2 cm chuẩn rùi mk làm rùi

cực cí lụ khu

1) \(25\%-1\dfrac{1}{2}+0,5\cdot\dfrac{12}{5}\\ =\dfrac{1}{4}-\dfrac{3}{2}+\dfrac{1}{2}\cdot\dfrac{12}{5}\\ =-\dfrac{5}{4}+\dfrac{6}{5}\\ =-\dfrac{1}{20}\)

2) \(\left(-3,2\right)\cdot\dfrac{-15}{64}+\left(0,8-2\dfrac{4}{15}\right):3\dfrac{1}{2}\)

\(=\dfrac{-16}{5}\cdot\dfrac{-15}{64}+\left(\dfrac{4}{5}-\dfrac{34}{15}\right):\dfrac{7}{2}\\ =\dfrac{3}{4}-\dfrac{22}{15}\cdot\dfrac{2}{7}\\ =\dfrac{3}{4}-\dfrac{44}{105}\\ =\dfrac{139}{420}\)

\(\dfrac{-7}{13}\) + \(\dfrac{-5}{3}\) = \(\dfrac{-21}{39}\) + \(\dfrac{-65}{39}\) = \(\dfrac{-86}{39}\)

-7/13+-5/3=-21/39+-65/39

=-86/39

\(6⋮\left(x-1\right)\)

\(\Rightarrow x-1\in\text{ư}\left(6\right)=1.2.3.6\)

\(\Rightarrow x-1=1\)

\(\Rightarrow x-1=2\)

\(\Rightarrow x-1=3\)

\(\Rightarrow x-1=6\)

\(x=1+1=2\)

\(x=1+2=3\)

\(x=1+3=4\)

\(x=1+6=7\)

=> x - 1 \(\in U\left[6\right]\in\left\{-6;-3;-2;-1;1;2;3;6\right\}\)

\(\Rightarrow x\in\left\{-5;-2;-1;0;2;3;4;7\right\}\)

\(\left(x+1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)

Vì \(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\ne0\)

=> x+1 =0

=> x =-1

THực hiện phép tính:
  117 : [ 2 x ( 4\(^2\)-9 ) + 3\(^2\). ( 15 - 10 ) ]

= 117 : [ 2 . ( 16 - 9 ) + 9 . 5]

= 117 : 2 . 8 + 45

= 58,5 .8 + 45

= 292,5 + 45

= 337,5 
Tk và kb hộ mình nha m.n! thanks 

Bằng 337,5 nhé bạn

Khỏi cần ngoặc nha Songoku Sky Fc11

2002 - 57 - 2002

= 2002 - 2002 - 57 = -57

k cho bạn Trần Nhật Quỳnh nha bạn ấy làm đúng rồi

2002 - 57 - 2002

= 2002 - 2002 - 57

= 0 - 57

= -57

Ta có : 

\(\frac{a}{b}< \frac{a+c}{b+c}\)

\(\Leftrightarrow\)\(a\left(b+c\right)< b\left(a+c\right)\)

\(\Leftrightarrow\)\(ab+ac< ab+bc\)

\(\Leftrightarrow\)\(ac< bc\)

\(\Leftrightarrow\)\(a< b\)

\(\Leftrightarrow\)\(\frac{a}{b}< 1\) ( đpcm ) 

Và trường hợp này chỉ xảy ra khi \(\frac{a}{b}< 1\) và \(a,b,c\inℕ^∗\)

Chúc bạn học tốt ~ 

Theo đề bài ta có:

\(\frac{a}{b}< \frac{a+c}{b+c}\)

\(\Rightarrow a\left(b+c\right)< b\left(a+c\right)\)

\(\Rightarrow ab+ac< ab+bc\)

\(\Rightarrow ac< bc\)

\(\Rightarrow a< b\)

Vậy nếu \(\frac{a}{b}< 1\)thì \(\frac{a}{b}< \frac{a+c}{b+c}\)( ĐPCM )

P/s: ĐPCM: Điều phải chứng minh

a)\(\dfrac{25}{42}-\dfrac{20}{36}=\dfrac{900}{1512}-\dfrac{840}{1512}=\dfrac{60}{1512}=\dfrac{5}{126}\)

b)\(\dfrac{9}{50}-\dfrac{13}{75}-\dfrac{1}{6}=\dfrac{4050}{22500}-\dfrac{3900}{22500}-\dfrac{3750}{22500}=\dfrac{-3600}{22500}=\dfrac{-4}{25}\)

\(1^{21719248263531785614752315761240587704250}=1\)

goood!