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

\(\Leftrightarrow2x^2-2x-x^2-2x+4=0\)

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

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

\(\Leftrightarrow x-2=0\)

\(\Leftrightarrow x=2\)

Vậy \(S=\left\{2\right\}\)

a) (x-2)³-x(x+1)(x-1)+6x(x-3)

= x³-6x²+12x-8-x(x²-1)+6x²-18x

=x³-6x²+12x-8-x³+x+6x²-18x

=(x³-x³)+(-6x²+6x²)+(12x+x-18x)-8

=0+0-5x-8

=-5x-8

b xem lại đề

\(P=4^{25}\cdot\left(\dfrac{1}{2}-\dfrac{1}{3}\right)\cdot\left(\dfrac{1}{2}-\dfrac{1}{5}\right)\cdot\left(\dfrac{1}{2}-\dfrac{1}{7}\right)\cdot...\cdot\left(\dfrac{1}{2}-\dfrac{1}{101}\right)\)

\(P=4^{25}\cdot\left(\dfrac{1}{2}\cdot\left(\dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{7}-...+\dfrac{1}{101}\right)\right)\)

\(P=4^{25}\cdot\left(\dfrac{1}{2}\cdot\left(\dfrac{1}{101}-\dfrac{1}{3}\right)\div2+1\right)\)

\(P=4^{25}\cdot\left(\dfrac{1}{2}\cdot\dfrac{1}{50}\right)\)

\(P=4^{25}\cdot\dfrac{1}{100}\)

3 + 3 =6 

k mik nhé mik k lại cho

3 + 3 = 6.

Mk k bạn rồi đó nha.

3 + 3 =6 

~Hok tốt~

Thay x vào đa thức ta có:
2009^4 + 3x2009^3 -8
=2009^3 x 2009 + 3x2009^3 -8
=2009^3(2009+3) -8
=2009^3 x 2012 -8
bấm máy tính sẽ ra: 16314275298740
k cho mình nha!!!

c) \(\dfrac{x+4}{20}=\dfrac{5}{x+4}\)

⇔\(\left(x+4\right)\left(x+4\right)=100\)

⇔\(\left(x+4\right)^2=10^2\)

⇔\(\left[{}\begin{matrix}x+4=10\\x+4=-10\end{matrix}\right.\)

⇔\(\left[{}\begin{matrix}x=6\\x=-14\end{matrix}\right.\)

\(c,ĐK:x\ne-4\\ PT\Leftrightarrow\left(x+4\right)^2=100\\ \Leftrightarrow\left[{}\begin{matrix}x+4=10\\x+4=-10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\left(tm\right)\\x=-14\left(tm\right)\end{matrix}\right.\\ d,ĐK:x\ne-2;x\ne-3\\ PT\Leftrightarrow\left(x-1\right)\left(x+3\right)=\left(x-2\right)\left(x+2\right)\\ \Leftrightarrow x^2+2x-3=x^2-4\\ \Leftrightarrow2x=-1\Leftrightarrow x=-\dfrac{1}{2}\left(tm\right)\)

P(x),Q(x) là gì vậy em ?

giúp mik đi ạ đang gấp

x+1<x+2<x+3<x+4 ( với mọi x)

\(\dfrac{1}{100}\) < \(\dfrac{1}{99}\)<\(\dfrac{1}{3}\) <\(\dfrac{1}{2}\)

=>\(\dfrac{x+1}{100}\)+\(\dfrac{x+2}{99}\) <\(\dfrac{x+3}{3}\)+\(\dfrac{x+4}{2}\) là đúng