\(187-\left\{\left[497-\left(8x+11\right):x\right]:3-78\right\}=150\)

\(\left[497-\left(8x+11\right):x\right]:3-78=37\)

\(\left[497-\left(8x+11\right):x\right]:3=115\)

\(497-\left(8x+11\right):x=345\)

\(8+\frac{11}{x}=152\)

\(\frac{11}{x}=144\)

\(x=\frac{11}{144}\)

x . 3 + 497 = 498,02

=> 3x = 498,02 - 497

=> 3x = 1,02

=> x = 1,02 : 3

=> x = 0,34

~Study well~

#Thạc_Trân

X*3 + 497 = 498,02

X*3          = 498,02 - 497

X*3          = 1,02

X             = 1,02 : 3

X             = 0,34

 Vậy x = 0,34
 

 X * 3 + 497 = 498,02

 X * 3         = 498,02 -497

 X * 3         = 1,02

X               =1,02:3

X              =0,34

Vậy X=0,34

5672-97-5672

=5672-5672-97

=0-97

=-97

(5672-97)-5672

= 5672 -97 - 5672

= ( 5672 - 5672)  - 97

= 0 - 97

= - 97

(5672-97)-5672

=5672-97-5672

=(5672-5672)-97

=0-97

=-97

Bài 1:

Ta có: \(\frac{497}{-499}=-\frac{497}{499}>-\frac{499}{499}=-1\left(1\right)\)

\(-\frac{2345}{2341}< -\frac{2341}{2341}=-1\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\frac{497}{-499}>-\frac{2345}{2341}\)

Bài 2:

\(\frac{x+5}{2005}+\frac{x+6}{2004}=\frac{x+7}{2003}+3=0\)

\(\Rightarrow\frac{x+5}{2005}+\frac{x+6}{2004}+\frac{x+7}{2003}+3=0\)

\(\Rightarrow\frac{x+5}{2005}+1+\frac{x+6}{2004}+1+\frac{x+7}{2003}+1=0\)

\(\Rightarrow\frac{x+2010}{2005}+\frac{x+2010}{2004}+\frac{x+2010}{2003}=0\)

\(\Rightarrow\left(x+2010\right)\times\left(\frac{1}{2005}+\frac{1}{2004}+\frac{1}{2003}\right)=0\)

Vì \(\left(\frac{1}{2005}+\frac{1}{2004}+\frac{1}{2003}\right)\ne0\Rightarrow x+2010=0\)

\(\Rightarrow x=0-2010=-2010\)

Vậy x = -2010

Đáp án: 26838