\(=\dfrac{1}{5}\left(\dfrac{5}{4\cdot9}+\dfrac{5}{9\cdot14}+...+\dfrac{5}{1999\cdot2004}\right)\)

\(=\dfrac{1}{5}\left(\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{14}+...+\dfrac{1}{1999}-\dfrac{1}{2004}\right)\)

\(=\dfrac{1}{5}\left(\dfrac{1}{4}-\dfrac{1}{2004}\right)=\dfrac{1}{5}\cdot\dfrac{125}{501}=\dfrac{25}{501}\)

Sửa \(\dfrac{1}{9.4}\rightarrow\dfrac{1}{9.14}\) nha

\(=\dfrac{1}{5}\left(\dfrac{5}{4.9}+\dfrac{5}{9.14}+...+\dfrac{5}{1999.2004}\right)\)

\(=\dfrac{1}{5}\left(\dfrac{1}{4}-\dfrac{1}{2004}\right)=\dfrac{1}{5}.\dfrac{125}{501}=\dfrac{25}{501}\)

\(=2\left(\dfrac{1}{4.9}+\dfrac{1}{9.14}+.....+\dfrac{1}{1999.2004}\right)\)

\(=2\left(\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{14}+...+\dfrac{1}{1999}-\dfrac{1}{2004}\right)\)

\(=2\left(\dfrac{1}{4}-\dfrac{1}{2004}\right)=2\cdot\dfrac{125}{501}=\dfrac{250}{501}\)

36

9x4=36

36

\(1+9.4=1+36=37\)

1 + 9 x 4

= 1 + 36​

​= 37

1+9.4

=1+36

=37

k mk nha

\(9x^4-10x^2+1=0\\ \Rightarrow\left(9x^4-9x^2\right)-\left(x^2-1\right)=0\\ \Rightarrow9x^2\left(x^2-1\right)-\left(x^2-1\right)=0\\ \Rightarrow\left(x-1\right)\left(x+1\right)\left(9x^2-1\right)=0\\ \Rightarrow\left(x-1\right)\left(x+1\right)\left(3x-1\right)\left(3x+1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=1\\x=-1\\x=\dfrac{1}{3}\\x=-\dfrac{1}{3}\end{matrix}\right.\)

Đặt x^2 = t ( t>= 0 ) 

9t^2 - 10t + 1 = 0 

ta có : a + b + c = 9 - 10 + 1 = 0 

=> t = 1 ; t = 1/9 

theo cách đặt x = 1 ; x = 1/3 

\(\Leftrightarrow\left(9x^2-1\right)\left(x^2-1\right)=0\)

hay \(x\in\left\{\dfrac{1}{3};-\dfrac{1}{3};1;-1\right\}\)

a: Ta có: \(-\left(-3x^2\right)^3+4x-9-27x^6\)

\(=27x^6-27x^6+4x-9\)

=4x-9

=-1

a: \(50x^5-8x^3\)

\(=2x^3\left(25x^2-4\right)\)

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

b: \(x^4-5x^2-4y^2+10y\)

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

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

c: \(36a^2+12a+1-b^2\)

\(=\left(6a+1\right)^2-b^2\)

\(=\left(6a+1-b\right)\left(6a+1+b\right)\)

d: \(x^3+y^3-xy^2-x^2y\)

\(=\left(x+y\right)\left(x^2-xy+y^2\right)-xy\left(x+y\right)\)

\(=\left(x+y\right)\left(x^2-2xy+y^2\right)\)

\(=\left(x+y\right)\cdot\left(x-y\right)^2\)

e: Ta có: \(4x^2+4x-3\)

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

\(=2x\left(2x+3\right)-\left(2x+3\right)\)

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

f: Ta có: \(9x^4+16x^2-4\)

\(=9x^4+18x^2-2x^2-4\)

\(=9x^2\left(x^2+2\right)-2\left(x^2+2\right)\)

\(=\left(x^2+2\right)\left(9x^2-2\right)\)

g: Ta có: \(-6x^2+5xy+4y^2\)

\(=-6x^2+8xy-3xy+4y^2\)

\(=-2x\left(3x-4y\right)-y\left(3x-4y\right)\)

\(=\left(3x-4y\right)\left(-2x-y\right)\)

h: Ta có: \(\left(x^2+4x\right)^2+8\left(x^2+4x\right)+15\)

\(=\left(x^2+4x\right)^2+3\left(x^2+4x\right)+5\left(x^2+4x\right)+15\)

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

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

\(\frac{5}{9}x\frac{4}{7}+\frac{5}{9}x\frac{3}{7}\)

\(=\frac{5}{9}x\left(\frac{4}{7}+\frac{3}{7}\right)\)

\(=\frac{5}{9}x1\)

\(=\frac{5}{9}\)

\(\frac{5}{9}\) x \(\frac{4}{7}\)+ \(\frac{5}{9}\)+ \(\frac{3}{7}\)

= \(\frac{5}{9}\)x ( \(\frac{4}{7}\)+ \(\frac{3}{7}\))

= \(\frac{5}{9}\) x 1 

= \(\frac{5}{9}\)

5/9*4/7+5/9*3/7

=20/63+5/21

=5/9.

\(\dfrac{4}{5}\times\dfrac{16}{9}-\dfrac{7}{9}\times\dfrac{4}{5}-\dfrac{4}{5}\)
\(=\dfrac{4}{5}\times\left(\dfrac{16}{9}-\dfrac{7}{9}-1\right)\)
\(=\dfrac{4}{5}\times0\)
\(=0\)
#Đang Bận Thở

4/5 x 16/9 - 7/9 x 4/5 - 4/5

= 4/5 x ( 16/9 - 7/9 - 1 )

= 4/5 x 0 = 0

=\(\dfrac{4}{5}\times\left(\dfrac{16}{9}-\dfrac{7}{9}-1\right)\)

=\(\dfrac{4}{5}\times0\) = 0