Các bn giúp mình nhanh đk ko. Cảm ơn các bạn nhiều nha.

bạn cứ chia ra trường hợp là đc

a: =>-140x+1050-132+84x=100

=>-56x=100+132-1050=-818

=>x=409/28

b: \(\Leftrightarrow82-287x-15x+60=20\)

=>-302x+142=20

=>-302x=-122

=>x=61/151

c: =>165x-66+12x-32=15

=>177x=113

=>x=113/177

d: =>140-16x-77+22x=19

=>6x+63=19

=>6x=-44

=>x=-22/3

a/ \(\left(-19\right)+5+\left(-8\right)+19+\left(-3\right).\left(-2\right)^3\)

\(\left(-19\right)+5+\left(-8\right)+19+\left(-3\right).\left(-8\right)\)

\(\left(-19\right)+5+\left(-8\right)+19+24\)

\(\left(-14\right)+11+24\)

\(\left(-3\right)+24=21\)

b/ \(\left(-5\right)^3-\left(3^3+4\right).\left(-2\right)+\left(3-27\right):4\)

\(\left(-125\right)-\left(27+4\right).\left(-2\right)+\left(3-27\right):4\)

\(\left(-125\right)-\left(31\right).\left(-2\right)+\left(-24\right):\text{4}\)

\(\left(-125\right)-\left(-62\right)+\left(-6\right)\)

\(\left(-63\right)+\left(-6\right)=-69\)

c/ \(\left[504-\left(5^2.8+70\right):3^2+6\right]:\left(-20\right)\)

\(\left[504-\left(25.8+70\right):9+6\right]:\left(-20\right)\)

\(\left[504-270:9+6\right]:\left(-20\right)\)

\(\left[504-30+6\right]:\left(-20\right)\)

\(480:\left(-20\right)=-24\)

d/ \(\left(-7\right)-\left[\left(-19\right)+21\right].\left(-3\right)-\left[32+\left(-7\right)\right]\)

\(\left(-7\right)-2.\left(-3\right)-25\)

\(\left(-7\right)-\left(-6\right)-25\)

\(\left(-1\right)-25=-26\)

\(\)

a, |2x+1| - 19 = -7

|2x+1| = 12

TH1: 2x + 1 = 12

2x = 11

x = \(\frac{11}{2}\)

TH2: 2x + 1 = -12

2x = -13

x = \(\frac{-13}{2}\)

b, -28 - 7.|-3x+15| = -70

-7.|-3x+15| = -42

|-3x+15| = 6

TH1: -3x + 15 = 6

-3x = -9

x = 3

TH2: -3x + 15 = -6

-3x = -21

x = 7

d, 12 - 2( -x + 3)² = -38

-2( -x + 3)² = -50

( -x + 3)² = 25

( -x + 3)² = +- 5²

-x + 3 = +- 5

TH1: -x + 3 = 5

-x = 2

x = -2

TH2: -x + 3 = -5

-x = -8

x = 8

e, -20 + 3(3x+1)³ = -101

3(3x+1)³ = -81

(3x+1)³ = -27

(3x+ 1)³ = -3³

3x + 1 = -3

3x = -4

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

Chúc bạn học tốt !

hello

a, |2x+1|-19= -7

a: Đặt x/5=y/7=k

=>x=5k; y=7k

Ta có: xy=70

nên \(35k^2=70\)

\(\Leftrightarrow k^2=2\)

Trường hợp 1: \(k=\sqrt{2}\)

\(\Leftrightarrow x=5\sqrt{2};y=7\sqrt{2}\)

Trường hợp 2: \(k=-\sqrt{2}\)

\(\Leftrightarrow x=-5\sqrt{2};y=-7\sqrt{2}\)

b: Ta có: 5x=3y

nên x/3=y/5

Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{2x-y}{2\cdot3-5}=11\)

Do đó: x=33; y=55

c: 7x=5y=140

=>x=20; y=28

d: Ta có: 2x=3y

nên x/3=y/2

Áp dụng tính chất của dãy tỉ số bằng nhau, ta được;

\(\dfrac{x}{3}=\dfrac{y}{2}=\dfrac{x-2y}{3-2\cdot2}=\dfrac{-7}{-1}=7\)

Do đó: x=21; y=14

Bài 1:

+) Có: \(2^{12}\equiv1\left(mod13\right)\)

\(\left(2^{12}\right)^5\equiv1^5\equiv1\left(mod13\right)\)

=> \(2^{60}\cdot2^{10}\equiv1\cdot10\equiv10\left(mod13\right)\) (*)

+) Có: \(3^{12}\equiv1\left(mod13\right)\)

\(\left(3^{12}\right)^5\equiv1^5\equiv1\left(mod13\right)\)

\(\Rightarrow3^{60}\cdot3^{10}\equiv1\cdot3\equiv3\left(mod13\right)\) (**)

Từ (*); (**)

=> \(2^{70}+3^{70}\equiv10+3\equiv13\left(mod13\right)\)

hay \(2^{70}+3^{70}⋮13\left(đpcm\right)\)

Bài 2 : Làm tương tự '-,,,,

a/

$\frac{7}{4}+\frac{9}{16}=\frac{28}{16}+\frac{9}{16}=\frac{28+9}{16}=\frac{37}{16}$

b/

$\frac{-70}{27}+\frac{35}{9}=\frac{-70}{27}+\frac{105}{27}=\frac{35}{27}$

c/

$\frac{-4}{-5}+\frac{-2}{3}=\frac{4}{5}-\frac{2}{3}=\frac{12}{15}-\frac{10}{15}=\frac{2}{15}$

d/ Đề thiếu. Bạn xem lại.

i/

$\frac{-7}{2}+(\frac{-5}{-4}-\frac{4}{-9}$

$=\frac{-7}{2}+\frac{5}{4}+\frac{4}{9}$

$=\frac{-14}{4}+\frac{5}{4}+\frac{4}{9}$

$=\frac{-9}{4}+\frac{4}{9}=\frac{-81}{36}+\frac{16}{36}=\frac{-65}{36}$

k/

$\frac{1}{4}+\frac{1}{3}-\frac{1}{2}$

$=\frac{3}{12}+\frac{4}{12}-\frac{6}{12}=\frac{1}{12}$

m/

$\frac{-1}{9}-\frac{1}{7}-\frac{3}{4}$

$=-(\frac{1}{9}+\frac{1}{7}+\frac{3}{4})$

$=-(\frac{28}{252}+\frac{36}{252}+\frac{189}{252})$

$=\frac{-253}{252}$

n/

$\frac{11}{12}-(\frac{-3}{18}+\frac{5}{6}$

$=\frac{11}{12}+\frac{1}{6}+\frac{5}{6}$

$=\frac{11}{12}+1=\frac{23}{12}$

o/

$\frac{20}{17}+\frac{-11}{2}-\frac{30}{4}$

$=\frac{20}{17}-\frac{11}{2}-\frac{15}{2}$

$=\frac{20}{7}-13=\frac{-71}{7}$

v/

$\frac{1}{2}-\frac{3}{4}+\frac{5}{6}$

$=\frac{6}{12}-\frac{9}{12}+\frac{10}{12}$

$=\frac{6-9+10}{12}=\frac{7}{12}$

e/

$\frac{-2}{3}-\frac{1}{7}+(\frac{-6}{2})$

$=-(\frac{2}{3}+\frac{1}{7})-3$

$=-\frac{17}{21}-3=-\frac{80}{21}$

f/

$\frac{-4}{5}+\frac{3}{10}-\frac{2}{-15}$

$=\frac{-24}{30}+\frac{9}{30}+\frac{4}{30}$

$=\frac{-11}{30}$

g/

$=\frac{20}{21}+\frac{133}{21}-\frac{36}{21}$

$=\frac{117}{21}=\frac{39}{7}$

h/

$=\frac{1}{4}-\frac{14}{4}+\frac{3}{11}$

$=\frac{-13}{4}+\frac{3}{11}=\frac{-131}{44}$

y/

$=-(\frac{9}{4}+\frac{1}{7}+\frac{12}{7})$

$=-(\frac{9}{4}+\frac{13}{7})=-\frac{115}{28}$

\(\frac{2.6.1}{33.24}=\frac{12.1}{33.2.12}=\frac{1}{66}\)

\(\frac{21.4}{9.7.5.3}=\frac{7.3.4}{9.5.7.3}=\frac{4}{45}\)

P/s: Hoq chắc ((:

AB//CD

nên \(\widehat{B}+\widehat{C}=180^0\)

mà \(\widehat{B}-\widehat{C}=70^0\)

nên \(\widehat{B}=\dfrac{180^0+70^0}{2}=125^0\)

\(\widehat{C}=180^0-\widehat{B}=180^0-125^0=55^0\)

Ta có AB//CD

nên \(\widehat{D}+\widehat{A}=180^0\)

\(\Leftrightarrow\widehat{A}\cdot\dfrac{2}{7}+\widehat{A}=180^0\)

\(\Leftrightarrow\widehat{A}=180^0:\dfrac{9}{7}=140^0\)

\(\widehat{D}=\dfrac{2}{7}\cdot140^0=40^0\)