\(10^n\)có 1 chữ số 1 và n chữ số 0 nên tổng các chữ số của \(10^n+8\)bằng 9, do vậy nó chia hết cho 9

a) \(\left|3x-\frac{1}{2}\right|+\left|\frac{1}{2}y+\frac{3}{5}\right|=0\)

\(\Rightarrow\left|3x-\frac{1}{2}\right|=0\)                                \(\Rightarrow\left|\frac{1}{2}y+\frac{3}{5}\right|=0\)

\(\Rightarrow3x-\frac{1}{2}=0\)                                      \(\Rightarrow\frac{1}{2}y+\frac{3}{5}=0\)

\(3x=\frac{1}{2}\)                                                          \(\frac{1}{2}y=\frac{-3}{5}\)

\(x=\frac{1}{2}:3\)                                                             \(y=\left(\frac{-3}{5}\right):\frac{1}{2}\)

\(x=\frac{1}{6}\)                                                                  \(y=\frac{-6}{5}\)

KL: x = 1/6; y = -6/5

b) \(\left|\frac{3}{2}x+\frac{1}{9}\right|+\left|\frac{1}{5}y-\frac{1}{2}\right|\le0\)

mà \(\left|\frac{3}{2}x+\frac{1}{9}\right|>0;\left|\frac{1}{5}y-\frac{1}{2}\right|>0\)

\(\Rightarrow\left|\frac{3}{2}x+\frac{1}{9}\right|+\left|\frac{1}{5}y-\frac{1}{2}\right|>0\)

=> trường hợp |3/2x +1/9| + |1/5y -1/2| < 0 không thế xảy ra

\(\Rightarrow\left|\frac{3}{2}x+\frac{1}{9}\right|+\left|\frac{1}{5}y-\frac{1}{2}\right|=0\)

rùi bn lm tương tự như phần a nhé!

a) \(\frac{3}{7}x-\frac{1}{35}=\frac{3}{5}\)

\(\frac{3}{7}x=\frac{3}{5}+\frac{1}{35}\)

\(\frac{3}{7}x=\frac{22}{35}\)

\(x=\frac{49}{35}=1,4\)

b) \(1,5-x:\frac{1}{2}=\frac{1}{4}\)

\(x:\frac{1}{2}=1,5-\frac{1}{4}\)

\(x:\frac{1}{2}=\frac{5}{4}\)

\(x=\frac{5}{4}.\frac{1}{2}\)

\(x=\frac{5}{8}\)

Vậy ..

Nhìu thế

Câu a:

7.(\(x-4\)) = \(7^4:7^3\)

7.(\(x-4\) )=7

(\(x-4\) )=7 : 7

(\(x-4\) )=1

\(x\) = 1+4

\(x\) = 5

Vậy \(x=5\)

Giải:

Vì:

\(\left\{{}\begin{matrix}\left|3x-\dfrac{1}{2}\right|\ge0\\\left|\dfrac{1}{2}y+\dfrac{3}{5}\right|\ge0\end{matrix}\right.\)

Nên dấu "=" xảy ra khi và chỉ khi:

\(\left\{{}\begin{matrix}\left|3x-\dfrac{1}{2}\right|=0\\\left|\dfrac{1}{2}y+\dfrac{3}{5}\right|=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}3x-\dfrac{1}{2}=0\\\dfrac{1}{2}y+\dfrac{3}{5}=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}3x=\dfrac{1}{2}\\\dfrac{1}{2}y=-\dfrac{3}{5}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{6}\\y=-\dfrac{6}{5}\end{matrix}\right.\)

Vậy ...

b) \(\left|\dfrac{3}{2}x+\dfrac{1}{9}\right|+\left|\dfrac{1}{5}y-\dfrac{1}{2}\right|\le0\)

Vì:

\(\left\{{}\begin{matrix}\left|\dfrac{3}{2}x+\dfrac{1}{9}\right|\ge0\\\left|\dfrac{1}{5}y-\dfrac{1}{2}\right|\ge0\end{matrix}\right.\)

Dấu "=" xảy ra, khi và chỉ khi:

\(\left\{{}\begin{matrix}\left|\dfrac{3}{2}x+\dfrac{1}{9}\right|=0\\\left|\dfrac{1}{5}y-\dfrac{1}{2}\right|=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3}{2}x+\dfrac{1}{9}=0\\\dfrac{1}{5}y-\dfrac{1}{2}=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3}{2}x=-\dfrac{1}{9}\\\dfrac{1}{5}y=\dfrac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{2}{27}\\y=\dfrac{5}{2}\end{matrix}\right.\)

Vậy ...