\(\left(x-5\right)^{2020}+\left(y-x+1\right)^{2022}=0\left(1\right)\)

Ta có \(\left\{{}\begin{matrix}\left(x-5\right)^{2020}\ge0,\forall x\\\left(y-x+1\right)^{2022}\ge0,\forall x;y\end{matrix}\right.\)

\(\left(1\right)\Rightarrow\left\{{}\begin{matrix}\left(x-5\right)^{2020}=0\\\left(y-x+1\right)^{2022}=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x-5=0\\y-x+1=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=5\\y-5+1=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=5\\y=4\end{matrix}\right.\)

( 2x - 5 )2020 + ( 5y + 1 )2022 ≤ 0

Ta có : ( 2x - 5 )2020 ≥ 0 ∀ x

            ( 5y + 1 )2022 ≥ 0 ∀ y

=> ( 2x - 5 )2 + ( 5y + 1 )2022 ≥ 0 ∀ x, y

Kết hợp với đề bài => Chỉ xảy ra trường hợp ( 2x - 5 )2020 + ( 5y + 1 )2022 = 0

Khi đó \hept{2�−5=05�+1=0⇔\hept{�=52�=−15\hept{2x−5=05y+1=0​⇔\hept{x=25​y=−51​​
 

\(a.\left[\left(8x-12\right):4\right]\cdot3^3=3^6\\ \left(8x-12\right):4=3^6:3^3\\ \left(8x-12\right):4=27\\ 8x-12=27\cdot4\\ 8x-12=108\\ 8x=108+12\\ 8x=120\\ x=15\\ b.41-2^{x+1}=9\\ 2^{x+1}=32\\ 2^{x+1}=2^5\\=> x+1=5\\ =>x=4\)

a,[(8x-12):4].3³=3⁶

(8x-12):4=3⁶:3³

(8x-12):4=3³

(8x-12):4=9

8x-12=9.4

8x-12=36

8x=36+12

8x=48

x=6

Vậy x=6

b,41-2^x+1=9

2^x+1=41-9

2^x+1=32

2^x+1=2⁵

^x+1=5

^x=5-1

^x=4

^{Vậy x=4}

^{Chúc bạn học tốt:33}

a) \(\dfrac{7}{4}\text{⇔}1\dfrac{3}{4}\text{⇔}1,75\text{⇔}\dfrac{175}{100}\text{⇔}175\%\)

b) \(\dfrac{12}{5}\text{⇔}2\dfrac{2}{5}\text{⇔}2,4\text{⇔}\dfrac{240}{100}\text{⇔}240\%\)

THANK YOU NHAT VAN!

x có dạng 14k (k\(\in\)Z;k<3)

14

\(\left(5-x\right)\left(x-2\right)+\left(x-7\right)\left(x+7\right)=\left(3x-1\right)^2-\left(3x-2\right)\left(3x+2\right)\\ \Leftrightarrow-x^2+7x-10+x^2-49=9x^2-6x+1-9x^2+4\\\Leftrightarrow7x-59=-6x+5\\ \Leftrightarrow13x=44\\ \Leftrightarrow x=\dfrac{64}{13} \)