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

a) \(\dfrac{25}{37}\times\dfrac{18}{29}+\dfrac{18}{29}\times\dfrac{12}{37}\)

\(=\dfrac{18}{29}\times\left(\dfrac{25}{37}+\dfrac{12}{37}\right)\)

\(=\dfrac{18}{29}\times\dfrac{37}{37}\)

\(=\dfrac{18}{29}\times1\)

\(=\dfrac{18}{29}\)

b) \(\dfrac{31}{85}\times\dfrac{11}{19}+\dfrac{31}{85}\times\dfrac{12}{19}-\dfrac{42}{19}\times\dfrac{31}{85}\)

\(=\dfrac{31}{85}\times\left(\dfrac{11}{19}+\dfrac{12}{19}-\dfrac{42}{19}\right)\)

\(=\dfrac{31}{85}\times\dfrac{-19}{19}\)

\(=\dfrac{31}{85}\times-1\)

\(=-\dfrac{31}{85}\)

c) \(\dfrac{16}{53}:\dfrac{17}{9}-\dfrac{16}{53}:\dfrac{17}{8}\)

\(=\dfrac{16}{53}:\left(\dfrac{9}{17}-\dfrac{8}{17}\right)\)

\(=\dfrac{16}{53}:\dfrac{1}{17}\)

\(=\dfrac{16}{901}\)

c) \(\dfrac{1}{5}\times\dfrac{12}{31}\times\dfrac{4}{3}+\dfrac{19}{31}\times\dfrac{4}{15}\)

\(=\dfrac{4}{15}\times\dfrac{12}{31}+\dfrac{19}{31}\times\dfrac{4}{15}\)

\(=\dfrac{4}{15}\times\left(\dfrac{12}{31}+\dfrac{19}{31}\right)\)

\(=\dfrac{4}{15}\times\dfrac{31}{31}\)

\(=\dfrac{4}{15}\times1\)

\(=\dfrac{4}{15}\)

a: =18/29*(25/37+12/37)

=18/29

b: =31/85(11/19+12/19-42/19)

=-31/85

c; =16/53(9/17+8/17)=16/53

d: =4/15(12/31+19/31)=4/15

Tìm số nguyên x, biết:
1) -16 + 23 + x = - 16

7+x=-16

    x=-16-7

    x=-23
2) 2x – 35 = 15

2x=15+35

2x=50

  x=50:2

  x=25
3) 3x + 17 = 12

3x=12-17

3x=-5

  x=-5/3
4) (2x – 5) + 17 = 6

2x-5=6-17

2x-5=-11

2x=-11+5

2x=-6

  x=-6:2

  x=-3
5) 10 – 2(4 – 3x) = -4

2(4-3x)=10-(-4)

2(4-3x)=14

4-3x=14:2

4-3x=7

3x=4-7

3x=-3

  x=-3:3

  x=-1
6) - 12 + 3(-x + 7) = -18

3(-x+7)=-18-(-12)

3(x+7)=-6

x+7=-6:3

x+7=-2

    x=-2-7

    x=-9

tự đi mà làm

Giúp mình với mình đang vội!!

cứ bình tĩnh bạn ơi

$\frac{4^{10}{.9}^6+3^{12}{.8}^5}{}$

$\frac{6^{13}.4-2^{16}{.3}^{12}}{}$

$=$

$\frac{{\left(2^2\right)}^{10}.{\left(3^2\right)}^6+3^{12}.{\left(2^3\right)}^5}{{\left(2.3\right)}^{13}{.2}^2-2^{16}{.3}^{12}}$

$=$

$\frac{2^{20}{.3}^{12}+3^{12}{.2}^{15}}{}$

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\

\(\frac{3}{4}.x+\frac{12}{16}=\frac{3}{4}\)

\(\frac{3}{4}.x+\frac{3}{4}=\frac{3}{4}\)

\(\frac{3}{4}.x=\frac{3}{4}-\frac{3}{4}\)

\(\frac{3}{4}.x=\frac{0}{4}=0\)

\(x=0:\frac{3}{4}\)

\(x=0\)

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