C

Diện tích xung quan là 2 x 8 x ( 15 + 10 ) = 400

c

Bài 5: 

a) Ta có: \(x⋮5\)

\(\Leftrightarrow x\in B\left(5\right)\)

\(\Leftrightarrow x\in\left\{....;-15;-10;-5;0;5;10;15;...\right\}\)

mà -12<x<12

nên \(x\in\left\{-10;-5;0;5;10\right\}\)

Vậy: \(x\in\left\{-10;-5;0;5;10\right\}\)

b) Ta có: \(36⋮x\)

\(\Leftrightarrow x\inƯ\left(36\right)\)

\(\Leftrightarrow x\in\left\{-36;-18;-12;-9;-6;-4;-3;-2;-1;1;2;3;4;6;9;12;18;36\right\}\)

mà x<0

nên \(x\in\left\{-36;-18;-12;-9;-6;-4;-3;-2;-1\right\}\)

Vậy: \(x\in\left\{-36;-18;-12;-9;-6;-4;-3;-2;-1\right\}\)

Bài 6: 

a) Ta có: \(-24\cdot x=72\)

\(\Leftrightarrow x=72:\left(-24\right)\)

\(\Leftrightarrow x=-72:24=-3\)

Vậy: x=3

b) Ta có: \(-5\cdot\left|x\right|=-30\)

\(\Leftrightarrow\left|x\right|=6\)

hay \(x\in\left\{6;-6\right\}\)

Vậy: \(x\in\left\{6;-6\right\}\)

c) Ta có: \(12\cdot x=\left(-36\right)\)

\(\Leftrightarrow x=-36:12\)

\(\Leftrightarrow x=-3\)

Vậy: x=-3

d) Ta có: \(-8\cdot\left|x\right|=-32\)

\(\Leftrightarrow\left|x\right|=4\)

hay \(x\in\left\{4;-4\right\}\)

Vậy: \(x\in\left\{4;-4\right\}\)

\(\frac{17}{2}-\left|2x-\frac{5}{2}\right|=-\frac{7}{6}\)

\(\left|2x-\frac{5}{2}\right|=\frac{17}{2}-\frac{-7}{6}\)

\(\left|2x-\frac{5}{2}\right|=\frac{51}{6}+\frac{7}{6}\)

\(\left|2x-\frac{5}{2}\right|=\frac{29}{3}\)

\(2x-\frac{5}{2}=\frac{29}{3}\)hoặc \(2x-\frac{5}{2}=\frac{-29}{3}\)

Trường hợp 1:

\(2x-\frac{5}{2}=\frac{29}{3}\)

\(2x=\frac{29}{3}+\frac{5}{2}\)

\(2x=\frac{73}{6}\)

\(x=\frac{73}{6}:2\)

\(x=\frac{73}{12}\)

Trường hợp 2:

\(2x-\frac{5}{2}=\frac{-29}{3}\)

\(2x=\frac{-29}{3}+\frac{5}{2}\)

\(2x=\frac{-43}{6}\)

\(x=\frac{-43}{6}:2\)

\(x=\frac{-43}{12}\)

Vậy \(x=\frac{73}{12}\)hoặc \(x=\frac{-43}{12}\)

17/2 - |2x-5/2| = -7/6

         |2x-5/2|= 17/2 - (-7/6)

         |2x-5/2|= 29/3

2x-5/2= 29/3      hoặc     2x-5/2= -29/3

Tự tính 2 kết quả

17/2-|2x-5/2|=-7/6

=>-|2x-5/2|=-7/6-17/2    (chuyển vế đổi dấu)

=>-|2x-5/2|=-29/3          (thực hiện pép tính vế phải)

=>.|2x-5/2|=29/3             (bỏ dấu trừ cả hai vế)

như ta đã biết với a bất kì ta luôn có |a|=-a khi a<0     |a|=a khi a>0 hoặc a=0

xét trường hợp 2x-5/2>=0 =>x>=1.25

ta có |2x-5/2|=2x-5/2      (vì lớn hơn hoặc bằng 0 nên nó bằng chính nó)

khi đó 2x-5/2=29/3=>2x=73/6=>x=73/12     (giải phương trình)

tương tự 2x-5/2<0=>x<1.25

ta có |2x-5/2|=-(2x-5/2)=5/2-2x

khi đó 5/2-2x=29/3=>2x=5/2-29/3=-43/6=>x=-43/12   (như trên)

vậy x=73/12 và x=-43/12

1.99 + 2.98 + 3.97 + ..... + 49.51 + 50.50

= 1..99 + 2.(99 - 1) + 3.(99 - 2) + ..... + 49.(99 - 48) + 50.(99 - 49)

= 1.99 + 2.99 - 1.2 + 3.99 - 2.3 + ........ + 49.99 - 48.49 + 50.99 - 49.50

= (1.99 + 2.99 + 3.99 + ..... + 49.99 + 50.99) - ( 1.2 + 2.3 + ........ + 49.50)

= 99(1 + 2 + 3 + ..... + 50) - ( 1.2 + 2.3 + ........ + 49.50)

= 

\(d,\dfrac{5}{7}+\dfrac{9}{23}+-\dfrac{12}{7}+\dfrac{14}{23}\)

\(=\left(\dfrac{5}{7}+\dfrac{-12}{7}\right)+\left(\dfrac{9}{23}+\dfrac{14}{23}\right)\)

\(=-\dfrac{7}{7}+\dfrac{23}{23}\)

\(=\left(-1\right)+1=0\)

\(e,\dfrac{3}{17}+-\dfrac{5}{13}+-\dfrac{18}{35}+\dfrac{14}{17}+\dfrac{17}{-35}+-\dfrac{8}{13}\)

\(=\left(\dfrac{3}{17}+\dfrac{14}{17}\right)+\left(\dfrac{-5}{13}+\dfrac{-8}{13}\right)+\left(\dfrac{-18}{35}+\dfrac{-17}{35}\right)\)

\(=\dfrac{17}{17}+\dfrac{-13}{13}+-\dfrac{35}{35}\)

\(=1+\left(-1\right)+\left(-1\right)=0+\left(-1\right)=-1\)

\(f,\dfrac{-3}{8}. \dfrac{1}{6}+\dfrac{3}{-8}.\dfrac{5}{6}+\dfrac{-10}{16}\)

\(=\dfrac{-3}{8}.\dfrac{1}{6}+\dfrac{-3}{8}.\dfrac{5}{6}+\dfrac{-10}{16}\)

\(=\dfrac{-3}{8}.\left(\dfrac{1}{6}+\dfrac{5}{6}\right)+-\dfrac{10}{16}\)

=\(\dfrac{-3}{8}.1+\dfrac{-10}{16}\)

\(=\dfrac{-3}{8}+\dfrac{-10}{16}\)

\(=\dfrac{-6}{16}+\dfrac{-10}{16}=\dfrac{-16}{16}=-1\)

\(g,\dfrac{-4}{11}.\dfrac{5}{15}.\dfrac{11}{-4}=\dfrac{-4}{11}.\dfrac{5}{15}.\dfrac{-11}{4}\)

\(=\left(\dfrac{-4}{11}.\dfrac{-11}{4}\right).\dfrac{5}{15}\)

\(=1.\dfrac{5}{15}=1.\dfrac{1}{3}=\dfrac{1}{3}\)

\(h,\dfrac{7}{36}-\dfrac{8}{-9}+\dfrac{-2}{3}=\dfrac{7}{36}-\dfrac{-8}{9}+\dfrac{-2}{3}\)

\(=\dfrac{7}{36}-\dfrac{-32}{36}+\dfrac{-24}{36}=\dfrac{7-\left(-32\right)+\left(-24\right)}{36}\)

\(=\dfrac{15}{56}=\dfrac{5}{12}\)

Tick mình nha ^^

d.\(\dfrac{5}{7}\)+\(\dfrac{9}{23}\)+\(\dfrac{12}{7}\)+\(\dfrac{14}{23}\)=(\(\dfrac{5}{7}\)+\(\dfrac{12}{7}\))+(\(\dfrac{9}{23}\)+\(\dfrac{14}{23}\))

                            =\(\dfrac{17}{7}\)+ 1 = \(\dfrac{24}{7}\)

e.\(\dfrac{3}{17}\)+\(\dfrac{-5}{13}\)+\(\dfrac{-18}{35}\)+\(\dfrac{14}{17}\)+\(\dfrac{17}{-35}\)+\(\dfrac{-8}{13}\)

=(\(\dfrac{3}{17}\)+\(\dfrac{14}{17}\))+(\(\dfrac{-5}{13}\)+\(\dfrac{-8}{13}\))+(\(\dfrac{-18}{35}\)+\(\dfrac{17}{-35}\))

= 1+ (-1) + (-1) = -1

f. \(\dfrac{-3}{8}\).\(\dfrac{1}{6}\)+\(\dfrac{3}{-8}\).\(\dfrac{5}{6}\)+\(\dfrac{-10}{16}\)=\(\dfrac{-3}{8}\)(\(\dfrac{1}{6}\)+\(\dfrac{5}{6}\)) + \(\dfrac{-5}{8}\)

                                   =\(\dfrac{-3}{8}\)+\(\dfrac{-5}{8}\)= -1

g. \(\dfrac{-4}{11}\).\(\dfrac{5}{15}\).\(\dfrac{11}{-4}\)=\(\dfrac{5}{15}\)

h.\(\dfrac{7}{36}\)-\(\dfrac{8}{-9}\)+\(\dfrac{-2}{3}\)= \(\dfrac{7}{36}\)-\(\dfrac{-32}{36}\)+\(\dfrac{-24}{36}\)

                       =\(\dfrac{-49}{36}\)

d: \(\dfrac{5}{7}+\dfrac{9}{23}+\dfrac{-12}{7}+\dfrac{14}{23}=1-1=0\)

e: \(\dfrac{3}{17}+\dfrac{-5}{13}+\dfrac{-18}{35}+\dfrac{14}{17}+\dfrac{17}{-35}+\dfrac{-8}{13}=-1-1+1=-1\)

a. x : 13/16 = 5/-8

    x : 13/16 = -5/8

    x             = -5/8 . 13/16

    x             = -65/128

b.

x . \(\dfrac{-14}{28}\) = \(\dfrac{6}{-9}\) - \(\dfrac{2}{15}\)

x . \(\dfrac{-14}{28}\) = \(\dfrac{-9}{6}\) - \(\dfrac{2}{15}\)

x . \(\dfrac{-14}{28}\) = \(\dfrac{-49}{30}\)

x            = \(\dfrac{-49}{30}\) : \(\dfrac{-14}{28}\)

x            = \(\dfrac{-49}{30}\) . \(\dfrac{-28}{14}\)

x            = \(\dfrac{-49}{15}\)

(x-1)(x-2)=0

=>x-1=0

x=0+1

x=1

=>x-2=0

x=0+2

x=2

x-1=0 suy ra x=1

x-2=0 suy ra x=2

(x-1)(x-2)=0

=> x-1=0 hoặc x-2=0

=> x=1.     ,      x=2

\(\frac{x}{40}+\frac{x}{30}=\frac{3}{4}\)

\(< =>\frac{70x}{1200}=\frac{900}{1200}\)

\(< =>\frac{7x}{120}=\frac{9}{120}\)

\(< =>x=\frac{9}{7}\)

Bài làm

\(\frac{x}{40}+\frac{x}{30}=\frac{3}{4}\)

\(\Leftrightarrow\frac{3x}{120}+\frac{4x}{120}=\frac{90}{120}\)

\(\Rightarrow7x=90\)

\(\Leftrightarrow x=\frac{90}{7}\)

Vậy \(x=\frac{90}{7}\)là nghiệm phương trình. 

@๖ۣۜNhi's Ghostッ: dòng thứ 3 của bạn là 90 mới đúng nhé :>