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

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.

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

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

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

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

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

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

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

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

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

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

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

a: Ta có: \(12=2^2\cdot3;18=2\cdot3^2;21=3\cdot7\)

=>BCNN(12;18;21)=\(2^2\cdot3^2\cdot7=4\cdot9\cdot7=36\cdot7=252\)

x⋮12; x⋮18; x⋮21

=>x∈BC(12;18;21)

=>x∈B(252)

mà x<400

nên x=252

b: 40 chia x dư 4

=>40-4⋮x và x>4

=>36⋮x và x>4(1)

45 chia x dư 3

=>45-3⋮x và x>3

=>42⋮x và x>3(2)

50 chia x dư 2

=>50-2⋮x và x>2

=>48⋮x và x>2(3)

\(36=2^2\cdot3^2;42=2\cdot3\cdot7;48=2^4\cdot3\)

=>ƯCLN(36;42;48)=\(2\cdot3=6\)

Từ (1),(2),(3) suy ra x∈ƯC(36;42;48) và x>4

=>x∈Ư(6) và x>4

=>x=6

c: x chia 3 dư 1

=>x-1⋮3

=>x-1+3⋮3

=>x+2⋮3(1)

x chia 4 dư 2

=>x-2⋮4

=>x-2+4⋮4

=>x+2⋮4(2)

x chia 5 dư 3

=>x-3⋮5

=>x-3+5⋮5

=>x+2⋮5(3)

\(3=3;4=2^2;5=5\)

=>BCNN(3;4;5)=\(3\cdot2^2\cdot5=4\cdot3\cdot5=60\)

Từ (1),(2),(3) suy ra x+2∈BC(3;4;5)

=>x+2∈B(60)

=>x+2∈{60;120;180;240;...}

=>x∈{58;118;178;238;...}

mà x<200

nên x∈{58;118;178}

x=2

x=500

câu 1 x = 2

câu 2 x = 500