Ta có :1.2+2.4+3.6+4.8+5.10/3.4+6.8+9.12+12.16+15.20=[1.2(1+4+9+...+25)]/[3.4(1+4+9+16)]

=(1.2)/(3.4)=2/12=1/6

CHÚC BẠN HỌC TỐT!!

cho tớ nhé!!!!

=2+8+18+32+50/12+48+108+192+300

=110/660

=0.166666667\

a) 1 + 1/4 + 1/8 + 1/16

= 16/16 + 4/16 + 2/16 + 1/16

= 23/16

b) 2 - 1/8 - 1/12 - 1/16

= 96/48 - 6/46 - 4/48 - 3/48

= 83/48

c) 4/99 × 18/5 : 12/11 + 3/5

= 8/55 : 12/11 + 3/5

= 2/15 + 3/5

= 2/15 + 9/15

= 11/15

d) (1 - 3/4) × (1 + 1/3) : (1 - 1/3)

= 1/4 × 4/3 : 2/3

= 1/3 : 2/3

= 2

1 + 1/4 + 1/8 + 1/16

= 16 + 4 + 2 + 1 / 16

=23/16

2-1/8-1/12-1/16

= 96 - 6 - 4 - 3 / 48

= 83 / 48

4/99 x 18/5 : 12 /11 + 3/5

= 4/99 x 18/5 x 11/12 + 3/5

= 2/15 + 3/5

= 11/15

(1 - 3/4) x (1+1/3) : (1-1/3) 

= 1/4 x 4/3 : 2/3

= 1/4 x 4/3 x 3/2

= 1/2

\(16-8:4\times2+12\)

\(=16-2\times2+12\)

\(=16-4+12\)

\(=12+12\)

\(=24\)

Chúc bạn học tốt:>

  16 - 8 : 4 \(\times\) 2 + 12 

= 16 - 2 \(\times\) 2 + 12

=  16 - 4 + 12

=     12 + 12

= 24

16-8:4x2+12

=16-2x2+12

=16-4+12

=14+12

=26

a/ \(\dfrac{3-x}{12}=\dfrac{2x+2}{8}\)

\(< =>\dfrac{2\left(3-x\right)}{24}=\dfrac{3\left(2x+2\right)}{24}\)

\(< =>6-2x-6x-6=0\)

\(< =>-8x=0\)

\(< =>x=0\)

Vậy tập nghiệm.....

b/ \(\dfrac{x+3}{x-4}+\dfrac{x-3}{x+4}=\dfrac{2\left(x^2+12\right)}{x^2-16}\)

Tìm ĐKXĐ của pt là: \(x\ne\pm4\)  (làm tắt, bạn làm rõ ra nhé)

\(\dfrac{x+3}{x-4}+\dfrac{x-3}{x+4}=\dfrac{2\left(x^2+12\right)}{x^2-16}\)

\(< =>\dfrac{\left(x+3\right)\left(x+4\right)}{\left(x-4\right)\left(x+4\right)}+\dfrac{\left(x-3\right)\left(x-4\right)}{\left(x-4\right)\left(x+4\right)}=\dfrac{2\left(x^2+12\right)}{\left(x+4\right)\left(x-4\right)}\)

\(< =>x^2+3x+4x+12+x^2-3x-4x+12-2x^2-24=0\)

\(< =>0x=0\)

=> x có vô số nghiệm

Vậy ....

a) `(3-x)/12=(2x+2)/8`

`<=> (3-x)/12 =(x+1)/4`

`<=> 3-x=3(x+1)`

`<=>3-x=3x+3`

`<=> x=0`

Vậy `S={0}`.

b) ĐK: `x \ne \pm 4`

`(x+3)/(x-4)+(x-3)/(x+4)=(2(x^2+12))/(x^2-16)`

`<=> (x+3)(x+4)+(x-3)(x-4)=2(x^2+12)`

`<=> x^2+7x+12+x^2-7x+12=2x^2+24`

`<=> 0x=0`

Vậy PT có nghiệm với mọi x thỏa mãn điều kiện.

\(\dfrac{x+3}{x-4}+\dfrac{x-3}{x+4}=\dfrac{2\left(x^2+12\right)}{x^2-16}\)

⇔\(\dfrac{\left(x+3\right)\left(x+4\right)}{\left(x-4\right)\left(x+4\right)}+\dfrac{\left(x-3\right)\left(x-4\right)}{\left(x-4\right)\left(x+4\right)}=\dfrac{2\left(x^2+12\right)}{\left(x-4\right)\left(x+4\right)}\)

⇔\(\dfrac{x^2+4x+3x+12}{\left(x-4\right)\left(x+4\right)}+\dfrac{x^2-4x-3x+12}{\left(x-4\right)\left(x+4\right)}=\dfrac{2x^2+24}{\left(x-4\right)\left(x+4\right)}\)

⇔\(\dfrac{x^2+7x+12}{\left(x-4\right)\left(x+4\right)}+\dfrac{x^2-7x+12}{\left(x-4\right)\left(x+4\right)}=\dfrac{2x^2+24}{\left(x-4\right)\left(x+4\right)}\)

⇒ \(x^2+7x+12+x^2-7x+12=2x^2+24\)

⇔ \(2x^2+24=2x^2+24\)

⇔ \(2x^2-2x^2=24-24\)

⇔ x=0

\(\dfrac{4}{3.5}+\dfrac{8}{5.9}+\dfrac{12}{9.15}+...+\dfrac{32}{x\left(x+16\right)}=\dfrac{16}{15}\)

\(2.\left(\dfrac{2}{3.5}+\dfrac{4}{5.9}+\dfrac{6}{9.15}+..+\dfrac{16}{X.\left(X+16\right)}\right)=\dfrac{16}{15}\)

\(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{15}+...+\dfrac{1}{X}-\dfrac{1}{X+16}=\dfrac{8}{15}\)

\(\dfrac{1}{X+16}=\dfrac{1}{3}-\dfrac{8}{15}\)

\(\dfrac{1}{X+16}=\dfrac{-1}{5}\)

\(X+16=-5\)

\(X=-21\)

Giúp mình với mình đang vội!!

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