15 x 360 + 40 x 61 x 9 + 180 x 62 - 210 x 12

= 15 x 360 + (40 x 9) x 61 + 180 x 62 - 180 x \(\frac{7}{6}\) x 12

 = 15 x 360 + 360 x 61 + 180 x 62 - 180 x 14

= (15 + 61) x 360 + 180 x (62 - 14)

= 76 x 360 + 180 x 48

= 76 x 2 x 180 + 180 x 48

= 152 x 180 + 180 x 48

= (152 + 48) x 180 

= 200 x 180

= 36000

15 x 360 + 40 x 61 x 9 + 180 x 62  - 210 x 12

= 5400 + 40 x 61 x 9 + 180 x 62 - 210 x 12

= 5400 + 21960 - 2520

= 27360 - 2520

= 24840

AI K MK MK K LẠI

=36000 kb với mk nha

Nguyễn Hồng Vân

15 x 360 + 40 x 61 x 9 + 180 x 62 - 210 x 12

= 15 x 360 + 360 x 61 + 360 x 31 - 360 x 7

= ( 15 + 61 + 31 - 7 ) x 360

= 100 x 360 = 36000 

=5400+21960+11160-2520

=27090+8640

=35730

15 x 360 + 40 x 61 x 9 + 180 x 62 - 210 x 12

= 15 x 360 + 360 x 61 + 360 x 31 - 360 x 7

= ( 15 + 61 + 31 - 7 ) x 360

= 100 x 360

= 36000

a) 6 . x - 15 = 15

6 . x = 15 + 15

6 . x = 30

x = 30 : 6

x = 5

b) 2 . x + 3 = 9

2 . x = 9 - 3

2 . x = 6

x = 6 : 2

x = 3

c) 17 - 6 : x = 15

6 : x = 17 - 15

6 : x = 2

x = 6 : 2

x = 3

d) x : 3 + 12 = 14

x : 3 = 14 - 12

x : 3 = 2

x = 2 . 3

x = 6

a) 6 . x -15 = 15 

⇒ 6 .x = 15 + (-15)

⇒ 6 .x = 0

⇔ x = 0

Vậy x = 0.

b) 2.x + 3 = 9 

⇒2.x = 9 -3 

⇒  2.x = 6

⇔ x = 3

Vậy x = 3.

\(\frac{3}{3.6.9}+\frac{3}{6.9.12}+\frac{3}{9.12.15}+\frac{3}{12.15.18}=\frac{3}{6}\left(\frac{6}{3.6.9}+\frac{6}{6.9.12}+\frac{6}{9.12.15}+\frac{6}{12.15.18}\right)\)

= \(\frac{1}{2}\left(\frac{1}{3.6}-\frac{1}{6.9}+\frac{1}{6.9}-\frac{1}{9.12}+\frac{1}{9.12}-\frac{1}{12.15}+\frac{1}{12.15}-\frac{1}{15.18}\right)\)

\(=\frac{1}{2}\left(\frac{1}{3.6}-\frac{1}{15.18}\right)=\frac{1}{2}.\frac{14}{270}=\frac{7}{270}\)

a) \(\dfrac{x}{12}=\dfrac{6}{9}\)

\(\Rightarrow x=\dfrac{12\cdot6}{9}=8\)

b) \(\dfrac{9}{15}=x+\dfrac{11}{35}\)

\(\Rightarrow x=\dfrac{9}{15}-\dfrac{11}{35}\)

\(\Rightarrow x=\dfrac{2}{7}\)

c) \(x+\dfrac{6}{5}=\dfrac{x}{2}\)

\(\Rightarrow2\left(x+\dfrac{6}{5}\right)=x\)

\(\Rightarrow2x+\dfrac{12}{5}=x\)

\(\Rightarrow2x-x=-\dfrac{12}{5}\)

\(\Rightarrow x=-\dfrac{12}{5}\)

\(a,\dfrac{x}{12}=\dfrac{6}{9}\\ \Leftrightarrow x=12\cdot\dfrac{6}{9}\\ \Leftrightarrow x=8\\ b,\dfrac{9}{15}=x+\dfrac{11}{35}\\ \Leftrightarrow x=\dfrac{9}{15}-\dfrac{11}{35}\\ \Leftrightarrow x=\dfrac{2}{7}\\ c,x+\dfrac{6}{5}=\dfrac{x}{2}\\ \Leftrightarrow\dfrac{x}{2}=-\dfrac{6}{5}\\ \Leftrightarrow x=-\dfrac{12}{5}\)

a) 8

b) 2/7

c) -12/5

Phương pháp giải:

Muốn tìm một số hạng ta lấy tổng trừ đi số hạng kia.

Lời giải chi tiết:

a) x + 8 = 10

    x = 10−8

    x = 2

b) x + 5 = 17

    x = 17 − 5

    x = 12

c) 2 + x = 12

    x = 12 − 2

    x = 10

d) 7 + x = 10

    x = 10−7

    x = 3

e) x + 4 = 15

    x = 15 − 4

    x = 11

\(a)\)\(6-\left(15+15\right)=x-\left(15-6\right)\)

\(\Leftrightarrow6-30=x-9\)

\(\Leftrightarrow-24=x-9\)

\(\Leftrightarrow x=-24+9\)

\(\Leftrightarrow x=-15\)

Vậy\(x=-15\)

\(b)\)\(x+\frac{4}{12}=\frac{3}{-9}\)

\(\Leftrightarrow x+\frac{1}{3}=-\frac{1}{3}\)

\(\Leftrightarrow x=-\frac{1}{3}-\frac{1}{3}\)

\(\Leftrightarrow x=-\frac{2}{3}\)

Vậy\(x=\frac{-2}{3}\)

Linz

a, \(6-\left(15+15\right)=x-\left(15-6\right)\)

\(\Leftrightarrow6-30=x-9\Leftrightarrow15=x+30\Leftrightarrow x=-15\)

b, \(\frac{x+4}{12}=\frac{3}{-9}\)

\(\Leftrightarrow-9x-36=36\Leftrightarrow-9x=72\Leftrightarrow x=-8\)

\(\frac{x+12}{7}+\frac{x+4}{15}+\frac{x+6}{13}=\frac{x+8}{11}+\frac{x+10}{9}+\frac{x+12}{7}\)

=>\(\left(\frac{x+12}{7}+1\right)+\left(\frac{x+4}{15}+1\right)+\left(\frac{x+6}{13}+1\right)=\left(\frac{x+8}{11}+1\right)+\left(\frac{x+10}{9}+1\right)+\left(\frac{x+12}{7}+1\right)\)

=> \(\frac{x+19}{7}+\frac{x+19}{15}+\frac{x+19}{13}=\frac{x+19}{11}+\frac{x+19}{9}+\frac{x+19}{7}\)

=> \(\frac{x+19}{7}+\frac{x+19}{15}+\frac{x+19}{13}-\frac{x+19}{11}-\frac{x+19}{9}-\frac{x+19}{7}=0\)

\(\Rightarrow\left(x+19\right)\left(\frac{1}{7}+\frac{1}{15}+\frac{1}{13}-\frac{1}{11}-\frac{1}{9}-\frac{1}{7}\right)=0\)

=> x + 19 = 0 Vì \(\frac{1}{7}+\frac{1}{15}+\frac{1}{13}-\frac{1}{11}-\frac{1}{9}-\frac{1}{7}\ne0\)

=> x = - 19

Bài làm:

Ta có: \(\frac{x+12}{7}+\frac{x+4}{15}+\frac{x+6}{13}=\frac{x+8}{11}+\frac{x+10}{9}+\frac{x+12}{7}\)

\(\Leftrightarrow\left(\frac{x+12}{7}+1\right)+\left(\frac{x+4}{15}+1\right)+\left(\frac{x+6}{13}+1\right)-\left(\frac{x+8}{11}+1\right)-\left(\frac{x+10}{9}+1\right)-\left(\frac{x+12}{7}+1\right)=0\)

\(\Leftrightarrow\frac{x+19}{7}+\frac{x+19}{15}+\frac{x+19}{13}-\frac{x+19}{11}-\frac{x+19}{9}-\frac{x+19}{7}=0\)

\(\Leftrightarrow\left(x+19\right)\left(\frac{1}{7}+\frac{1}{15}+\frac{1}{13}-\frac{1}{11}-\frac{1}{9}-\frac{1}{7}\right)=0\)

\(\Leftrightarrow\left(x+19\right)\left(\frac{1}{15}+\frac{1}{13}-\frac{1}{11}-\frac{1}{9}\right)=0\)

Mà \(\hept{\begin{cases}\frac{1}{15}< \frac{1}{11}\\\frac{1}{13}< \frac{1}{9}\end{cases}\Rightarrow}\frac{1}{15}+\frac{1}{13}-\frac{1}{11}-\frac{1}{9}< 0\)

\(\Rightarrow x+19=0\)

\(\Rightarrow x=-19\)

\(\frac{x+12}{7}+\frac{x+4}{15}+\frac{x+6}{13}=\frac{x+8}{11}+\frac{x+10}{9}+\frac{x+12}{7}\)

\(\Leftrightarrow\left(\frac{x+12}{7}+1\right)+\left(\frac{x+4}{15}+1\right)+\left(\frac{x+6}{13}+1\right)=\left(\frac{x+8}{11}+1\right)+\left(\frac{x+10}{9}+1\right)+\left(\frac{x+12}{7}+1\right)\)

\(\Leftrightarrow\frac{x+12+7}{7}+\frac{x+4+15}{15}+\frac{x+6+13}{13}=\frac{x+8+11}{11}+\frac{x+10+9}{9}+\frac{x+12+7}{7}\)

\(\Leftrightarrow\frac{x+19}{7}+\frac{x+19}{15}+\frac{x+19}{13}=\frac{x+19}{11}+\frac{x+19}{9}+\frac{x+19}{7}\)

\(\Leftrightarrow\frac{x+19}{7}+\frac{x+19}{15}+\frac{x+19}{13}-\frac{x+19}{11}-\frac{x+19}{9}-\frac{x+19}{7}=0\)

\(\Leftrightarrow\left(x+19\right)\left(\frac{1}{7}+\frac{1}{15}+\frac{1}{13}-\frac{1}{11}-\frac{1}{9}-\frac{1}{7}\right)=0\)

Vì \(\frac{1}{7}+\frac{1}{15}+\frac{1}{13}-\frac{1}{11}-\frac{1}{9}-\frac{1}{7}\ne0\)

\(\Rightarrow x+19=0\Leftrightarrow x=-19\)

a) Vì x vừa là bội của 15 vừa là bội của 9 nên x cũng là bội của BCNN(15; 9) = 45

Do đó x ϵ B(45) hay x ϵ {...; -90; -45; 0; 45; 90; 135; 180; 225; 270; ...}

Mà 135 ≤ x < 230 và x là số tự nhiên nên x ϵ {135; 180; 225}

b) Vì x khi chia cho 12; 21 và 28 đều dư 3 nên x - 3 là bội của 12; 21 và 28.

Do đó x - 3 cũng là bội của BCNN(12; 21; 28) = 84

Suy ra (x - 3) ϵ B(84) hay (x - 3) ϵ {...; -84; 0; 84; 168; 252; ...}

Do đó x ϵ {...; -81; 3; 87; 171; 255; ...}

Mà x < 180 và x là số tự nhiên nên x ϵ {3; 87; 171}