=-37,06

a)\(\frac{X}{5}=\frac{5}{6}+\frac{-19}{30}\)

   \(\frac{X}{5}=\frac{1}{5}\)

Vậy \(X=1\)

b)\(X-\frac{41}{5}=\frac{-2}{3}\)

                   \(X=\frac{-2}{3}+\frac{41}{5}\)

                   \(X=\frac{113}{15}\)

Vậy \(X=\frac{113}{15}\)

c)\(\frac{31}{5}-X=\frac{11}{3}+\frac{7}{10}\)

    \(\frac{31}{5}-X=\frac{131}{30}\)

                   \(X=\frac{31}{5}-\frac{131}{30}\)

                    \(X=\frac{11}{6}\)

Vậy \(X=\frac{11}{6}\)

d)\(\frac{9}{X}=\frac{2}{5}+\frac{-7}{20}\)

   \(\frac{9}{X}=\frac{1}{20}\)

     \(X=9:\frac{1}{20}\)

     \(X=180\)

Vậy \(X=180\)

Hc tốt

Câu 1 : 

a, \(\frac{3\left(2x+1\right)}{4}-\frac{5x+3}{6}=\frac{2x-1}{3}-\frac{3-x}{4}\)

\(\Leftrightarrow\frac{6x+3}{4}+\frac{3-x}{4}=\frac{2x-1}{3}+\frac{5x+3}{6}\)

\(\Leftrightarrow\frac{5x+6}{4}=\frac{9x+1}{6}\Leftrightarrow\frac{30x+36}{24}=\frac{36x+4}{24}\)

Khử mẫu : \(30x+36=36x+4\Leftrightarrow-6x=-32\Leftrightarrow x=\frac{32}{6}=\frac{16}{3}\)

tương tự 

\(\frac{19}{4}-\frac{2\left(3x-5\right)}{5}=\frac{3-2x}{10}-\frac{3x-1}{4}\)

\(< =>\frac{19.5}{20}-\frac{8\left(3x-5\right)}{20}=\frac{2\left(3-2x\right)}{20}-\frac{5\left(3x-1\right)}{20}\)

\(< =>95-24x+40=6-4x-15x+5\)

\(< =>-24x+135=-19x+11\)

\(< =>5x=135-11=124\)

\(< =>x=\frac{124}{5}\)

Bài 1:

a) \(x.\dfrac{3}{4}=\dfrac{9}{14}\)

\(\Rightarrow x=\dfrac{9}{14}:\dfrac{3}{4}\)

\(\Rightarrow x=\dfrac{6}{7}\)

b) \(x:\dfrac{5}{9}=\dfrac{3}{10}\)

\(\Rightarrow x=\dfrac{3}{10}.\dfrac{5}{9}\)

\(\Rightarrow x=\dfrac{1}{6}\)

help me

  Bài 1: 

1; ( - 35) : (-7) = 5

2; (- 42) : 21 =   - 2

3; 45 : (-9) = -5

4; 18 : 9 = 2

5; (- 30) : (- 15) = 2

6; 0 : 18 = 0

7; 0 : (-13) = 0

8; 44 : (-4) = - 11

9; - 55 : 11 = - 5

10; 46 : 23 = 2

Mn giúp mh với đg cần gấp lắm ln🥺🥺

`Answer:`

a. \(x^3+6x^2+12=19\)

\(\Leftrightarrow x^3+6x^2+12x-19=0\)

\(\Leftrightarrow x^3-x^2+7x^2-7x+19x-19=0\)

\(\Leftrightarrow x^2.\left(x-1\right)+7x\left(x-1\right)+19\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^2+7x+19\right)=0\)

Ta có \(x^2+7x+19=x^2+2x.3,5+12,25+6,75=\left(x+3,5\right)^2+6,75>0\)

\(\Rightarrow x-1=0\Leftrightarrow x=1\)

b. \(5\left(x+9\right)^2.\left(x-4\right)^3-10\left(x+9\right)^3.\left(x-4\right)^2=0\)

\(\Leftrightarrow5\left(x+9\right)^2.\left(x-4\right)^2.[x-4-2\left(x+9\right)]=0\)

\(\Leftrightarrow\left(x+9\right)^2.\left(x-4\right)^2.\left(x-4-2x-18\right)=0\)

\(\Leftrightarrow\left(x+9\right)^2.\left(x-4\right)^2.\left(-x-22\right)=0\)

\(\Leftrightarrow\left(x+9\right)^2=0\) hoặc \(\left(x-4\right)^2=0\) hoặc \(-x-22=0\)

\(\Leftrightarrow x+9=0\) hoặc \(x-4=0\) hoặc \(-x=22\)

\(\Leftrightarrow x=-9\) hoặc \(x=4\) hoặc \(x=-22\)

c. \(\left(2x+3\right)^2+\left(x-2\right)^2-2\left(2x+3\right)\left(x-2\right)\)

\(=\left(2x+3\right)^2-2\left(2x+3\right)\left(x-2\right)+\left(x-2\right)^2\)

\(=\left(2x+3-x+2\right)^2\)

\(=\left(x+5\right)^2\)

( Mik làm mấy phần mà bạn dưới chưa làm)

11) xy+x+y=9

\(\Leftrightarrow\) xy+x+y+1=9+1

\(\Leftrightarrow\left(xy+x\right)+\left(y+1\right)\)=10

\(\Leftrightarrow x\left(y+1\right)+\left(y+1\right)=10\)

\(\Leftrightarrow\) (x+1)(y+1)=10=1.10=10.1=-1.-10=-10.-1=2.5=5.2=-2.-5=-5.-2

\(\Rightarrow\) TH1: x+1=1 ; y+1=10

\(\Leftrightarrow x=0;y=9\)

TH2: x+1=10;y+1=1

\(\Leftrightarrow\)x=9;y=0

TH3: x+1=-1;y+1=-10

\(\Leftrightarrow\) x=-2;y=-11

...........

Vậy:........

( Bạn tự làm nốt chứ dài quá, mik chỉ hướng dẫn cách làm bài thôi)

1) -x = -7

=> x = 7

2) - x = 17

=> x = - 17

3) |x| = 17

=> x = ±17

4) -(-x) = |-17|

=> x = 17

5) - 19 - x = 17

=> - x = 17 + 19

=> x = - 36

6) - 19 - x = - 17

=> - x = - 17 + 19

=> -x = 2

=> x = - 2

7) - 5 - (10 - x) = 7

=> - 5 - 10 + x = 7

=> - 15 + x = 7

=> x = 7 + 15

=> x = 22

8) |x + 3| + 7 = 12

=> |x + 3| = 12 - 7

=> |x + 3| = 5

=> x + 3 = 5 hoặc x + 3 =- 5

=> x = 2 hoặc x = - 8

9) 2 - |x - 2| = x

=> - |x - 2| - x = - 2

TH1: x >= 2

- (x - 2) - x = - 2

=> - x + 2 - x =- 2

=> - 2x = - 4

=> x = 2 (nhận)

TH2: x < 2

-[-(x - 2)] - x = - 2

=> x - 2 - x = - 2

=> 0x = 0 (vô số nghiệm)