Vì (2x-1)^2016 >= 0; (y-2/5)^2016 >= 0 ;|x+y-z| >= 0

=> (2x-1)^2016+(y-2/5)^2016+|x+y-z| >= 0

=> (2x-1)^2016=0 ; (y-2/5)^2016=0 ; |x+y-z|=0

=> 2x-1=0 ; y-2/5=0 ; x+y-z=0

=> x=1/2 ; y=2/5 ; 9/10-z=0

=> x=1/2 ; y=2/5 ; z=9/10

À bn ơi, t nhớ đề hình như là (2x-1)^2016+(y-2/5)^2016+|x+y+z| đó, bn thử kiểm tra lại đề xem viết đúng chưa. Còn cách làm cũng như vậy nha :)

x=9+2-2

x=9

\(\left(2x+1\right)^2=625\)

<=>  \(\orbr{\begin{cases}2x+1=25\\2x+1=-25\end{cases}}\)

<=>  \(\orbr{\begin{cases}x=12\\x=-13\end{cases}}\)

Vậy...

\(\left(2x+1\right)^2=625\)

\(\Leftrightarrow\left(2x+1\right)^2=25^2\)

\(\Rightarrow2x+1=25\)

\(\Leftrightarrow x=12\)

P/s tham khảo nha

\(\left(2x+1\right)^2=625\)

\(\left(2x+1\right)^2=25^2\)

\(\Rightarrow\orbr{\begin{cases}2x+1=25\\2x+1=-25\end{cases}\Leftrightarrow\orbr{\begin{cases}2x=24\\2x=-26\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=12\\x=-13\end{cases}}}\)

Vậy \(\orbr{\begin{cases}x=12\\x=-13\end{cases}}\)

\(2x+12-3x=-21\)

\(x.\left(2+12-3\right)=-21\)

\(x.11=-21\)

\(x=-\frac{21}{11}\)

\(235-288:\left[4\times\left(48-72\right)\right]\)

\(=235-288:4\times\left(48-72\right)\)

\(=235-72\times\left(-24\right)\)

\(=235-\left(-1728\right)\)

\(=235+1728\)

\(=1963\)

1/ kq: 238.

2/ kq: 33.

I don't now

or no I don't

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

sorry

\(\left(2x-3\right)^2-\left(2x-3\right)=0\)

\(\Rightarrow\left(2x-3\right)\left(2x-3-1\right)=0\)

\(\Rightarrow\left(2x-3\right)\left(2x-4\right)=0\)

\(\Rightarrow\hept{\begin{cases}2x-3=0\\2x-4=0\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x=\frac{3}{2}\\x=2\end{cases}}\)

(2x+1)²=25²

2x+1=25

2x=25-1=24

x=12

t*** mình nhé

\(\left(2x+1\right)^2=625\)

\(\Rightarrow\orbr{\begin{cases}2x+1=25\\2x+1=-25\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}2x=24\\2x=-26\end{cases}}\Rightarrow\orbr{\begin{cases}x=12\\x=-13\end{cases}}\)

(2x+1)²=625

=> 2x+1=25

=> 2x=24

=> x=12