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

a. |x + 8| = 6

TH1: x + 8 = -6

x = -14

TH2: x + 8 = 6

x = -2

b. 1 < |x - 2| < 4

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

TH1: |x - 2| = 2

\(\Leftrightarrow\orbr{\begin{cases}x-2=-2\\x-2=2\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=4\end{cases}}\)

TH2: |x - 2| = 3

\(\Leftrightarrow\orbr{\begin{cases}x-2=-3\\x-2=3\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=5\end{cases}}\)

c. |x - a| = a

TH1: x - a = -a

x = 0

TH2: x - a = a

x = 2a

a, =) x+6= -6 hc 6 =) +) x+6=6=)x=0                                                                                                                                                                                                        +) x+6= -6 =)x=-1                                                                                                                                                            mk lm dc mỗi í a thôi à :) chúc pn hok tốt

= 3 nữa nha

a)|2x-5|=13

2x-5=13=>x=9

2x-5=-13=>x=-4

b)3|x+1|+1=28

3|x+1|=28-1

3|x+1|=27

|x+1|=27:3

|x+1|=9

x+1=9=>x=8

x+1=-9=>x=-10

tick nha

a)(x+1)+(x+3)+...+(x+97)+(x+99)=0

x.50+2500=0

x.50=0-2500

x.50=-2500

x=-2500:5

x=-500

\(1)|5-2x|=|x+4|\)

\(\Leftrightarrow\orbr{\begin{cases}5-2x=x+4\\5-2x=-x-4\end{cases}\Leftrightarrow\orbr{\begin{cases}-2x-x=4-5\\-2x+x=-4-5\end{cases}\Leftrightarrow}\orbr{\begin{cases}-3x=-1\\-x=-9\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{1}{3}\\x=9\end{cases}}}\)

Vậy \(x=\frac{1}{3};x=9\)

\(2)|x-1|=|2x+5|\)

\(\Leftrightarrow\orbr{\begin{cases}x-1=2x+5\\x-1=-2x-5\end{cases}\Leftrightarrow\orbr{\begin{cases}x-2x=5+1\\x+2x=-5+1\end{cases}\Leftrightarrow}\orbr{\begin{cases}-x=4\\3x=-4\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-4\\x=-\frac{4}{3}\end{cases}}}\)

Vậy \(x=-4;x=-\frac{4}{3}\)

\(3)|x+1|+|x+2|+|x+3|=0\left(1\right)\)

Ta có: \(|x+1|\ge0\forall x;|x+2|\ge0\forall x;|x+3|\ge0\forall x\)

\(\Leftrightarrow|x+1|+|x+2|+|x+3|\ge0\forall x\)

\(\left(1\right)\Leftrightarrow|x+1|+|x+2|+|x+3|=0\)

\(\Leftrightarrow\left(x+1\right)+\left(x+2\right)+\left(x+3\right)=0\)

\(\Leftrightarrow x+1+x+2+x+3=0\)

\(\Leftrightarrow\left(x+x+x\right)+\left(1+2+3\right)=0\)

\(\Leftrightarrow3x+6=0\)

\(\Leftrightarrow3x=-6\)

\(\Leftrightarrow x=-6:3\)

\(\Leftrightarrow x=-2\)

Vậy x=-2

1: x = 1/3 , x=9

2: x = 4 , x = -4/3

3: x=2

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............................

a) ĐKXĐ : \(x\ne0;x\ne5\)

b) \(A=\frac{x}{x-5}-\frac{10}{x}-\frac{75}{5x-x^2}\)

\(A=\frac{x}{x-5}-\frac{10}{x}+\frac{75}{x^2-5x}\)

\(A=\frac{x^2-10\left(x-5\right)+75}{x\left(x-5\right)}=\frac{x^2-10x+25}{x\left(x-5\right)}=\frac{\left(x-5\right)^2}{x\left(x-5\right)}=\frac{x-5}{x}\)

c) \(A=2\Leftrightarrow\frac{x-5}{x}=2\Leftrightarrow x-5=2x\Leftrightarrow x=-5\) (t/m)

d) \(A=\frac{x-5}{x}=1-\frac{5}{x}\)

\(A\in Z\Leftrightarrow\) \(x\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\). Kết hợp đkxđ => \(x\in\left\{-5;-1;1\right\}\) tmđb

đề là gì vậy bạn?

\(\left(x-2\right)\left(x^2+2x+4\right)\)

\(=\left(x-2\right)\left(x^2+x.2+2^2\right)\)

\(=x^3-2^3\)

\(\text{ -2005 < l x +5 l _< 1}\text{ -2005 < l x +5 l _< 1}\) -2005 < l x +5 l \(\le\)1

xét  l x +5 l \(\ge\)0

mà theo đề bài thì l x +5 l \(\le\)1

nên  l x +5 l = 1 hoặc 0

nếu  l x +5 l  = 1

=)   x +5 = 1

=) x = 1 - 5 = -4

nếu  l x +5 l = 0

=)   x +5 = 0

=) x = 0 - 5 = -5

=) \(x\in\left\{-5;-4\right\}\) 

đăng kí kênh của V-I-S nha !

2)

xy - 10 + 5x - 2y = -115

x(y + 5) - 2(y + 5) = -115

(x - 2)(y + 5) = -115

\(\Rightarrow\left(x-2\right);\left(y+5\right)\inƯ\left(115\right)=\left\{1;5;23;115\right\}\)

Vì \(x;y\in Z\)nên ta xét bảng sau:

tới đây dễ rồi nha, bạn tự làm

bạn phước lộc thiếu giá trị 2 âm nữa nha .