a ) 4 . ( x² + 1 ) = 0

            x² + 1   = 0 : 4

            x² + 1   = 0

                   x²  = 0 - 1

                   x²  = - 1

                   x²  = - 1² => x = - 1

Vậy x = - 1

Thế còn phần b

b ) - 2018 . ( X + 2019 ) = 0²⁰²⁰

    - 2018 . ( x + 2019 )  = 0

                   x + 2019   = 0 : ( - 2018 )

                   x + 2019   = 0

                              x   = 0 - 2019

                               x  = - 2019

Vậy x = - 2019

\(\left(24-4y\right)^{2018}+\left|x^2-4\right|^{2019}\le0\left(1\right)\)

Vì \(\hept{\begin{cases}\left(24-4y\right)^{2018}\ge0;\forall x,y\\\left|x^2-4\right|^{2019}\ge0;\forall x,y\end{cases}}\)\(\Rightarrow\left(24-4y\right)^{2018}+\left|x^2-4\right|^{2019}\ge0;\forall x,y\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\hept{\begin{cases}\left(24-4y\right)^{2018}=0\\\left|x^2-4\right|^{2019}=0\end{cases}}\)

                      \(\Leftrightarrow\hept{\begin{cases}y=6\\x=\pm2\end{cases}}\)

Vậy \(\left(x,y\right)\in\left\{\left(2;6\right);\left(-2;6\right)\right\}\)

có sai đề ko bạn ? mik thấy nó ko theo quy luật nào hết

Hình như đề sai rồi bn ơi
Mk đoán như z mới đúng:

x + ( x + 1 ) + ( x + 2 ) + ... + ( x + 2018 ) + ( x + 2019 ) = 2019

Hình như là sai đề rồi .Mình xin lỗi

a) \(\left(x+1\right)+\left(x+3\right)+\left(x+5\right)+..+\left(x+99\right)=0\)

Tổng các số hạng là;

\(\left(99+1\right):2=50\)(số hạng)

\(\left(x+1\right)+\left(x+3\right)+\left(x+5\right)+..+\left(x+99\right)=0\)

\(\Leftrightarrow50x+\left(1+3+..+99\right)=0\)

\(\Leftrightarrow50x+\frac{\left(99+1\right).50}{2}=0\)

\(\Leftrightarrow50x+2500=0\)

\(\Leftrightarrow50x=-2500\)

\(\Leftrightarrow x=\frac{-2500}{50}=-50\)

b) \(\left(x-3\right)+\left(x-2\right)+\left(x-1\right)+..+10+11=11\)

\(\left(x-3\right)+\left(x-2\right)+\left(x-1\right)+..+10=0\)

gọi các số hạng từ ( x-3) đến 10 là n 

Ta có; \(\left[10+\left(x-3\right)\right].n:2=0\)

\(\Rightarrow\left(x+7\right).n=0\)

Vì \(n\ne0\)

Nên \(x+7=0\)

\(\Rightarrow x=-7\)

Ta có; \(x+\left(x+1\right)+\left(x+2\right)+..+2018+2019=2019\)

\(x+\left(x+1\right)+\left(x+2\right)+..+2018=0\)

Tổng số hạng là;

\(\left(2018-x\right):1+1=2019-x\)

Trung bình cộng: \(\frac{2018+x}{2}\)

Do đó ta đc; \(\left(2019-x\right).\frac{2018+x}{2}=0\)

\(\Rightarrow\orbr{\begin{cases}2019-x=0\\2018+x=0\end{cases}\Rightarrow\orbr{\begin{cases}x=2019\left(vl\right)\\x=-2018\end{cases}}}\)

Vậy x= -2018

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

Tổng các số hạng là: (99+1):2=50 (số hạng)

=> (x+1)+(x+3)+...+(x+99)=0 <=> 50.x+(1+3+5+...+99) = 0

<=> 50.x+\frac{\left(99+1\right).50}{2}2(99+1).50​=0 <=> 50.x+2500=0 => x=-2500/50=-50

Bài 1 :

\(3x+5=2\left(x-\frac{1}{4}\right)\)

\(\Leftrightarrow3x+5=2x-\frac{1}{2}\)

\(\Leftrightarrow5+\frac{1}{2}=2x-3x\)

\(\Leftrightarrow\frac{11}{2}=-x\)

\(\Leftrightarrow\frac{-11}{2}=x\)

Vậy \(x=\frac{-11}{2}\)

Bài 2:

a, \(\left|x+\frac{19}{5}\right|+\left|y+\frac{2018}{2019}\right|+\left|z-3\right|=0\)

Vì \(\hept{\begin{cases}\left|x+\frac{19}{5}\right|\ge0\\\left|y+\frac{2018}{2019}\right|\ge0\\\left|z-3\right|\ge0\end{cases}}\)

       Mà \(\left|x+\frac{19}{5}\right|+\left|y+\frac{2018}{2019}\right|+\left|z-3\right|=0\)

\(\Rightarrow+,\left|x+\frac{19}{5}\right|=0\)

\(\Leftrightarrow x+\frac{19}{5}=0\)

\(\Leftrightarrow x=\frac{-19}{5}\)

\(\Rightarrow+,\left|y+\frac{2018}{2019}\right|=0\)

\(\Leftrightarrow y+\frac{2018}{2019}=0\)

\(\Leftrightarrow y=\frac{-2018}{2019}\)

\(\Rightarrow+,\left|z-3\right|=0\)

\(\Leftrightarrow z-3=0\)

\(\Leftrightarrow z=3\)

Vậy \(\hept{\begin{cases}x=\frac{-19}{5}\\y=\frac{-2018}{2019}\\z=3\end{cases}}\)

b, Ta có : \(\left|x-\frac{1}{2}\right|+\left|2y+4\right|+\left|z-5\right|\ge0\)

Vì : \(\hept{\begin{cases}\left|x-\frac{1}{2}\right|\ge0\\\left|2y+4\right|\ge0\\\left|z-5\right|\ge0\end{cases}}\)

Mà : \(\left|x-\frac{1}{2}\right|+\left|2y+4\right|+\left|z-5\right|\ge0\)

\(\Rightarrow+,\left|x-\frac{1}{2}\right|\ge0\)

\(\Rightarrow x\inℚ\)

\(\Rightarrow+,\left|2y+4\right|\ge0\)

\(\Rightarrow y\inℚ\)

\(\Rightarrow+,\left|z-5\right|\ge0\)

\(\Rightarrow z\inℚ\)

Vậy chỉ cần \(\hept{\begin{cases}x\inℚ\\y\inℚ\\z\inℚ\end{cases}}\)thì thỏa mãn.

234*(-26)+134*26

a)\(2019-\left|x-2019\right|=x\)

\(\Rightarrow2019-x=\left|x-2019\right|\)

=>\(\left|x-2019\right|=-\left(x-2019\right)\)

=>\(x-2019\le0\)

=>\(x\le2019\)

b) Vì \(\left(2x-1\right)^{2018}\ge0\forall x\)

        \(\left(y-\frac{2}{5}\right)^{2018}\ge0\forall y\)

\(\left|x+y-z\right|\ge0\forall x,y,z\)

=> \(\left(2x-1\right)^{2018}+\left(y-\frac{2}{5}\right)^{2018}\)\(+\left|x+y-z\right|\ge0\forall x,y,z\)

mà \(\left(2x-1\right)^{2018}+\left(y-\frac{2}{5}\right)^{2018}\)\(+\left|x+y-z\right|=0\)

\(\Leftrightarrow\hept{\begin{cases}2x-1=0\\y-\frac{2}{5}=0\\x+y-z=0\end{cases}}\)=>\(\Leftrightarrow\hept{\begin{cases}x=\frac{1}{2}\\y=\frac{2}{5}\\z=\frac{9}{10}\end{cases}}\)

a, Ta có:

\(\left|x-2019\right|=\orbr{\begin{cases}x-2019\ge0\Rightarrow x\ge2019\\-x+2019< 0\Rightarrow x< 2019\end{cases}}\)

Xét x<2019 thì |x-2019|=-x+2019

Khi đó: 2019-(-x+2019)=x

\(\Leftrightarrow\)-x+2019=2019-x

\(\Leftrightarrow\)-x+2019+x=2019

\(\Leftrightarrow\)0x+2019=2019

\(\Leftrightarrow\)0x=0     (thỏa mãn)

Xét 2019\(\le\)x thì |x-2019|=x-2019

Khi đó 2019-(x-2019)=x

\(\Leftrightarrow\)2019-x+2019=x

\(\Leftrightarrow\)4038-x=x

\(\Leftrightarrow\)4038=2x

\(\Leftrightarrow\)x=2019(thỏa mãn)

Vậy .......................................................!!!