Bạn ghi lại đề bằng công thức đi bạn

\(\left(\dfrac{2}{3}x-\dfrac{1}{3}\right)^5=\dfrac{1}{243}\)

\(\)\(\left(\dfrac{2}{3}x-\dfrac{1}{3}\right)^5=\left(\dfrac{1}{3}\right)^5\)

⇒\(\left[{}\begin{matrix}\dfrac{2}{3}x-\dfrac{1}{3}=\dfrac{1}{3}\\\dfrac{2}{3}x-\dfrac{1}{3}=-\dfrac{1}{3}\end{matrix}\right.\)

⇒\(\left[{}\begin{matrix}\dfrac{2}{3}x=\dfrac{2}{3}\\\dfrac{2}{3}x=0\end{matrix}\right.\)

⇒\(\left[{}\begin{matrix}x=1\\x=0\end{matrix}\right.\)

bạn ghi lại đề được không ?

\(Q=x^3-7x^4xy+xy^3-11\)

Bậc của đa thức là \(6\)

viết bằng công thức ở chỗ \(\sum\) đi bạn

Bạn bảo cái gì cơ

\(P\left(x\right)=x^2+5x^4-3x^3+x^2+4x^4+3x^3-x+5\)

\(=\left(5x^4+4x^4\right)+\left(-3x^3+3x^3\right)+\left(x^2+x^2\right)-x+5\)

\(=9x^4+2x^2-x+5\)

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

\(=\left(-5x^4-x^4\right)+4x^3+\left(-x^2-x^2\right)+\left(x+3x\right)-1\)

\(=-6x^4+4x^3-2x^2+4x-1\)

P(x)+Q(x)

\(=9x^4+2x^2-x+5-6x^4+4x^3-2x^2+4x-1\)

\(=3x^4+4x^3+3x+4\)

Đặt P(x)+Q(x)=0

=>\(3x^4+4x^3+3x+4=0\)

=>\(x^3\left(3x+4\right)+\left(3x+4\right)=0\)

=>\(\left(3x+4\right)\left(x^3+1\right)=0\)

=>\(\left[\begin{array}{l}3x+4=0\\ x^3+1=0\end{array}\right.\Longrightarrow\left[\begin{array}{l}x=-\frac43\\ x=-1\end{array}\right.\)

3⁴⁰.8⁴¹- 8/4⁴⁰. 6^40-1=?

vầy hả

\(a=4^5.9^4-2.\dfrac{6^9}{2^{10}}.3^8+6^8.20\)

Đề là như vầy đúng ko bn?

$\textbf{A)}$

$\dfrac{4^5\cdot9^4-2\cdot6^9}{2^{10}\cdot3^8+6^8\cdot20}$

$=\dfrac{(2^2)^5(3^2)^4-2(2\cdot3)^9}{2^{10}\cdot3^8+2^2\cdot5(2\cdot3)^8}$

$=\dfrac{2^{10}3^8-2^{10}3^9}{2^{10}3^8+2^{10}\cdot5\cdot3^8}$

$=\dfrac{2^{10}3^8(1-3)}{2^{10}3^8(1+5)}$

$=\dfrac{-2}{6}$

$=-\dfrac{1}{3}.$

5ˣ⁺¹ - 5ˣ = 100 . 25²⁹

5ˣ.(5 - 1) = 100 .(5²)²⁹

5ˣ.4 = 100.5⁵⁸

5ˣ = 100 . 5⁵⁸ : 4

5ˣ = 25 .5⁵⁸

5ˣ = 5².5⁵⁸

5ˣ = 5⁶⁰

x = 60

\(5^{x+1}-5^x=100\cdot25^{29}\\\Rightarrow 5^x\cdot5-5^x=2^2\cdot5^2\cdot(5^2)^{29}\\\Rightarrow 5^x\cdot(5-1)=4\cdot 5^2\cdot5^{58}\\\Rightarrow 4\cdot5^x=4\cdot 5^{60}\\\Rightarrow5^x=5^{60}\\\Rightarrow x=60\\Vậy:x=60\)