Bài 1:

|x-2|=4-x

ĐK: \(4-x\ge0\Leftrightarrow x\le4\)

Ta có: \(\orbr{\begin{cases}x-2=4-x\\x-2=x-4\end{cases}\Rightarrow\orbr{\begin{cases}2x=6\\0=2\left(loại\right)\end{cases}\Rightarrow}}x=3\left(tm\right)\)

Vậy x = 3 

Bài 2:

a, sao có z

b, Vì \(\hept{\begin{cases}\left|2017-x\right|\ge0\\\left|y-x+2018\right|\ge0\end{cases}\Rightarrow\left|2017-x\right|+\left|y-x+2018\right|\ge0}\)

Mà |2017-x|+|y-x+2018|=0

\(\Rightarrow\hept{\begin{cases}\left|2017-x\right|=0\\\left|y-x+2018\right|=0\end{cases}\Rightarrow\hept{\begin{cases}x=2017\\y-2017+2018=0\end{cases}\Rightarrow}\hept{\begin{cases}x=2017\\y=1\end{cases}}}\)

Vậy x=2017,y=1

c, giống b

Bài 2 cũng có z bạn ạ Làm luôn hộ mình câu b

b) ta thấy /2017-x/>=0

/y-x+2018/>= 0

=> /2017-x/+/y-x+2018/>=0

dấu = xảy ra khi 2017-x=0 => x=2017

                     và y-x+2018=0 => y= 1

vậy (x;y)=(2017;1)

Bài 1

a) \(x=x^5\)

\(x^5-x=0\)

\(x\left(x^4-1\right)=0\)

\(x=0\) hoặc \(x^4-1=0\)

* \(x^4-1=0\)

\(x^4=1\)

\(x=1\)

Vậy x = 0; x = 1

b) \(x^4=x^2\)

\(x^4-x^2=0\)

\(x^2\left(x^2-1\right)=0\)

\(x^2=0\) hoặc \(x^2-1=0\)

*) \(x^2=0\)

\(x=0\)

*) \(x^2-1=0\)

\(x^2=1\)

\(x=1\)

Vậy \(x=0\); \(x=1\)

c) \(\left(x-1\right)^3=x-1\)

\(\left(x-1\right)^3-\left(x-1\right)=0\)

\(\left(x-1\right)\left[\left(x-1\right)^2-1\right]=0\)

\(x-1=0\) hoặc \(\left(x-1\right)^2-1=0\)

*) \(x-1=0\)

\(x=1\)

*) \(\left(x-1\right)^2-1=0\)

\(\left(x-1\right)^2=1\)

\(x-1=1\) hoặc \(x-1=-1\)

**) \(x-1=1\)

\(x=2\)

**) \(x-1=-1\)

\(x=0\)

Vậy \(x=0\); \(x=1\); \(x=2\)

\(x^{2020}=x\Leftrightarrow x^{2020}-x=0\Leftrightarrow x\left(x^{2019}-1\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\x^{2019}-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x^{2019}=1\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)

\(1+2+2^2+2^3+....+2^{2019}+2^{2020}\)

\(A=\left(1+2+2^2\right)+\left(2^3+2^4+2^5\right)+....+\left(2^{2016}+2^{2017}+2^{2018}\right)+2^{2019}+2^{2020}\)

\(A=\left(1+2+2^2\right)+2^3\left(1+2+2^2\right)+.....+2^{2016}\left(1+2+2^2\right)+2^{2019}+2^{2020}\)

\(A=7+2^3.7+2^6.7+2^9.7+....+2^{2016}.7+2^{2019}+2^{2020}\)

\(\text{Ta có:}2^{2019}+2^{2020}=8^{673}+8^{673}.2\equiv1+1.2\left(\text{mod 7}\right)\equiv3\left(\text{mod 7}\right)\Rightarrow A\text{ chia 7 dư 3}\)

( 2y - z - x ) mũ 2020 nhé :vv

_{Hướng dẫn ông viết mũ nhá =) Ông hãy nhìn bên trên phần mình đăng bài + trả lời . Bẹn có thể thấy các kí tự khó hiểu vl :v Như : Hình ảnh  , Tex , ... Hãy nhìn X2 và X2 ấn vô đó , lak ấn đc :VVV} 

#Mật

X₂ , X² nha =) 

#Mật

b. 1404 : [118 - (4x + 6)] = 27

118 - (4x + 6) = 52

4x + 6 = 66

4x = 60

x = 15

d) \(5x^2-3x=0\)

\(\Leftrightarrow x\left(5x-3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\5x-3=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{3}{5}\end{cases}}\)

e) \(3\left(x-1\right)+4\left(x-1\right)^2=0\)

\(\Leftrightarrow\left(x-1\right)\left[3-4.\left(x-1\right)\right]=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\3-4\left(x-1\right)=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=1\\4\left(x-1\right)=3\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=1\\x-1=\frac{3}{4}\Rightarrow x=\frac{7}{4}\end{cases}}\)

f) \(2\left(x-2\right)^2=\left(x-2\right)\)

\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\2\left(x-2\right)-1=0\end{cases}}\)

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

g) \(\left(x-2020\right)^4=\left(x-2020\right)^2\)

\(\Leftrightarrow\orbr{\begin{cases}\left(x-2020\right)^2=0\\\left(x-2020\right)^2-1=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=2020\\x=2019,x=2021\end{cases}}\)

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Ta có: \(5^{4x}:5^5=5^{2020}:5^{2019}\)

    \(\Leftrightarrow5^{4x-5}=5^1\)

    \(\Leftrightarrow4x-5=1\)

    \(\Leftrightarrow4x=6\)

    \(\Leftrightarrow x=\frac{3}{2}\left(L\right)\)

Vậy \(x=\varnothing\)

\(\sqrt{x-1}=5.\left(2018+2019+2020\right)^0\)

\(\sqrt{x-1}^2=5^2\)

\(x-1=25\)

\(x=25+1\)

\(\Rightarrow x=26\)

Mình làm hơi tắt, để mình làm lại nhé!

\(\sqrt{x-1}=5.\left(2018+2019+2020\right)^0\)

\(\sqrt{x-1}=5\)

\(\sqrt{x-1}^2=5^2\)

\(x-1=25\)

\(x=25+1\)

\(\Rightarrow x=26\)