Toán

12:

\(B=\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{4}\right)\left(1-\dfrac{1}{5}\right)...\left(1-\dfrac{1}{2004}\right)\)

\(B=\left(\dfrac{2}{2}-\dfrac{1}{2}\right)\left(\dfrac{3}{3}-\dfrac{1}{3}\right)\left(\dfrac{4}{4}-\dfrac{1}{4}\right)...\left(\dfrac{2004}{2004}-\dfrac{1}{2004}\right)\)

\(B=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot\dfrac{4}{5}...\cdot\dfrac{2003}{2004}\)

\(B=\dfrac{1\cdot2\cdot3\cdot...\cdot2002\cdot2003}{2\cdot3\cdot4\cdot...\cdot2003\cdot2004}\)

\(B=\dfrac{1}{2004}\)

`12) B=(1-1/2)(1-1/3)(1-1/4)(1-1/5)....(1-1/2003)(1-1/2004)`

    `B=1/2 . 2/3 . 3/4 . 4/5 .... 2002/2003 . 2003/2004`

    `B= [1.2.3.4......2002.2003]/[2.3.4.5.....2003.2004]`

    `B=1/2004`

`7)`

`a)[254xx399-145]/[254+399xx253]`

`=[(253+1)xx399-145]/[254+399xx253]`

`=[253xx399+399-145]/[254+399xx253]`

`=[254+253xx399]/[254+253xx399]=1`

`b)[5932+6001xx5931]/[5932xx6001-69]`

`=[5932+6001xx5931]/[(5931+1)xx6001-69]`

`=[5932+6001xx5931]/[5931xx6001+6001-69]`

`=[5932+6001xx5931]/[5932+6001xx5931]=1`

`a)` Vì `O` là trung điểm của `AC;BD`

    `=>{(\vec{OA}=-\vec{OC}),(\vec{OB}=-\vec{OD}):}`

Ta có: `\vec{OA}+\vec{OB}+\vec{OC}+\vec{OD}`

    `=-\vec{OC}-\vec{OD}+\vec{OC}+\vec{OD}=0`

`b)` Vì `E` là trung điểm `AD=>\vec{EA}=-\vec{ED}`

Ta có: `\vec{EA}+\vec{EB}+2\vec{EC}`

`=\vec{EA}+\vec{EA}+\vec{AB}+2\vec{ED}+2\vec{DC}`

`=-2\vec{ED}+\vec{AB}+2\vec{ED}+2\vec{AB}=3\vec{AB}`

`c)` Ta có: `\vec{EB}+2\vec{EA}+4\vec{ED}`

`=\vec{EB}-2\vec{ED}+4\vec{ED}`

`=\vec{EB}+2\vec{ED}`

`=\vec{EA}+\vec{AB}+2\vec{ED}`

`=-\vec{ED}+\vec{AB}+2\vec{ED}`

`=\vec{AB}+\vec{EC}+\vec{CD}` (Mà `\vec{AB}=-\vec{CD}`)

`=\vec{EC}`

Ta có:

\(a=36=2^2\cdot3^2\)

\(b=60=3\cdot2^2\cdot5\)

\(c=90=3^2\cdot2\cdot5\)

\(\Rightarrow\text{Ư}CLN\left(a,b,c\right)=\text{Ư}CLN\left(36;60;90\right)=3\cdot2=6\)

\(\Rightarrow BCNN\left(a,b,c\right)=BCNN\left(36,60,90\right)=3^2\cdot5\cdot2^2=180\)

\(x^2=-5\)

Giá trị của một số chính phương luôn luôn giữa giá trị dương \(\ge0\)

Mà: \(-5< 0\)

\(\Rightarrow x^2=-5\) (vô lí)

⇒ \(x\in\varnothing\)

a: Xét ΔAHD vuông tại H và ΔAED vuông tại E có

AD chung

góc HAD=góc EAD

=>ΔAHD=ΔAED

=>AH=AE và DH=DE

=>AD là trung trực của HE

b: Xét ΔFAC có

AK,CH là đường cao

AK cắt CH tại D

=>D là trực tâm

=>FD vuông góc AC

a: =>x-2/5=3/4:1/3=3/4*3=9/4

=>x=9/4+2/5=45/20+8/20=53/20

b: =>x-2/3=7/3:4/5=7/3*5/4=35/12

=>x=35/12+2/3=43/12

c: 1/3(x-2/5)=4/5

=>x-2/5=4/5*3=12/5

=>x=12/5+2/5=14/5

d: =>2/3x-1/3-1/4x+1/10=7/3

=>5/12x-7/30=7/3

=>5/12x=7/3+7/30=77/30

=>x=77/30:5/12=154/25

e: \(\Leftrightarrow x\cdot\dfrac{3}{7}-\dfrac{2}{7}+\dfrac{1}{2}-\dfrac{5}{4}x+\dfrac{5}{2}=0\)

=>\(x\cdot\dfrac{-23}{28}=\dfrac{2}{7}-3=\dfrac{-19}{7}\)

=>x=19/7:23/28=76/23

f: =>1/2x-3/2+1/3x-4/3+1/4x-5/4=1/5

=>13/12x=1/5+3/2+4/3+5/4=257/60

=>x=257/65

i: =>x^2-2/5x-x^2-2x+11/4=4/3

=>-12/5x=4/3-11/4=-17/12

=>x=17/12:12/5=85/144

`A = |6 : 3/5 - 1 1/6 . 6/7| : |4 1/5 . 10/11 + 5 2/11|`

`=|10 - 1| : |42/11 + 57/11|`

`=9 : 9`

`=1`

=| 6 . 5/3 - 7/6 . 6/7 | : | 21/5 . 10/11 + 57/11 |
=| 10 - 1 | : | 42/11 + 57/11 |
= 9 : |99/11|
= 9 : 9
=1