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\(30\in\left(3x-2\right)\)
\(\Rightarrow\left(3x-2\right)\in B\left(30\right)=\left\{0;30;60;90...\right\}\)
\(\Rightarrow x\in\left\{\dfrac{2}{3};\dfrac{32}{3};\dfrac{62}{3};\dfrac{92}{2}...\right\}\)
Ta có:
\(xy-2x=0\rightarrow x\left(y-2\right)=0\)\(\left(x,y\in N\right)\\\)
\(\rightarrow x=0\)hoặc \(y-2=0\rightarrow y=2\)
theo đề bài ta có :
\(\left|x+3\right|=\left|x-5\right|\) Đk : \(x\in Z\)
mà x+3 > x-5
\(\Rightarrow x\in\varnothing\)( vô nghiệm )
=>-1+(3-5)+(7-9)+...+[(x-4)-(x-2)]+x=600
=>-1+(-2)+(-2)+...+(-2)+x=600 (\(\frac{x-1}{4}\) số hạng -2)
=>-1+(-2)\(\frac{x-1}{4}\)+x=600
=>\(\frac{-4+\left(-2\right)\left(x-1\right)+4x}{4}=600\)
=>-4+(-2)(x-1)+4(x-1)+4=600.4
=>(-2+4)(x-1)=2400+4-4
=>2(x-1)=2400
=>x-1=2400:2
=>x-1=1200
=>x=1201
bài 1: x.(x+7) = 0
Th1:x=0 Th2:x+7=0
=>x=-7
bài 2 (x+12).(x-3)= 0
Th1:x+12=0 Th2:x-3=0
=>x=-12 =>x=3
bài 3 (-x+5).(3-x)=0
Th1 (-x)+5=0 Th2:3-x=0
=>-x=-5 =>x=3
bài 4 x.(2+x).(7-x)=0
Th1:x=0 Th3:7-x=0
Th2:2+x=0 =>x=7
=>x=-2
bài 5 (x-1).(x+2).(-x-3)=0
Th1:x-1=0 Th2:x+2=0
=>x=1 =>x=-2
Th3:-x-3=0
=>-x=-3
`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) 2(x + 3) = 5(1 - x) - 2
<=> 2x + 6 = 5 - 5x - 2
<=> 2x + 6 = 3 - 5x
=> 2x - 5x = 6 + 3
=> -3x = 9
=> x = 9 : (-3)
=> x = -3
\(24\in\left(x-3\right)\)
\(\Rightarrow\left(x-3\right)\in B\left(24\right)=\left\{0;24;48;72;...\right\}\)
\(\Rightarrow x\in\left\{3;27;51;75;...\right\}\)