(1-1/2x1/2)x(1-1/3x1/3)x(1-1/4x1/4)x…x(1-1/2007x1/2007)
= ( 1 - 1/4 ) x ( 1 - 1/9 ) x ( 1- 1/16) x .....x ( 1 - 1/4028049)
= 3/4 x 8/9 x 15/16 x .....x 4028048 / 4028049
= 3 x 8 x 15 x .....x 4028048 / 4 x 9 x 16 x ......x 4028049
= 1 x 3 x 2 x 4 x 3 x 5 x .....x 2006 x 2008 / 2x2 x 3 x3 x 4 x 4 x .....x 2007 x 2007
= ( 1 x 2 x 3 x .....x 2006) x ( 3 x 4 x 5 x .....x 2008) / ( 2 x 3 x 4 x ....x 2007) x ( 2 x 3 x 4 x ....x 2007)
= 2008 / 2007 x 2
= 1004 / 2007

a,

=1/2x2/3x3/4x...x17/18x18/19x19/20

=1/20

b,

co dung ko day ban

`@Fù`

`=1/(1×2)+1/(2×3)+1/(3×4)+...+1/(72×73)`

`=1/1-1/2+1/2-1/3+...+1/72-1/73`

`=1-1/73`

`=72/73`

tích đúng mình giải cho

= 1/1x2 + 1/2x3 + 1/3x4 + 1/4x5 + 1/5 x 6

= (1 - 1/2) + (1/2 - 1/3) + (1/3 - 1/4) + (1/4 - 1/5) + (1/5 - 1/6)

= 1 - 1/6

= 5/6

nha,mơn nhìu

\(\frac{1}{1}x\frac{1}{2}+\frac{1}{2}x\frac{1}{3}+\frac{1}{3}x\frac{1}{4}+\frac{1}{4}x\frac{1}{5}+\frac{1}{5}x\frac{1}{6}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\)

\(=1-\frac{1}{6}=\frac{5}{6}\)

  1/2x1/3+1/3x1/4+1/4x1/5+1/5x1/6+1/6x1/7+1/7x1/8+1/8x1/9

=7/18

câu 2 đề sai

a) (*) m = 0 => x = \(\dfrac{7}{8}\) (loại)

(*) \(m\ne0\) Phương trình có nghiệm

\(\Delta=\left[2\left(m-4\right)\right]^2-4m\left(m+7\right)=-60m+64\ge0\Leftrightarrow m\le\dfrac{16}{15}\) 

Hệ thức Viet kết hợp 4x₁ + 3x₂ = 1

\(\Leftrightarrow\left\{{}\begin{matrix}x_1x_2=\dfrac{m+7}{m}\\x_1+x_2=\dfrac{8-2m}{m}\\x_1=2x_2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x_1x_2=\dfrac{m+7}{m}\\x_1=\dfrac{16-4m}{3m}\\x_2=\dfrac{8-2m}{3m}\end{matrix}\right.\)

\(\Leftrightarrow\dfrac{16-4m}{3m}.\dfrac{8-2m}{3m}=\dfrac{m+7}{m}\)

\(\Leftrightarrow2\left(8-2m\right)^2=9m\left(m+7\right)\)

\(\Leftrightarrow8m^2-64m+128=9m^2+63m\)

\(\Leftrightarrow m^2+127m-128=0\Leftrightarrow\left[{}\begin{matrix}m=1\\m=128\left(\text{loại}\right)\end{matrix}\right.\)<=> m = 1

a: Khi m=1/2 thì \(x^2-2x-\dfrac{1}{4}-4=0\)

\(\Leftrightarrow x^2-2x-\dfrac{17}{4}=0\)

\(\Leftrightarrow4x^2-8x-17=0\)

\(\Leftrightarrow\left(2x-2\right)^2=21\)

hay \(x\in\left\{\dfrac{\sqrt{21}+2}{2};\dfrac{-\sqrt{21}+2}{2}\right\}\)

b: \(\text{Δ}=\left(-2\right)^2-4\left(-m^2-4\right)\)

\(=4+4m^2+16=4m^2+20>0\)

Do đó: Phương trình luôn có hai nghiệm phân biệt