\(\frac{x^2+2x+1}{5x^3+5x^2}=\frac{\left(x+1\right)^2}{5x^2\left(x+1\right)}=\frac{x+1}{5x^2};\)

b, \(\frac{2x^2+2x}{x+1}=\frac{2x\left(x+1\right)}{x+1}=2x\)

\(a,\frac{x^2+2x+1}{5x^3+5x^2}=\frac{\left(x+1\right)^2}{5x^2\left(x+1\right)}=\frac{x+1}{5x^2}\)

\(b,\frac{2x^2+2x}{x+1}=\frac{2x\left(x+1\right)}{x+1}=2x\)

\(a,\frac{4x^3}{10x^2y}=\frac{2x}{5y}\)

\(b,\frac{10xy^5\left(2x-3y\right)}{12xy\left(2x-3y\right)}=\frac{5y^4}{6}\)

Hok Tốt~~

\(\frac{4x^3}{10x^2y}=\frac{2x}{5y}\)

\(\frac{10xy^5\left(2x-3y\right)}{12xy\left(2x-3y\right)}=\frac{5y^4}{4}\)

Tham khảo nhé~

( x - 1)² + x( x - 4) = 0

x² - 2x + 1 + x² - 4x = 0

2x² - 6x + 1 = 0

2x² - 6x = -1

2x ( x - 3) = -1

      => 2x = -1 => -1/2

hoặc    x - 3 = -1 => x = 2

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

\(x^2-2x+1+x^2-4x=0\)

\(2x^2-6x+1=0\)

\(2.\left[x^2-2.x.1,5+\left(1,5\right)^2\right]-3,5=0\)

\(2.\left(x-1,5\right)^2-3,5=0\)

\(2.\left[\left(x-1,5\right)^2-1,75\right]=0\)

\(\Leftrightarrow\left(x-1,5\right)^2-1,75=0\)

\(\left(x-1,5\right)^2-\left(\sqrt{1,75}\right)^2=0\)

\(\left(x-1,5-\sqrt{1,75}\right)\left(x-1,5+\sqrt{1,75}\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-1,5-\sqrt{1,75}=0\\x-1,5+\sqrt{1,75}=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1,5+\sqrt{1,75}\\x=1,5-\sqrt{1,75}\end{cases}}\)

Vậy \(\orbr{\begin{cases}x=1,5+\sqrt{1,75}\\x=1,5-\sqrt{1,75}\end{cases}}\)

\(P\left(x\right)=x^7-80x^6+80x^5-80x^4+80x^3-80x^2+80x+15\)

\(P\left(x\right)=x^7-\left(79+1\right)x^6+\left(79+1\right)x^5-\left(79+1\right)x^4+\left(79+1\right)x^3-\left(79+1\right)x^2+\left(79+1\right)x+15\)

\(P\left(79\right)=x^7-\left(x+1\right)x^6+\left(x+1\right)x^5-\left(x+1\right)x^4+\left(x+1\right)x^3-\left(x+1\right)x^2+\left(x+1\right)x+15\)

\(P\left(79\right)=x^7-x^7-x^6+x^6+x^5-x^5-x^4+x^4-x^3+x^3-x^2+x^2+x+15\)

\(P\left(79\right)=79+15\)

\(P\left(79\right)=94\)

Vậy \(P\left(79\right)=94\)

\(2n^2-n+2⋮2n+1\Leftrightarrow2n^2-2n^2-n-n+2⋮2n+1\)

\(\Leftrightarrow-2n+2⋮2n+1\Leftrightarrow1⋮2n+1\Leftrightarrow n\in\left\{-1;0\right\}\)

\(a,x^3-13x=0\)

\(x.\left(x^2-13\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=0\\x^2=13\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=\sqrt{13}\end{cases}}}\)

\(b,2-25x^2=0\)

\(\Rightarrow25x^2=2\Rightarrow x^2=\frac{2}{25}\Rightarrow x=\sqrt{\frac{2}{25}}\)

\(c,x^2-x+\frac{1}{4}=0\)

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

a, x ³ - 13 x = 0

=> x ( x ² - 13 ) = 0

=> \(\orbr{\begin{cases}x=0\\x^2=13\end{cases}\Rightarrow[\begin{cases}x=0\\x=\sqrt{13}\\x=-\sqrt{13}\end{cases}}\)

b, 2 - 25 x ² = 0

=> 25 x ² = 2

=> x ² = 0,08

=> \(\orbr{\begin{cases}x=\frac{\sqrt{2}}{5}\\x=\frac{-\sqrt{2}}{5}\end{cases}}\)

x, x ² - x + \(\frac{1}{4}\)= 0 

=> \(\left(x-\frac{1}{2}\right)^2=0\)

=> \(x-\frac{1}{2}=0\)

=> \(x=\frac{1}{2}\)

a, 3x ³ - 3x = 0

=> 3x ( x ² - 1 ) = 0

=> \(\orbr{\begin{cases}3x=0\\x^2-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x^2=1\end{cases}\Rightarrow[}\begin{cases}x=0\\x=1\\x=-1\end{cases}}\)

b, x ( x - 2 ) + ( x - 2 ) = 0

=> ( x - 2 ) ( x + 1 ) = 0

=> \(\orbr{\begin{cases}x-2=0\\x+1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=2\\x=-1\end{cases}}}\)

c, 5x ( x - 2000 ) - x + 2000 = 0

=> ( x - 2000 ) ( 5x - 1 ) = 0

=> \(\orbr{\begin{cases}x-2000=0\\5x-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=2000\\x=\frac{1}{5}\end{cases}}}\)

ai tra loi nhanh cau hoi nay ho minh voi minh k cho

ai tra loi nhanh cau hoi nay ho minh voi minh k cho