Lời giải:

$(x-12)+14=2^3.3=8.3=24$

$x-12=24-14=10$

$x=12+10=22$

Đáp án C

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$2021-5(x+4)=1^{2022}=1$

$5(x+4)=2021-1=2020$

$x+4=404$

$x=400$

Đáp án B.

a)  (a-b)-(c-b)-a=a-b-c+b-a=(a-a)+(b-b)-c=0+0-c=-c

1)

a) (a-b) - (c-b) - a= a - b - c + b - a = -c

b) -(300 - 400) + (300 - 400) + 100 = 100

2) a) 4x - 20 =  50 - ( 30 -3x)

        4x - 20 = 50 - 30 + 3x

        4x - 3x = 50 - 30 + 20= 40

        x          = 40

b) 100 - (-5x) = 40 - (-4x +10)

    100 + 5x    = 40 + 4x -10

    5x - 4x       = 40 - 10 -100

    x                = -70

c) (-50) + (-7x) = 100 - 8x

     -7x + 8x      = 100 + 50

      x                = 150

Bài 1 : 15 = 3.5;        35 = 5.7  ; 50 = 2.5²

\(\Rightarrow\)ƯCLN (15, 35, 50) = 5 

\(\Rightarrow\)BCNN = 2.3.5².7 = 1050

Bài 2 : 

Ta có: 2.3.5 = 30

Vậy A = {300, 330, 360, 390} có 4 phần tử

\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x.\left(x+1\right):2}=\frac{399}{400}\)

\(2.\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x.\left(x+1\right)}\right)=\frac{399}{400}\)

\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x.\left(x+1\right)}=\frac{399}{400}:2\)

\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{399}{400}.\frac{1}{2}\)

\(\frac{1}{2}-\frac{1}{x+1}=\frac{399}{800}\)

\(\frac{1}{x+1}=\frac{1}{2}-\frac{399}{800}\)

\(\frac{1}{x+1}=\frac{400}{800}-\frac{399}{800}\)

\(\frac{1}{x+1}=\frac{1}{800}\)

\(=>x+1=800\)

\(=>x=800-1=799\)

Vậy x = 799

Ủng hộ mk nha ^_-

ok cảm ơn nha

\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\left(x+1\right):2}=\frac{399}{400}\)

\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{399}{400}\)

\(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{399}{400}\)

\(2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{399}{400}\)

\(2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{399}{400}\)

\(2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{399}{400}\)

\(\frac{1}{2}-\frac{1}{x+1}=\frac{399}{200}\)

\(\frac{1}{x+1}=\frac{-299}{200}\)

\(x+1=\frac{-200}{299}\)

\(x=\frac{-499}{299}\)

Câu 14:

a. $6.2^2-36:3^2=6.4-36:9=24-4=20$

b. $19.48+52.19-400=19(48+52)-400=19.100-400=1900-400=1500$