c) (5x – 36) : 18 = 13

(5x – 36) = 13.18

(5x – 36) = 234

5x = 234 + 36

5x = 270

x = 270:5

x = 54

Vậy x = 54

\(a,9x-8=838\)

\(9x=838+8\)

\(9x=846\)

\(x=846:9=94\)

\(b,\left(x+30\right)-75=125\)

\(x+30=125+75\)

\(x+30=200\)

\(x=200-30=170\)

\(c,x-72:36=18\)

\(x-2=18\)

\(x=18+2=20\)

#Urushi

a) \(5\times x-123=12\)

\(\Rightarrow5\times x=135\)

\(\Rightarrow x=27\)

b) \(x+3x+5x+7x=96\)

\(\Rightarrow16x=96\)

\(\Rightarrow x=6\)

a) \(5\times x-123=12\)

\(5x=12+123\)

\(5x=135\)

\(x=135:5\)

\(x=27\)

________

b) \(x+3x+5x+7x=96\)

\(x\left(1+3+5+7\right)=96\)

\(x.16=96\)

\(x=96:16\)

\(x=6\)

=>x-5=9

hay x=14

\(a,36:\left(x-5\right)=2^2\\ 36:\left(x-5\right)=4\\ x-5=9\\ x=14\)

a) 2(2x - 8) = 24

2x - 8 = 12

2x = 20

x = 10

b)3.(2^x - 26) = 18

2^x - 26 = 6

2^x = 32

x = 5

Hoctot

x=94 nhé bạn còn cách trình bày thì bạn tự vận óc làm nhé

a: x∈Ư(48)

=>x∈{1;2;3;4;6;8;12;16;24;48}

mà x>10

nên x∈{12;16;24;48}

b: x∈Ư(18)

=>x∈{1;2;3;6;9;18}

mà x∈B(3)

nên x∈{3;6;9;18}

c: x∈Ư(36)

=>x∈{1;2;3;4;6;9;12;18;36}

mà x>=12

nên x∈{12;18;36}

d: x∈B(12)

=>x∈{12;24;...}

mà 30<=x<=100

nên x∈{36;48;60;72;84;96}

e: x∈Ư(28)

x∈Ư(21)

Do đó: x∈ƯC(21;28)

=>x∈Ư(7)

=>x∈{1;7}

f: 1-x∈Ư(17)

=>1-x∈{1;-1;17;-17}

=>x∈{0;2;18}

g: x-1∈Ư(28)

=>x-1∈{1;-1;2;-2;4;-4;7;-7;14;-14;28;-28}

=>x∈{2;0;3;-1;5;-3;8;-6;15;-13;29;-27}

mà x là số tự nhiên

nên x∈{0;2;3;5;8;15;29}

h: x+2∈Ư(2x+5)

=>2x+5⋮x+2

=>2x+4+1⋮x+2

=>1⋮x+2

=>x+2∈{1;-1}

=>x∈{-1;-3}

mà x là số tự nhiên

nên x∈∅

i: 2x+3 là bội của 2x-1

=>2x+3⋮2x-1

=>2x-1+4⋮2x-1

=>4⋮2x-1

mà 2x-1 lẻ

nên 2x-1∈{1;-1}

=>2x∈{2;0}

=>x∈{1;0}

a) \(18-\left(2x+5\right)=9\)

\(2x+5=18-9\)

\(2x+5=9\)

\(2x=9-5\)

\(2x=4\)

\(x=2\)

a) \(18-\left(2x+5\right)=9\)

\(\Rightarrow2x+5=18-9=9\)

\(\Rightarrow2x=9-5=4\Rightarrow x=4:2=2\)

b) \(23x-4=32\Rightarrow23x=32+4=36\Rightarrow x=\dfrac{36}{23}\)

c) \(\left(3x+2\right)^2=64\)

\(\Rightarrow\left[{}\begin{matrix}3x+2=8\\3x+2=-8\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{10}{3}\end{matrix}\right.\)

d) \(x\left(2x-12\right)=0\Rightarrow6x\left(x-2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)