\(a,\frac{2}{9}-\frac{7}{8}.x=\frac{1}{3}\)

\(\frac{7}{8}x=\frac{2}{9}-\frac{1}{3}=-\frac{1}{9}\)

\(x=-\frac{1}{9}:\frac{7}{8}=-\frac{8}{63}\)

\(b,\frac{4}{5}+\frac{5}{7}:x=\frac{1}{6}\)

\(\frac{5}{7}:x=\frac{1}{6}-\frac{4}{5}=-\frac{19}{30}\)

\(x=\frac{5}{7}:-\frac{19}{30}=-\frac{150}{133}\)

học tốt ~~~

\(a,\frac{2}{9}-\frac{7}{8}.x=\frac{1}{3}\)

\(\frac{7}{8}.x=\frac{-1}{9}\)

\(x=\frac{-8}{63}\)

\(b,\frac{4}{5}+\frac{5}{7}:x=\frac{1}{6}\)

\(\frac{5}{7}:x=\frac{-19}{30}\)

\(x=\frac{-150}{133}\)

nhanh mik k

Ta có:

|2x-1| < 5

\(\Rightarrow\)|2x| < 5 + 1 = 6

\(\Rightarrow\)|x| < 3

\(\Rightarrow\)|x| \(\in\){ 2 ; 1 ; 0 ; .................. }

\(\Rightarrow\)x \(\in\){ 2 ; -2 ; 1 ; -1 ; 0 ; .............. }

Vậy ............

CTV mà làm sai à!!~

ta có :

\(\left|a\right|\ge0\); \(\left|2x-1\right|< 5\)

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

+ \(\left|2x-1\right|=4\)

\(\Rightarrow\orbr{\begin{cases}2x-1=4\\2x-1=-4\end{cases}\Rightarrow\orbr{\begin{cases}x=2,5\\x=-1,5\end{cases}}}\) trường hợp này loại vì x \(\in\) Z

+ \(\left|2x-1\right|=3\)

\(\Rightarrow\orbr{\begin{cases}2x-1=3\\2x+1=-3\end{cases}\Rightarrow\orbr{\begin{cases}x=2\\x=-1\end{cases}}}\) (TM)

+ \(\left|2x-1\right|=2\)

\(\Rightarrow\orbr{\begin{cases}2x-1=2\\2x-1=-2\end{cases}\Rightarrow\orbr{\begin{cases}x=1,5\left(L\right)\\x=-0,5\left(L\right)\end{cases}}}\)

+ \(\left|2x-1\right|=1\)

\(\Rightarrow\orbr{\begin{cases}2x-1=1\\2x-1=-1\end{cases}\Rightarrow\orbr{\begin{cases}x=1\\x=0\end{cases}}}\)  (TM)

+ \(\left|2x-1\right|=0\)

\(\Rightarrow2x-1=0\)

\(\Rightarrow x=1,5\) Loại

Vậy x \(\in\) { 2; -1; 1; 0 }

\(A=2+2^3+2^5+2^7+2^9+...+2^{2009}\)

\(\Leftrightarrow\)\(4A=2^3+2^5+2^7+2^9+2^{11}+...+2^{2011}\)

\(\Leftrightarrow\)\(4A-A=\left(2^3+2^5+2^7+...+2^{2011}\right)-\left(2+2^3+2^5+...+2^{2009}\right)\)

\(\Leftrightarrow\)\(3A=2^{2011}-2\)

\(\Leftrightarrow\)\(A=\frac{2^{2011}-2}{3}\)

Ta có : 

\(A=2+2^3+2^5+...+2^{2009}\)

\(4A=2^3+2^5+2^7+...+2^{2011}\)

\(4A-A=\left(2^3+2^5+2^7+...+2^{2011}\right)-\left(2+2^3+2^5+...+2^{2009}\right)\)

\(3A=2^{2011}-2\)

\(A=\frac{2^{2011}-2}{3}\)

Vậy \(A=\frac{2^{2011}-2}{3}\)

Câu b) dễ hơn nữa làm tương tư câu a) nhưng B nhân cho 2 

Câu c) thì C nhân cho 5

Câu d) thì D nhân cho 169

Bài 1:

Ta có:

(4n-5) \(⋮\)(n-3)

\(\Rightarrow\)(4n-12+17) \(⋮\)(n-3)

\(\Rightarrow\)4(n-3) + 17 \(⋮\)(n-3)

Mà 4(n-3) \(⋮\)(n-3) \(\Rightarrow\)17 \(⋮\)(n-3)

\(\Rightarrow\)n-3 \(\in\)Ư(17)

Ư(17) = { 1 ; -1 ; 17 ; -17 }

\(\Rightarrow\)n-3 \(\in\){ 1 ; -1 ; 17 ; -17 }

\(\Rightarrow\)n \(\in\){ 4 ; 2 ; 20 ; -14 }

Vậy n \(\in\){ 4 ; 2 ; 20 ; -14 }

Bài 2:

a) 7270 - ( 182 + 327 )

= 7270 - 182 - 327

= 7088 - 327

= 6761

b) ( 258 - 89 ) - ( 111 - 90 )

= 258 - 89 - 111 + 90

= 169 - 111 + 90

= 58 + 90

= 148

Bài 1=2