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: 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 \(x\in\left\{1;2\right\}\)

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: \(15\cdot\left(-19-4\right)-19\cdot\left(15-4\right)\)

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

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

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

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

Đề thiếu rồi bạn

a] x=-6,-5,-4,-3,-2,

b] x=-2,-1,0,1,2

chúc học tốt

a) x ∈ {-6,-5,-4,-3,-2}

b) x ∈ {-2,-1,0,1,2}

a) x={-6,-5,-4,-3,-2}

b)x={-2,-1,0,1,2,3}

a: =>x-xy+y=0

=>x(1-y)+1-y-1=0

=>(x+1)(1-y)=1

=>(x+1)(y-1)=-1

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

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

b: 2x-xy-2y=3

=>x(2-y)-2y+4=7

=>x(2-y)+2(2-y)=7

=>(x+2)(y-2)=-7

=>\(\left(x+2;y-2\right)\in\left\{\left(1;-7\right);\left(-7;1\right);\left(-1;7\right);\left(7;-1\right)\right\}\)

=>\(\left(x;y\right)\in\left\{\left(-1;-5\right);\left(-9;3\right);\left(-3;9\right);\left(5;1\right)\right\}\)

c: =>x(4-y)+5y-20=-3

=>x(4-y)-5(4-y)=-3

=>(4-y)(x-5)=-3

=>(x-5)(y-4)=3

=>\(\left(x-5;y-4\right)\in\left\{\left(1;3\right);\left(3;1\right);\left(-1;-3\right);\left(-3;-1\right)\right\}\)

=>\(\left(x,y\right)\in\left\{\left(6;9\right);\left(8;5\right);\left(4;1\right);\left(2;3\right)\right\}\)

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

Tham Khảo :

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