\(x^{20}=x\)

⇒\(x^{20}-x=0\)

⇒\(x\left(x^{19}-1\right)=0\)

⇔\(x^{19}-1=0\) hoặc x=0

⇔x=1 hoặc x=0

\(x^{20}=x\\ \Leftrightarrow x^{20}-x=0\\ \Leftrightarrow x\left(x^{19}-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)

\(x^{20}=x\\ \Rightarrow x\in\left\{0;1\right\}\)vì \(x\in N\)

-36

tick cho minh

\(x+1538=3215\\ \Rightarrow x=3215-1538=1677\\ x-215=1578\\ \Rightarrow x=1578+215=1793\\ 345-x=182\\ \Rightarrow x=345-182=163\)

x+1538=3215⇒x=3215−1538=1677x−215=1578⇒x=1578+215=1793345−x=182⇒x=345−182=163

x+1538=3215⇒x=3215−1538=1677x−215=1578⇒x=1578+215=1793345−x=182⇒x=345−182=163

\(x-\dfrac{1}{8}=\dfrac{4}{x-2}\) hay \(x-\dfrac{1}{8}=\dfrac{4}{x}-2\) vậy bạn?

\(\dfrac{x-1}{8}=\dfrac{4}{x-2}\)

=>\(\dfrac{\left(x-1\right)\left(x-2\right)}{8.\left(x-2\right)}=\dfrac{4.8}{\left(x-2\right).8}\)

=>khử mẫu

=>(x-1)(x-2)=32

.......

Là \(\dfrac{x+1}{8}=\dfrac{4}{x-2}\) nhoa

\(a)6/4=9/4\) =)??

thiếu dữ liệu

a: =>x:6,5=57,7

hay x=373,75

b: =>x:2,4=45.4

hay x=108,96

\(x:6,5=\dfrac{577}{10}\Leftrightarrow x=\dfrac{7201}{20}\)

\(x:2,4=45,4\Leftrightarrow x=\dfrac{2724}{25}\)

X: 6,5-34,7=23

X: 6,5= 23+34,7

X: 6,5= 23+34,7

X:6,5= 57,7

X= 57,7x6,5

X= 375,05

X: 2,4=42+3,4

x:2,4+=45,4

x= 45,4x2,4

x= 108,96

\(\Leftrightarrow xy=63\)

\(\Leftrightarrow\left(x,y\right)\in\left\{\left(1;63\right);\left(3;21\right);\left(7;9\right);\left(-63;-1\right);\left(-21;-3\right);\left(-9;-7\right)\right\}\)

1) \(\Rightarrow x^2\left(x^{2004}-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x^{2004}=1\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)

2) \(\Rightarrow\left(x-5\right)^4\left[\left(x-5\right)^2-1\right]=0\)

\(\Rightarrow\left[{}\begin{matrix}x-5=0\\\left(x-5\right)^2=1\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=5\\x-5=1\\x-5=-1\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=5\\x=6\\x=4\end{matrix}\right.\)

\(\dfrac{x}{3}=x+y=20\Rightarrow x=60\Rightarrow60+y=20\Rightarrow y=-40\)

Ta có:

\(\dfrac{x}{3}=20\)

\(\Rightarrow\)\(x=60\)

Lại có:

\(x+y=20\)

\(\Rightarrow\)\(y=20-60\)

\(\Rightarrow\)\(y=-40\)

Vây x = 60 và y = - 40

\(\dfrac{x}{3}=x+y\)

\(\Leftrightarrow x-\dfrac{1}{3}x=-y\)

\(\Leftrightarrow y=-\dfrac{2}{3}x\)

Ta có: x+y=20

\(\Leftrightarrow x\cdot\dfrac{1}{3}=20\)

hay x=60

=> y=40