Phương pháp giải

Phân tích số đã cho thành tổng số chục và đơn vị.

Lời giải chi tiết:

57 = 50 + 7

98 = 90 + 8

61 = 60 + 1

74 = 70 + 4

47 = 40 + 7

\(x-\left\{57-\left[42+\left(-23-x\right)\right]\right\}=13-\left\{47+\left[25-\left(32-x\right)\right]\right\}\)

=>\(x-\left\{57-42+23+x\right\}=13-47-25+32-x\)

=>\(x-\left(x+38\right)=-27-x\)

=>\(-27-x=-38\)

=>x+27=38

=>\(x=38-27=11\)

\(x-\left\{57-\left[42+\left(-23-x\right)\right]\right\}=13-\left\{47+\left[25-\left(32-x\right)\right]\right\}\)

\(\Leftrightarrow x-57+42-23-x=13-47-25+32-x\)

\(\Leftrightarrow x-x-57+42-23=-27-x\)

\(\Leftrightarrow-57+42-23=-27-x\)

\(\Leftrightarrow-38=-27-x\)

\(\Leftrightarrow x=\left(-27\right)-\left(-38\right)\)

\(\Leftrightarrow x=11\)

Vậy \(x=11\) là giá trị cần tìm

\(x-\left(57-\left(42+\left(-23-x\right)\right)\right)=13-\left(47+\left(25-\left(32-x\right)\right)\right)\)

\(\Leftrightarrow\) \(x-\left(57-\left(42-23-x\right)\right)=13-\left(47+\left(25-32+x\right)\right)\)

\(\Leftrightarrow\) \(x-\left(57-42+23+x\right)=13-\left(47+25-32+x\right)\)

\(\Leftrightarrow\) \(x-57+42-23-x=13-47-25+32-x\)

\(\Leftrightarrow\) \(-38=-27-x\) \(\Leftrightarrow\) \(-x=-38+27=-11\Leftrightarrow x=11\)

vậy \(x=11\)

 x-(57-(42+(-23-x)=13-47+25-23-x

 x-57-42-23-x=13-47+25-23-x

 x-122-x=-32-x

 x-x-122=-32-x

 -32-x=-122

        x=-32-(-122)

        x=90

 Vậy x=90

\(x-(57-(42+(-23-x)=13-47+25-13-x\)

\(\Rightarrow x-57-42+\left(-23\right)-x=13-47+25-23-x\)

\(\Rightarrow x-x+x=13-47+25-23+23+42+57\)

\(\Rightarrow x=90\)

\(x-\left\{57-\left[42+\left(-23-x\right)\right]\right\}=13-\left\{47+\left[25-\left(32-x\right)\right]\right\}\)

\(\Leftrightarrow x-57+42-23-x=13-47-25+32-x\)

\(\Leftrightarrow x-x-57+42-23=-27-x\)

\(\Leftrightarrow-57+42-23=-27-x\)

\(\Leftrightarrow-38=-27-x\)

\(\Leftrightarrow x=-27-\left(-38\right)\)

\(\Leftrightarrow x=11\left(TM\right)\)

Vậy \(x=11\) là giá trị cần tìm

\(\Leftrightarrow\) \(x-\left(57-\left(42-23-x\right)\right)=13-\left(47+\left(25-32+x\right)\right)\)

\(\Leftrightarrow\) \(x-\left(57-42+23+x\right)=13-\left(47+25-32+x\right)\)

\(\Leftrightarrow\) \(x-57+42-23-x=13-47-25+32-x\)

\(\Leftrightarrow\) \(-38=-27-x\) \(\Leftrightarrow\) \(-x=-38+27=-11\Leftrightarrow x=11\)

vậy \(x=11\)

Lời giải:

$x-\frac{31}{15}.\frac{26}{47}=\frac{-57}{15}.\frac{31}{47}-\frac{31}{15}:\frac{-141}{31}$

$x=\frac{-57}{47}.\frac{31}{15}-\frac{31}{15}.\frac{-31}{141}+\frac{31}{15}.\frac{26}{47}$

$x=\frac{31}{15}(\frac{-57}{47}-\frac{-31}{141}+\frac{26}{47})$

$x=\frac{31}{15}.\frac{-62}{141}=\frac{-1922}{2115}$

\(a,\text{A nguyên }\Leftrightarrow2n+7⋮n+2\)

\(\text{Ta có : }n+2⋮n+2\)

\(\Rightarrow\left(2n+7\right)-\left(n+2\right)⋮n+2\)

\(\Rightarrow2n+7-n-2⋮n+2\)

\(\Rightarrow5⋮n+2\)

\(\Rightarrow n+2\in\text{Ư}\left(5\right)=\left\{1;-1;5;-5\right\}\)

\(\Rightarrow n\in\left\{-1;-3;3;-7\right\}\)

\(b,\text{Ta có : }A=\frac{2n+7}{n+2}\)

\(=\frac{n+2+n+2+3}{n+2}\)

\(=2+\frac{3}{n+2}\)

A tối giản <=> ƯCLN ( 2n + 7; n + 2 )  = 1

\(\Leftrightarrow\text{Ư}CLN\left(3;n+2\right)=1\)

\(\Leftrightarrow n+2\) \(⋮̸\)\(3\)

\(\Leftrightarrow n+2\ne3d\left(d\in N\right)\)

\(\Leftrightarrow n\ne3d-2\left(d\in N\right)\)

Vậy .... 

=>x-57+42-23-x=13-47-25+32-x

=>x-x-57+42-23=-27-x

=>-57+42-23=-27-x

=>-38=-27-x

=>x=-27-(-38)

=>x=11

x-57+42-23-x=13-47-25+32-x

x=13-47-25+32+23-42+57

x=11

tick nha bạn