=\dfrac{\left(\dfrac{n-1}{1}+1\right).\left(n+1\right)}{2}=\dfrac{n\left(n+1\right)}{2}\)

\(M=\dfrac{3}{1+2}+\dfrac{3}{1+2+3}+...+\dfrac{3}{1+2+...+2022}\)

\(=3\left(\dfrac{1}{1+2}+\dfrac{1}{1+2+3}+...+\dfrac{1}{1+2+...+2022}\right)\)

\(=3\left(\dfrac{1}{\dfrac{2.\left(2+1\right)}{2}}+\dfrac{1}{\dfrac{3.\left(3+1\right)}{2}}+...+\dfrac{1}{\dfrac{2022.\left(2022+1\right)}{2}}\right)\)

\(=3\left(\dfrac{2}{2.3}+\dfrac{2}{3.4}+...+\dfrac{2}{2022.2023}\right)\)

\(=3.2.\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2022.2023}\right)\)

\(=6.\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2022}-\dfrac{1}{2023}\right)\)

\(=6.\left(\dfrac{1}{2}-\dfrac{1}{2023}\right)\)

\(=6.\dfrac{2021}{4046}=3.\dfrac{2021}{2023}=\dfrac{6063}{2023}=\dfrac{18189}{6069}\)

\(\dfrac{10}{3}=\dfrac{20230}{6069}>\dfrac{18189}{6069}=M\)