a) Để A là số nguyên => 7 chia hết cho x-3 => x-3  thuộc Ư(7)= { 1;-1;7; -7}

                           Vậy để A nguyên x thuộc { 10; 4;2;-4} 

b) Để B là số nguyên => x+2 chia hết cho x-1 

                                     <=> (x-1) + 3 chia hết cho x-1

                                      vì x-1 chia hết cho x-1

                                  => 3 chia hết cho x-1 ( tính chất chia hết một tổng)

                                  => x-1 thuộc Ư(3)= { 1; -1;3;-3}

Ta có bảng sau :

                                                   Vậy x thuộc { 0; 2;-2;4}

a, \(2.\left|2.x-3\right|=\frac{1}{2}\)

\(\left|2.x-3\right|=\frac{1}{2}:2\)

\(\left|2.x-3\right|=\frac{1}{4}\)

\(\Rightarrow2.x-3=\frac{1}{4}\)va \(2.x-3=\frac{-1}{4}\)

\(2.x=\frac{1}{4}+3=\frac{13}{4}\)va \(2.x=\frac{-1}{4}+3=\frac{11}{4}\)

\(x=\frac{13}{4}:2\)va \(x=\frac{11}{4}:2\)

vay \(x\in\left\{\frac{13}{8};\frac{11}{8}\right\}\)

b, \(7,5-3.\left|5-2.x\right|=-4.5\)

\(\frac{3}{4}-3.\left|5-2.x\right|=\frac{-9}{2}\)

\(3.\left|5-2.x\right|=\frac{3}{4}-\frac{-9}{2}\)

\(3.\left|5-2.x\right|=\frac{21}{4}\)

\(\left|5-2.x\right|=\frac{21}{4}:3\)

\(\left|5-2.x\right|=\frac{7}{4}\)

\(\Rightarrow5-2.x=\frac{7}{4}\)va \(5-2.x=\frac{-7}{4}\)

\(2.x=5-\frac{7}{4}=\frac{13}{4}\)va \(2.x=5-\frac{-7}{4}=\frac{27}{4}\)

\(\Rightarrow x\in\left\{\frac{13}{8};\frac{27}{8}\right\}\)

c, \(\left|3.x-4\right|=\left|3.y+5\right|=0\)

\(\Rightarrow3.x-4=0\)va \(3.y+5=0\)

\(3.x=4\)va \(3.y=-5\)

\(\Rightarrow x\in\left\{\frac{4}{3};\frac{-5}{3}\right\}\)

cảm ơn

\(7,5-3\left|5-2x\right|=4,5\)

\(\Rightarrow3\left|5-2x\right|=12\)

\(\Rightarrow\left|5-2x\right|=4\)

\(\Rightarrow\orbr{\begin{cases}5-2x=4\\5-2x=-4\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}2x=1\\2x=9\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=\frac{9}{2}\end{cases}}\)

\(2\left|2x-3\right|=\frac{1}{2}\)

\(\Rightarrow\left|2x-3\right|=\frac{1}{4}\)

\(\Rightarrow\orbr{\begin{cases}2x-3=\frac{1}{4}\\2x-3=\frac{-1}{4}\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}2x=\frac{13}{4}\\2x=\frac{11}{4}\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=\frac{13}{8}\\x=\frac{11}{8}\end{cases}}\)

\(\left|3x-4\right|+\left|3y-5\right|=0\)

Ta có: \(\hept{\begin{cases}\left|3x-4\right|\ge0\\\left|3y-5\right|\ge0\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}\left|3x-4\right|=0\\\left|3y-5\right|=0\end{cases}\Rightarrow\hept{\begin{cases}3x-4=0\\3y-5=0\end{cases}}}\Rightarrow\hept{\begin{cases}3x=4\\3y=5\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{4}{3}\\y=\frac{5}{3}\end{cases}}\)

\(\left(x+\frac{3}{20}\right)^4=\left(\frac{4}{15}\right)^4\)

\(\Rightarrow x+\frac{3}{20}=\frac{4}{15}\)

\(x=\frac{4}{15}-\frac{3}{20}\)

\(x=\frac{1}{12}\)

---> 2x = 5/2 + 1/2

---> 2x = 3

---> x = 3/2

xin tiick

\(\rightarrow\)2x = \(\frac{5}{2}\) + \(\frac{1}{2}\)

\(\rightarrow\)2x = 3

\(\rightarrow\) \(\frac{3}{2}\)

          ~ Hok Tốt ~

63=3².7

14=2.7

70=2.5.7

=> ƯCLN (63;14;70) = 7

\(A=\left|x\right|+x\)

Trường hợp 1: \(x\ge0\Rightarrow\left|x\right|=x\Rightarrow A=x+x=2x\)

Trường hợp 2: \(x< 0\Rightarrow\left|x\right|=-x\Rightarrow A=-x+x=0\)

\(B=\left|x\right|-x\)

Trường hợp 1: \(x\ge0\Rightarrow\left|x\right|=x\Rightarrow A=x-x=0\)

Trường hợp 2: \(x< 0\Rightarrow x=\left|-x\right|\Rightarrow A=-x-x=-2x\)

\(C=\left|x-1\right|+x\)

Nếu: \(\orbr{\begin{cases}\left|x-1\right|=x-1\Rightarrow x-1\ge0hayx\ge1\\\left|x-1\right|=-\left(x-1\right)=-x+1\Rightarrow x-1< 0hayx< 1\end{cases}}\)

Trường hợp 1: \(x\ge1\Rightarrow\left|x-1\right|+x=x-1+x=2x-1\)

Trường hợp 2: \(x< 1\Rightarrow\left|x-1\right|+x=-x+1+x=1\)

x = 2,(142857) = 15/7

xin tiick

x=15/7