đoạn ống nước mới dài số m là:

    0,8+1,35 -\(\dfrac{2}{25}\)= 0,8+ 1,35- 0,08=2.07

Độ dài đoạn ống nước mới là:

Đáp số: $2,07m$

Ta có:

\(\dfrac{a-b}{a+b}=\dfrac{b-c}{b+c}\)

\(=>(a-b)(b+c)=(b-c)(a+b)\)

\(=>ab+ac-b^2-bc=ab+b^2-ac-bc\)

\(=>ab-ab+ac+ac=b^2+b^2-bc+bc\)

\(=>2ac=2b^2\)

\(=>ac=b^2\)

Hay \(b^2=ac\)

     `->Đpcm`

mong mn giúp đỡ

Lời giải:

$2^2S=1-\frac{1}{2^2}+\frac{1}{2^4}-.......+\frac{1}{2^{2000}}-\frac{1}{2^{2002}}$

$S=\frac{1}{2^2}-\frac{1}{2^4}+\frac{1}{2^6}-.....+\frac{1}{2^{2002}}-\frac{1}{2^{2004}}$

$4S+S=1-\frac{1}{2^{2004}}<1$

$5S<1$

$S< \frac{1}{5}=0,2$ (đpcm)

\(a)5 \times \dfrac{7}{5}=7\)

~~~~~~~~~~~~~~~~~~~~~~~~~~~~

\(b)\dfrac{3}{4} \times \dfrac{7}{9}+\dfrac{1}{4} \times \dfrac{7}{9}\)

\(=\dfrac{7}{9} \times (\dfrac{3}{4}+\dfrac{1}{4})\)

\(=\dfrac{7}{9} \times \dfrac{4}{4}=\dfrac{7}{9}\)

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

\(c)\dfrac{1}{7} \times \dfrac{5}{9}+\dfrac{5}{9} \times \dfrac{1}{7}+\dfrac{5}{9} \times \dfrac{3}{7}\)

\(=\dfrac{5}{9} \times (\dfrac{1}{7}+\dfrac{1}{7}+\dfrac{3}{7}\)

\(=\dfrac{5}{9} \times \dfrac{5}{7}=\dfrac{25}{63}\)

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

\(d)\dfrac{4 \times 11 \times 3}{4 \times 9 \times 121}\)

\(=\dfrac{4 \times 3 \times 11}{4 \times 3 \times 3 times 11 \times 11}\)

\(=\dfrac{1}{3 \times 11}=\dfrac{1}{33}\)

a, 5 x 7/5 = 7

b, 3/4 x 7/9 + 1/4 x 7/9

= 7/9 x ( 3/4 + 1/4 )

= 7/9 x 1

= 7/9

c, 1/7 x 5/9 + 5/9 x 1/7 + 5/9 x 3/7

= 1/7x ( 5/9+ 5/9) + 5/9 x 3/7

= 1/7 x 25/81 + 5/21

= 25/567 + 5/21

= 160/567

d, 4x11x3/4 x 9/121

= 44 x 3/4 x 9/121

= 33 x 9/121

= 27/11

57 + 725 + 605 - 5

= 782 + 600

= 1382

57 + 225 + 605 - 5

= 282 + 605 - 5

= 887 - 5

= 882

\(\dfrac{x-1}{2020}+\dfrac{x-2}{2021}=\dfrac{x-3}{2022}+\dfrac{x-4}{2023}\)

`=>`\(\dfrac{x-1}{2020}+1+\dfrac{x-2}{2021}+1=\dfrac{x-3}{2022}+1\dfrac{x-4}{2023}+1\)

`=>`\(\dfrac{x-1+2020}{2020}+\dfrac{x-2+2021}{2021}=\dfrac{x-3+2022}{2022}+\dfrac{x-4+2023}{2023}\)

`=>`\(\dfrac{x+2019}{2020}+\dfrac{x+2019}{2021}-\dfrac{x+2019}{2022}-\dfrac{x+2019}{2023}=0\)

`=>`\(\left(x+2019\right)\left(\dfrac{1}{2020}+\dfrac{1}{2021}+\dfrac{1}{2022}+\dfrac{1}{2023}\right)=0\)

`=>x+2019=0`

`=>x=-2019`

Vậy `x=-2019`

S = \(\dfrac{4n+5}{2n-3}\)  (n ϵ Z)

S = \(\dfrac{2\left(2n-3\right)+11}{\left(2n-3\right)}\)

S = 2 + \(\dfrac{11}{2n-3}\) 

S nguyên ⇔2n-3 ϵ Ư(11)  = {-11; -1; 1; 11}

⇔ n ϵ { -4; 1; 2; 7}

Ta có \(S=\dfrac{4n+5}{2n-3}=\dfrac{2\left(2n-3\right)+11}{2n-3}=2+\dfrac{11}{2n-3}\)

Để S lầ số nguyên =>\(\dfrac{11}{2n-3}\) nguyên

=> \(11⋮2n-3\) hay 2n-3 \(\in\) Ư(11)

  =>2n-3\(\in\) {1;11;-1;-11}

n\(\in\) {2;7;1;-4}