a) Ta có: \(\left|x+\dfrac{19}{5}\right|\ge0\forall x\in Q\)

\(\left|y+\dfrac{2017}{2018}\right|\ge0\forall y\in Q\)

\(\left|z-2019\right|\ge0\forall x\in Q\)

\(\Rightarrow\left|x+\dfrac{19}{5}\right|+\left|y+\dfrac{2017}{2018}\right|+\left|z-2019\right|\ge0\forall x,y,z\in Q\)

Dấu \("="\) xảy ra khi \(\left\{{}\begin{matrix}\left|x+\dfrac{19}{5}\right|=0\\\left|y+\dfrac{2017}{2018}\right|=0\\\left|z-2019\right|=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{-19}{5}\\y=\dfrac{-2017}{2018}\\z=2019\end{matrix}\right.\).

b) Lại có:

\(\left|x-\dfrac{9}{5}\right|\ge0\forall x\in Q\)

\(\left|y+\dfrac{3}{4}\right|\ge0\forall y\in Q\)

\(\left|z+\dfrac{7}{2}\right|\ge0\forall z\in Q\)

\(\Rightarrow\left|x-\dfrac{9}{5}\right|+\left|y+\dfrac{3}{4}\right|+\left|z+\dfrac{7}{2}\right|\ge0\forall x,y,zQ\)

Mà theo đề bài:

\(\left|x-\dfrac{9}{5}\right|+\left|y+\dfrac{3}{4}\right|+\left|z+\dfrac{7}{2}\right|\le0\forall\)

\(\Rightarrow\left|x-\dfrac{9}{5}\right|+\left|y+\dfrac{3}{4}\right|+\left|z+\dfrac{7}{2}\right|=0\)

\(\Rightarrow\left\{{}\begin{matrix}\left|x-\dfrac{9}{5}\right|=0\\\left|y+\dfrac{3}{4}\right|=0\\\left|z+\dfrac{7}{2}\right|=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{9}{5}\\y=\dfrac{-3}{4}\\z=\dfrac{-7}{2}\end{matrix}\right.\)

Vậy .....

a) \(\left|x+\dfrac{19}{5}\right|+\left|y+\dfrac{2017}{2018}\right|+\left|z-2019\right|=0\)

Ta có: \(\left|x+\dfrac{19}{5}\right|\ge0;\left|y+\dfrac{2017}{2018}\right|\ge0;\left|z-2019\right|\ge0\)

Để \(\left|x+\dfrac{19}{5}\right|+\left|y+\dfrac{2017}{2018}\right|+\left|z-2019\right|=0\) thì:

\(\left\{{}\begin{matrix}\left|x+\dfrac{19}{5}\right|=0\\\left|y+\dfrac{2017}{2018}\right|=0\\\left|z-2019\right|=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{-19}{5}\\y=\dfrac{-2017}{2018}\\z=2019\end{matrix}\right.\)

Vậy............................

b) Ta có: \(\left|x-\dfrac{9}{5}\right|\ge0;\left|y+\dfrac{3}{4}\right|\ge0;\left|z+\dfrac{7}{2}\right|\ge0\)

Mà \(\left|x-\dfrac{9}{5}\right|+\left|y+\dfrac{3}{4}\right|+\left|z+\dfrac{7}{2}\right|\le0\) thì:

\(\left|x-\dfrac{9}{5}\right|=\left|y+\dfrac{3}{4}\right|=\left|z+\dfrac{7}{2}\right|=0\)

\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{9}{5}\\y=\dfrac{-3}{4}\\z=\dfrac{-7}{2}\end{matrix}\right.\)

Vậy............................

d, \(\left(3x-2^4\right).7^3=2.7^4\)

\(\Rightarrow3x-2^4=2.7^4:7^3\)

\(\Rightarrow3x-16=2.7\\ \Rightarrow3x=14+16\\ \Rightarrow3x=30\Rightarrow x=10\)

Vậy.....

e, \(x-\left[42+\left(-28\right)\right]=-8\)

\(\Rightarrow x-14=-8\\ \Rightarrow x=6\)

Vậy.....

g, \(x-7=-5\)

\(\Rightarrow x=-5+7\Rightarrow x=2\)

Vậy.....

h, \(15-5\left(x+4\right)=-12-3\)

\(\Rightarrow15-5x-20=-15\)

\(\Rightarrow-5x=-15-15+20\)

\(\Rightarrow-5x=-10\Rightarrow x=2\)

Vậy.....

Chúc bạn học tốt!!!

d/ \(\left(3x-2^4\right)\cdot7^3=2\cdot7^4\)

\(\Rightarrow3x-16=\dfrac{2\cdot7^4}{7^3}=14\)

\(\Rightarrow3x=14+16=30\)

\(\Rightarrow x=\dfrac{30}{3}=10\)

e/ Đễ ==> tự lm thì tốt hơn nhé

g/ Đễ ==> tự lm thì tốt hơn nhé

h/ \(15-5\left(x+4\right)=-12-3\)

\(\Rightarrow15-5x-20=-15\)

\(\Rightarrow-5x=-15+20-15=-10\)

\(\Rightarrow x=\dfrac{-10}{-5}=2\)

i/ \(\left(7-x\right)-\left(25+7\right)=-25\)

\(\Rightarrow7-x-25-7=-25\)

\(\Rightarrow-x=-25-7+7+25\)

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

k/ \(\left|x+2\right|=0\Rightarrow x+2=0\Rightarrow x=-2\)

l/ \(\left|x-3\right|=7-\left(-2\right)\)

\(\Rightarrow\left|x-3\right|=9\)

\(\Rightarrow\left[{}\begin{matrix}x-3=9\\x-3=-9\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=12\\x=-6\end{matrix}\right.\)

m/ \(\left|x-5\right|=\left|-7\right|\Rightarrow\left|x-5\right|=7\)

\(\Rightarrow\left[{}\begin{matrix}x-5=7\\x-5=-7\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=12\\x=-2\end{matrix}\right.\)

i, \(\left(7-x\right)-\left(25+7\right)=-25\)

\(\Rightarrow7-x-25-7=-25\)

\(\Rightarrow-x=-25-7+25+7\)

\(\Rightarrow x=0\)

Vậy.....

k, \(\left|x+2\right|=0\)

\(\Rightarrow x+2=0\Rightarrow x=-2\)

Vậy.....

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

\(\Rightarrow\left|x-3\right|=-5\)

Với mọi giá trị của \(x\in Z\) ta có:

\(\left|x-3\right|\ge0\) mà \(-5< 0\) nên không tìm được giá trị nào của x thoả mãn \(\left|x-3\right|=-5\).

Vậy \(x\in\varnothing\)

m, \(\left|x-5\right|=\left|-7\right|\)

\(\Rightarrow\left|x-5\right|=7\\ \Rightarrow\left\{{}\begin{matrix}x-5=-7\\x-5=7\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=-2\\x=12\end{matrix}\right.\)

Vậy...,..

Chúc bạn học tốt!!!

Ta có |x+25|+|-25+y|=0

Mà \(\left|x+25\right|\ge0\forall x;\left|-y+25\right|\ge0\forall y\)

=> x+25=-y+25=0 => x=-25;y=25

Ta co : \(\left|x+25\right|\ge0\forall x\in Z\)

             \(\left|-y+5\right|\ge0\forall x\in Z\)

Mà : |x + 25| + |-y + 5| = 0 

Nên : \(\hept{\begin{cases}\left|x+25\right|=0\\\left|-y+5\right|=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x+25=0\\-y+5=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=-25\\y=5\end{cases}}\)

các bạn ơi giúp mình đi mà

a) X=-25; Y=5

Câu 2: 

a: Để A là số nguyên thì \(x-1\in\left\{1;-1;3;-3\right\}\)

hay \(x\in\left\{2;0;4;-2\right\}\)

b: Để -4/2x-1 là số nguyên thì \(2x-1\in\left\{1;-1;2;-2;4;-4\right\}\)

hay \(x\in\left\{1;0;\dfrac{3}{2};-\dfrac{1}{2};\dfrac{5}{2};-\dfrac{3}{2}\right\}\)

c: Để 3x+7/x-1là số nguyên thì \(3x-3+10⋮x-1\)

\(\Leftrightarrow x-1\in\left\{1;-1;2;-2;5;-5;10;-10\right\}\)

hay \(x\in\left\{2;0;3;-1;6;-4;11;-9\right\}\)

d: Để 4x-1/x-3 là số nguyên thì \(4x-12+11⋮x-3\)

\(\Leftrightarrow x-3\in\left\{1;-1;11;-11\right\}\)

hay \(x\in\left\{4;2;14;-8\right\}\)

a) x=-2

b) x=12; x=-2

c) x=12; x=-6

Lắm phần c,d , b quá

15 chia hết cho 2x+1 thì x= 1, x=4 và x=7 (nếu cả số âm nữa thì tự tìm nhé)

10 chia hết cho 3x+1 thì x=0, x=3 (nếu cả số âm nữa thì tự tìm nhé)

(7-x)-(25+7)=25 thì x=-36

6 chi hết cho x-1 thì x=2: x=3: x=4: x=7 (nếu cả số âm nữa thì tự tìm nhé)

5 chia hết cho x+1 thì x=0; x=4  (nếu cả số âm nữa thì tự tìm nhé)

e) x=0: x=1: x=3: x=9

f) x=1

g) x=0: x=2; x=4; x=14

z) x=0: x=1: x=4: x=9

vai cut