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

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

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

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

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

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

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

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

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

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

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

Đề bài yêu cầu j vậy?

mũ 2 tui ko bt

\(3.x^{4+5}.x^2.y^{2+2}.y^{4+2}.y^2\)

NTN đúng ko??

5^x+5^(x+2)=650

=>5^x+5^x. 5^2=650

=>5^x.(1+5^2)=650

=>5^x.26=650

=>5^x=650:26

=>5^x=25

=>5^x=5^2

=>x=2

Vậy x=2

\(5^{x+2}+5^x=650\)

\(\Leftrightarrow5^x.5^2+5^x=650\)

\(\Leftrightarrow5^x\left(5^2+1\right)=650\)

\(\Leftrightarrow5^x=25=5^2\)

\(\Leftrightarrow x=2\)

a) \(x:\left(\frac{-1}{2}\right)^3=\frac{-1}{2}\)

\(\Rightarrow x=\frac{-1}{2}.\left(\frac{-1}{2}\right)^3\)

\(\Rightarrow x=\left(\frac{-1}{2}\right)^4=\frac{1}{16}\)

b) \(\left(\frac{3}{4}\right)^5.x=\left(\frac{3}{4}\right)^7\)

\(\Rightarrow x=\left(\frac{3}{4}\right)^7:\left(\frac{3}{4}\right)^5\)

\(\Rightarrow x=\left(\frac{3}{4}\right)^2=\frac{9}{16}\)

Cảm ơn bn 

\(\dfrac{x}{5}=\dfrac{y}{4}=\dfrac{z}{2}\)

\(\Rightarrow\dfrac{x^3}{125}=\dfrac{y^3}{64}=\dfrac{z^3}{8}=\dfrac{x^3-y^3+z^3}{125-64+8}=\dfrac{69}{69}=1\)

\(\Rightarrow\left\{{}\begin{matrix}x=\sqrt[3]{125}=5\\y=\sqrt[3]{64}=4\\z=\sqrt[3]{8}=2\end{matrix}\right.\)