giup minh nha

m=2945 =>BT=3755

bạn xài cái này gõ công thức ra đi

a) \(A=\left[\dfrac{x+2}{x^2-x}+\dfrac{x-2}{x^2+x}\right].\dfrac{x^2-1}{x^2-x}\)

\(A=\left[\dfrac{x+2}{x\left(x-1\right)}+\dfrac{x-2}{x\left(x+1\right)}\right].\dfrac{x^2-1}{x^2+2}\)

\(A=\left[\dfrac{\left(x+2\right)\left(x+1\right)+\left(x-2\right)\left(x-1\right)}{x\left(x-1\right)\left(x+1\right)}\right].\dfrac{x^2-1}{x^2+2}\)

\(A=\left[\dfrac{x^2+2x+x+2+x^2-2x-x+2}{x\left(x-1\right)\left(x+1\right)}\right].\dfrac{x^2-1}{x^2+2}\)

\(A=\dfrac{2x^2+4}{x\left(x^2-1\right)}.\dfrac{x^2-1}{x^2+2}\)

\(A=\dfrac{2\left(x^2+2\right)\left(x^2-1\right)}{x\left(x^2-1\right)\left(x^2+2\right)}=\dfrac{2}{x}\)

b) Thay \(x=-200\) vào biểu thức \(A=\dfrac{2}{x}\) ta được :

\(A=\dfrac{2}{x}=\dfrac{2}{-200}=\dfrac{-2}{200}=\dfrac{-1}{100}\)

\(25x^2+16y^2=50xy\)

\(\Leftrightarrow\) \(\left(5x+4y\right)^2-40xy=50xy\)

\(\Leftrightarrow\) \(\left(5x+4y\right)^2=90xy\)

Mặt khác, ta cũng có:  \(25x^2+16y^2=50xy\)

\(\Leftrightarrow\)  \(\left(5x-4y\right)^2=10xy\)

Do đó:

\(P^2=\frac{\left(5x-4y\right)^2}{\left(5x+4y\right)^2}=\frac{10xy}{90xy}=\frac{1}{9}\)

Vậy,  \(P'=\frac{1+\frac{1}{9}}{1-\frac{1}{9}}=1\frac{1}{4}\)

1)

 \(25x^2-40xy+16y^2=10xy\Leftrightarrow\left(5x-4y\right)^2=10xy\)

\(25x^2+40xy+16y^2=10xy\Leftrightarrow\left(5x+4y\right)^2=90xy\)

\(P^2=\frac{1}{9}\Leftrightarrow Q=\frac{1+P^2}{1-P^2}=\frac{1+\frac{1}{81}}{1-\frac{1}{81}}=\frac{82}{80}=\frac{41}{40}\)

 ai nhanh mik k

=m x 3 + m + n x 4 + p x 2 + p x 2

=m x 3 + m x 1 + n x 4 + p x 2 + p x 2

=m x (3 + 1) + n x 4 + p x (2 + 2)

=m x 4 + n x 4 + p x 4

=(m + n + p) x 4

=2009 x 4

=8036

=m + n + p)