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a) \(\left|2x+1\right|-19=-7\)

\(\Rightarrow\left|2x+1\right|=12\)

\(\Rightarrow\orbr{\begin{cases}2x+1=12\\2x+1=-12\end{cases}\Rightarrow}\orbr{\begin{cases}2x=11\\2x=-13\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{11}{2}\\x=\frac{13}{2}\end{cases}}}\)

Vậy \(x\in\left\{\frac{11}{2};\frac{13}{2}\right\}\)

21 tháng 2 2020

a,\(/2x+1/-19=-7\)

\(=>/2x+1/=-7+19=12\)

\(=>\orbr{\begin{cases}2x+1=12\\2x+1=-12\end{cases}}\)

\(=>\orbr{\begin{cases}x=\frac{12-1}{2}=\frac{11}{2}\\x=\frac{-12-1}{2}=-\frac{13}{2}\end{cases}}\)

b,\(12-2\left(-x+3\right)^2=-38\)

\(=>2\left(-x+3\right)^2=12+38=50\)

\(=>\left(-x+3\right)^2=\frac{50}{2}=25=\pm5^2\)

\(=>\orbr{\begin{cases}-x+3=5\\-x+3=-5\end{cases}}=>\orbr{\begin{cases}-x=2\\-x=-8\end{cases}=>\orbr{\begin{cases}x=-2\\x=8\end{cases}}}\)

a: =>\(2x+7\in\left\{1;-1;2;-2;3;-3;4;-4;6;-6;12;-12\right\}\)

=>\(x\in\left\{-3;-4;-\dfrac{5}{2};-\dfrac{9}{2};-2;-5;-\dfrac{3}{2};-\dfrac{11}{2};-\dfrac{1}{2};-\dfrac{13}{2};\dfrac{5}{2};-\dfrac{19}{2}\right\}\)

b: =>x+2+5 chia hết cho x+2

=>\(x+2\in\left\{1;-1;5;-5\right\}\)

=>\(x\in\left\{-1;-3;3;-7\right\}\)

19 tháng 12 2021

Bài 13: 

a: =>20-x=15-8+13=20

hay x=0

23 tháng 12 2023

a: (x-2)(y-3)=5

=>\(\left(x-2\right)\cdot\left(y-3\right)=1\cdot5=5\cdot1=\left(-1\right)\cdot\left(-5\right)=\left(-5\right)\cdot\left(-1\right)\)

=>\(\left(x-2;y-3\right)\in\left\{\left(1;5\right);\left(5;1\right);\left(-1;-5\right);\left(-5;-1\right)\right\}\)

=>\(\left(x,y\right)\in\left\{\left(3;8\right);\left(7;4\right);\left(1;-2\right);\left(-3;2\right)\right\}\)

b: (2x-1)*(y-4)=-11

=>\(\left(2x-1\right)\cdot\left(y-4\right)=1\cdot\left(-11\right)=\left(-11\right)\cdot1=\left(-1\right)\cdot11=11\cdot\left(-1\right)\)

=>\(\left(2x-1;y-4\right)\in\left\{\left(1;-11\right);\left(-11;1\right);\left(-1;11\right);\left(11;-1\right)\right\}\)

=>\(\left(x,y\right)\in\left\{\left(1;-7\right);\left(-5;5\right);\left(0;15\right);\left(6;3\right)\right\}\)

c: xy-2x+y=3

=>\(x\left(y-2\right)+y-2=1\)

=>\(\left(x+1\right)\left(y-2\right)=1\)

=>\(\left(x+1\right)\cdot\left(y-2\right)=1\cdot1=\left(-1\right)\cdot\left(-1\right)\)

=>\(\left(x+1;y-2\right)\in\left\{\left(1;1\right);\left(-1;-1\right)\right\}\)

=>\(\left(x,y\right)\in\left\{\left(0;3\right);\left(-2;1\right)\right\}\)

a) Ta có: 12-5x=37

\(\Leftrightarrow5x=-25\)

hay x=-5

Vậy: x=-5

b) Ta có: 7-3|x-2|=-11

\(\Leftrightarrow3\left|x-2\right|=18\)

\(\Leftrightarrow\left|x-2\right|=6\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=6\\x-2=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-4\end{matrix}\right.\)

Vậy: \(x\in\left\{8;-4\right\}\)

c) Ta có: \(x+\dfrac{2}{8}=-\dfrac{15}{4}\)

\(\Leftrightarrow x=\dfrac{-15}{4}-\dfrac{2}{8}=\dfrac{-15}{4}-\dfrac{1}{4}\)

hay x=-4

Vậy: x=-4

28 tháng 6 2021

a, \(\Leftrightarrow5x=12-37=-25\)

\(\Leftrightarrow x=-\dfrac{25}{5}=-5\)

Vậy ...

b, \(\Leftrightarrow3\left|x-2\right|=7+11=18\)

\(\Leftrightarrow\left|x-2\right|=\dfrac{18}{3}=6\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-4\end{matrix}\right.\)

Vậy ...

c, \(\Leftrightarrow x=-\dfrac{15}{4}-\dfrac{2}{8}=-4\)

Vậy ..

 

 

4 tháng 4 2021

1,

a, \(\left(\dfrac{-4}{3}+\dfrac{1}{3}\right).\dfrac{5}{12}\)=-\(\dfrac{5}{12}\)

b, \(\dfrac{16}{5}+\left(\dfrac{-45}{14}\right):\dfrac{3}{28}\)

=\(\dfrac{-2}{15}\)

2,

a, 2x+19=25

=>x=3

b, \(-\dfrac{2}{9}x=\dfrac{1}{3}\)

=>x=\(\dfrac{-3}{2}\)

Bài 1: 

a) Ta có: \(\dfrac{-4}{3}\cdot\dfrac{5}{12}+\dfrac{1}{3}\cdot\dfrac{5}{12}\)

\(=\dfrac{5}{12}\cdot\left(\dfrac{-4}{3}+\dfrac{1}{3}\right)\)

\(=\dfrac{-5}{12}\)

b) Ta có: \(3\dfrac{1}{5}+\left(\dfrac{2}{7}-\dfrac{7}{2}\right):\dfrac{3}{28}\)

\(=\dfrac{16}{5}+\left(\dfrac{4}{14}-\dfrac{49}{14}\right):\dfrac{3}{28}\)

\(=\dfrac{16}{5}+\dfrac{-45}{14}\cdot\dfrac{28}{3}\)

\(=\dfrac{16}{5}-30=\dfrac{-134}{5}\)

Bài 10:

a: 2x-3 là bội của x+1

=>\(2x-3⋮x+1\)

=>\(2x+2-5⋮x+1\)

=>\(-5⋮x+1\)

=>\(x+1\in\left\{1;-1;5;-5\right\}\)

=>\(x\in\left\{0;-2;4;-6\right\}\)

b: x-2 là ước của 3x-2

=>\(3x-2⋮x-2\)

=>\(3x-6+4⋮x-2\)

=>\(4⋮x-2\)

=>\(x-2\inƯ\left(4\right)\)

=>\(x-2\in\left\{1;-1;2;-2;4;-4\right\}\)

=>\(x\in\left\{3;1;4;0;6;-2\right\}\)

Bài 14:

a: \(4n-5⋮2n-1\)

=>\(4n-2-3⋮2n-1\)

=>\(-3⋮2n-1\)

=>\(2n-1\inƯ\left(-3\right)\)

=>\(2n-1\in\left\{1;-1;3;-3\right\}\)

=>\(2n\in\left\{2;0;4;-2\right\}\)

=>\(n\in\left\{1;0;2;-1\right\}\)

mà n>=0

nên \(n\in\left\{1;0;2\right\}\)

b: \(n^2+3n+1⋮n+1\)

=>\(n^2+n+2n+2-1⋮n+1\)

=>\(n\left(n+1\right)+2\left(n+1\right)-1⋮n+1\)

=>\(-1⋮n+1\)

=>\(n+1\in\left\{1;-1\right\}\)

=>\(n\in\left\{0;-2\right\}\)

mà n là số tự nhiên

nên n=0

4 tháng 12 2023

thiếu bài 16