\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-3>0\\x-1< 0\end{matrix}\right.\\\left\{{}\begin{matrix}x-3< 0\\x-1>0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow1< x< 3\)

Tham Khảo :

a: Bạn ghi lại đề nha bạn

b: \(30\left(x+2\right)-6\left(x-5\right)-24x=100\)

=>\(30x+60-6x+30-24x=100\)

=>\(\left(30x-6x-24x\right)+\left(60+30\right)=100\)

=>0x=100-90=10(vô lý)

c: \(\left(x-7\right)\left(x+3\right)< 0\)

TH1: \(\left\{{}\begin{matrix}x-7>0\\x+3< 0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>7\\x< -3\end{matrix}\right.\)

=>\(x\in\varnothing\)

TH2: \(\left\{{}\begin{matrix}x-7< 0\\x+3>0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x< 7\\x>-3\end{matrix}\right.\)

=>-3<x<7

mà x nguyên

nên \(x\in\left\{-2;-1;0;1;2;3;4;5;6\right\}\)

d: -1<2x-1<4

=>\(-1+1< 2x< 4+1\)

=>0<2x<5

=>0<x<2,5

mà x nguyên

nên \(x\in\left\{1;2\right\}\)

a) Ta có: -7<x<-1

mà \(x\in Z\)

nên \(x\in\left\{-6;-5;-4;-3;-2\right\}\)

Vậy: \(x\in\left\{-6;-5;-4;-3;-2\right\}\)

b) Ta có: -3<x<3

mà \(x\in Z\)

nên \(x\in\left\{2;1;0;1;2\right\}\)

Vậy: \(x\in\left\{2;1;0;1;2\right\}\)

c) Ta có: \(-1\le x\le6\)

mà \(x\in Z\)

nên \(x\in\left\{-1;0;1;2;3;4;5;6\right\}\)

Vậy: \(x\in\left\{-1;0;1;2;3;4;5;6\right\}\)

d) Ta có: \(-5\le x< 6\)

mà \(x\in Z\)

nên \(x\in\left\{-5;-4;-3;-2;-1;0;1;2;3;4;5\right\}\)

Vậy: \(x\in\left\{-5;-4;-3;-2;-1;0;1;2;3;4;5\right\}\)

mik ko viết dc dấu nha mọi người thông cảm

mọi người giúp với 

a: x(x+5)=0

=>\(\left[{}\begin{matrix}x=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)

b: 2x(x+3)=0

=>x(x+3)=0

=>\(\left[{}\begin{matrix}x=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\end{matrix}\right.\)

c: \(\left(6-x\right)\left(x+10\right)=0\)

=>\(\left[{}\begin{matrix}6-x=0\\x+10=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6-0=6\\x=0-10=-10\end{matrix}\right.\)

d: \(\left(5x+20\right)\left(x^2+1\right)=0\)

=>\(5x+20=0\left(x^2+1>=1>0\forall x\right)\)

=>5x=-20

=>x=-4

câu 1:

a) 500-(300)-190+(-210)

= 500-300-190-210

= 200 - 210 -190

=-10 - 190

=-200

b) (-3)³ .5+12.(-6)

= -27.5 -72

=-135 - 72

=-207

c) 15.(-19-4)-19.(15-4)

= 15.(-23) - 19.11

=-345 - 209

=-554

câu 2: tìm x thuộc Z

a) 3x-2=3

=> 3x=3/2

=> x=1/2

b) x chia hết cho 5 và -7<x<11

=> x thuộc {-5;0;5;10}

Câu 1: 

a) Ta có: \(500-\left(300\right)-190+\left(-210\right)\)

\(=500-300-190-210\)

\(=\left(500-300\right)-\left(190+210\right)\)

\(=200-400=-200\)

b) Ta có: \(\left(-3\right)^3\cdot5+12\cdot\left(-6\right)\)

\(=\left(-3\right)^3\cdot5-3\cdot4\cdot3\cdot2\)

\(=-5\cdot3^3-3^2\cdot8\)

\(=3^2\cdot\left(-5\cdot3-8\right)\)

\(=9\cdot\left(-15-8\right)=9\cdot\left(-23\right)=-207\)

c) Ta có: \(15\cdot\left(-19-4\right)-19\cdot\left(15-4\right)\)

\(=-15\cdot19-15\cdot4-15\cdot19+19\cdot4\)

\(=-30\cdot19+4\cdot4\)

\(=-2\cdot\left(15\cdot19+2\cdot4\right)\)

\(=-2\cdot\left(285+8\right)=-586\)

a: =>5-x=-23

=>x=5+23=28

b: =>x-3-x+7-25+x=54

=>x-21=54

=>x=75

c: =>7-9x-2x+4=-5x-35+27-25=-5x-37

=>-11x+3=-5x-37

=>-6x=-40

=>x=20/3

a) \(10-x-5=\left(-5\right)-7-11\)

\(\Rightarrow5-x=-23\)

\(\Rightarrow x=5+23\)

\(\Rightarrow x=28\)

b) \(\left(x-3\right)-\left(x+17-24\right)-\left(25-x\right)=24-\left(-30\right)\)

\(\Rightarrow x-3-x-17+24-25+x=54\)

\(\Rightarrow x-21=54\)

\(\Rightarrow x=54+21\)

\(\Rightarrow x=75\)

c) \(\left(7-9x\right)-\left(2x-4\right)=-\left(5x+35\right)-\left(-27\right)-25\)

\(\Rightarrow7-9x-2x+4=-5x-35+27-25\)

\(\Rightarrow11-11x=-5x-33\)

\(\Rightarrow-11x+5x=-33-11\)

\(\Rightarrow-6x=-22\)

\(\Rightarrow x=\dfrac{22}{6}=\dfrac{11}{3}\)

a. 
10-x-5 = (-5) - 7 -11
=>5-x = 0
=>x=5
b
(x-3) - (x+17-24) - (25-x) = 24 - (-30)
=>x - 3 - x - 17 + 24 - 25 - x = 24 + 30
=>-x - 21 = 54
=>-x = 75
=>x = -75
c
(7 - 9x) - (2x - 4) = - (5x + 35) - (-27) - 25
=>7-9x - 2x + 4 = -5x - 35 + 27 - 35
=>11 - 11x + 5x = -43
=>16x = 11 + 43
=>16x = 54
=>x=4

a.

\(\Leftrightarrow x^2+3xy+\dfrac{9y^2}{4}=-\dfrac{3y^2}{4}+3y\)

\(\Leftrightarrow-\dfrac{3y^2}{4}+3y=\left(x+\dfrac{3y}{2}\right)^2\ge0\)

\(\Rightarrow-\dfrac{3y^2}{4}+3y\ge0\)

\(\Rightarrow3-\dfrac{3}{4}\left(y-2\right)^2\ge0\)

\(\Rightarrow\left(y-2\right)^2\le4\)

\(\Rightarrow-2\le y-2\le2\)

\(\Rightarrow0\le y\le4\)

\(\Rightarrow y=\left\{0;1;2;3;4\right\}\)

Lần lượt thế vào pt ban đầu:

Với \(y=0\Rightarrow x^2=0\Rightarrow x=0\)

Với \(y=1\Rightarrow x^2+3x=0\Rightarrow\left[{}\begin{matrix}x=0\\x=-3\end{matrix}\right.\)

Với \(y=2\Rightarrow x^2+6x+6=0\) ko có nghiệm nguyên ((loại)

Với \(y=3\Rightarrow x^2+9x+18=0\Rightarrow\left[{}\begin{matrix}x=-3\\x=-6\end{matrix}\right.\)

Với \(y=4\Rightarrow x^2+12x+36=0\Rightarrow x=-6\)

b. Kiểm tra lại đề, đề bài đúng như thế này thì không giải được (có vô số nghiệm)

Câu 1: 

a) Ta có: x-3 là ước của 13

\(\Leftrightarrow x-3\inƯ\left(13\right)\)

\(\Leftrightarrow x-3\in\left\{1;-1;13;-13\right\}\)

hay \(x\in\left\{4;2;16;-10\right\}\)(thỏa mãn)

Vậy: \(x\in\left\{4;2;16;-10\right\}\)

b) Ta có: \(x^2-7\) là ước của \(x^2+2\)

\(\Leftrightarrow x^2+2⋮x^2-7\)

\(\Leftrightarrow x^2-7+9⋮x^2-7\)

mà \(x^2-7⋮x^2-7\)

nên \(9⋮x^2-7\)

\(\Leftrightarrow x^2-7\inƯ\left(9\right)\)

\(\Leftrightarrow x^2-7\in\left\{1;-1;3;-3;9;-9\right\}\)

mà \(x^2-7\ge-7\forall x\)

nên \(x^2-7\in\left\{1;-1;3;-3;9\right\}\)

\(\Leftrightarrow x^2\in\left\{8;6;10;4;16\right\}\)

\(\Leftrightarrow x\in\left\{2\sqrt{2};-2\sqrt{2};-\sqrt{6};\sqrt{6};\sqrt{10};-\sqrt{10};2;-2;4;-4\right\}\)

mà \(x\in Z\)

nên \(x\in\left\{2;-2;4;-4\right\}\)

Vậy: \(x\in\left\{2;-2;4;-4\right\}\)

Câu 2: 

a) Ta có: \(2\left(x-3\right)-3\left(x-5\right)=4\left(3-x\right)-18\)

\(\Leftrightarrow2x-6-3x+15=12-4x-18\)

\(\Leftrightarrow-x+9+4x+6=0\)

\(\Leftrightarrow3x+15=0\)

\(\Leftrightarrow3x=-15\)

hay x=-5

Vậy: x=-5