Bạn ơi xem lại đề là 1/(2x+1).(2x+3) hay là 2/(2x+1).(2x+3) nhé .Tại mình thử giải với 2/(2x+1).(2x+3) thì mới ra kết quả

1. 3y = 0

=> y = 0

2. 1+x = 0

<+ x = -1

3.

\(1-2t=0\)

\(\Leftrightarrow2t=1\)

\(\Leftrightarrow\dfrac{1}{2}\)

4. 2x +x + 3 =0

\(\Leftrightarrow3x+3=0\)

\(\Leftrightarrow x=-3\)

5.

\(25x-20=0\)

\(\Leftrightarrow25x=20\)

\(\Leftrightarrow x=\dfrac{4}{5}\)

7.

2x-3 = x+5

<=> 2x - x = 5+3

<=> x = 8

8.

x-8=2x+3

<=> x - 2x = 3+8

<=> -x = 11

<=> x = -11

9. 17-2x = 3x-5

<=> -2x-3x = -5-17

<=> -5x = -22

<=> x = \(\dfrac{22}{5}\)

10.

2x+x+22=0

<=> 3x+22=0

<=> 3x = -22

<=> x = \(\dfrac{-22}{3}\)

Mấy bài kia tự giải tương tự nhá!!!

11.

x-3=3x+5

<=> x-3x=5+3

<=> -2x=8

<=> x = -4

12. 15x + 1/2 = 0

<=> 15x = -1/2

<=> x = -1/30

13.

3x+1 = 7x-11

<=> 3x-7x = -11-1

<=> -4x=-12

<=> x = 3

14.

5-3x = 6x + 7

<=> -3x-6x = 7-12

<=> -9x = -5

<=> x = 5/9

15.

11-2x=x-1

<=> -2x-x = -1-11

<=> -3x = -12

<=> x = 4

16.

15 - 8x = 9-5x

<=> -8x+5x = 9 - 15

<=> -3x = -6

<=> x=2

17.

25-3x=4x-5

<=> -3x-4x = -5-25

<=> -7x = -30

<=> x = 30/7

18.

0,25x+1,5=0

<=> 0,25x = -1,5

<=> x = -6

19.

3x-5/6 = 1/2

<=> 3x - 5/6 = 3/6

<=> 3x-5=3

<=> 3x = 8

<=> x = 8/3

20.

-2x + 14 =0

<=> -2x = -14

<=> x = 7

\(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{1}{x.\left(2x+1\right)}=\dfrac{1}{10}\)

\(\Leftrightarrow\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+...+\dfrac{1}{2x.\left(2x+1\right)}=\dfrac{1}{20}\)

\(\Leftrightarrow\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{2x.\left(2x+1\right)}=\dfrac{1}{20}\)

\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{2x}-\dfrac{1}{2x+1}=\dfrac{1}{20}\)

\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{2x+1}=\dfrac{1}{20}\)

\(\Leftrightarrow\dfrac{1}{2x+1}=\dfrac{9}{20}\)

\(\Leftrightarrow2x+1=\dfrac{20}{9}\Leftrightarrow x=\dfrac{11}{18}\)

Em giải như XYZ olm em nhé

Sau đó em thêm vào lập luận sau:

\(x\) = \(\dfrac{11}{18}\)

Vì \(\in\) N* 

Vậy \(x\in\) \(\varnothing\)

Cảm ơn mn ạ.

a) (x + 2) . (x + 3) - (x - 2) . (x + 5) = 6                

=> (x . x + 3x + 2x + 2 . 3) - (x . x + 5x - 2x - 2 . 5) = 6

=> (x² + 5x + 6) - (x² + 3x - 10) = 6                        

=> x² + 5x + 6 - x² - 3x + 10 = 6

=> 2x +16 = 6       => 2x = -10        => x = -5 

b) (3x + 2) . (2x + 9) - (x + 2) . (6x + 1) = (x + 1) - (x - 6)     

  => (3x . 2x + 3x . 9 + 2 . 2x + 2 . 9) - (x . 6x + 1x + 2 . 6x + 2 .1) = x + 1 - x + 6

=> (6x² + 31x + 18) - (6x² + 13x + 2) = 7             

=> 6x² + 31x + 18 - 6x² - 13x - 2 = 7

=> 18x + 16 = 7  => 18x = 9  => x = 0,5

c) 3 . (2x - 1) . (3x - 1) - (2x - 3) . (9x - 1) = 0

=> 3(2x . 3x - 2x -3x + 1) - (2x . 9x - 2x -3 . 9x + 3) = 0

=> 3(6x² - 5x +1) - (18x² - 29x + 3) = 0

=> (18x² -15x + 1) -(18x² - 29x +3) = 0

=> 18x² - 15x +1 -18x² + 29x - 3 = 0

=> 14x = 0  => x = 0

a)(x+2)(x+3)-(x-2)(x+5)=6

x(x+3)+2(x+3)-x(x+5)+2(x+5)=6

x²+3x+2x+6-x²-5x+2x+10=6

(x²-x²)+(3x+2x-5x+2x)+(10+6)=6

2x+16=6

2x=6-16

2x=-10

x=-10/2

x=-5. Vậy x=-5

b)3x(2x+9)+2(2x+9)-x(6x+1)-2(6x+1)=x+1-x+6

6x²+27x+4x+18-6x²-x-12x-2=7

(6x²-6x²)+(27x+4x-x-12x)+(18-2)=7

18x+16=7

18x=7-16

x=-9/18=-1/2. Vậy x=-1/2

c)[3(3x-1)](2x-1)-(2x-3)(9x-1)=0

(9x-3)(2x-1)-(2x-3)(9x-1)=0

9x(2x-1)-3(2x-1)-2x(9x-1)+3(9x-1)=0

18x²-9x-6x+3-18x²+2x+27x-3=0

(18x²-18x²)+(27x+2x-6x-9x)+(3-3)=0

14x=0

x=0/14

x=0. Vậy x=0

a) (x + 2) . (x + 3) - (x - 2) . (x + 5) = 6                    => (x . x + 3x + 2x + 2 . 3) - (x . x + 5x - 2x - 2 . 5) = 6

=> (x² + 5x + 6) - (x² + 3x - 10) = 6                          

=> x² + 5x + 6 - x² - 3x + 10 = 6

=> 2x +16 = 6  => 2x = -10    => x = -5 

b) (3x + 2) . (2x + 9) - (x + 2) . (6x + 1) = (x + 1) - (x - 6)

=> (3x . 2x + 3x . 9 + 2 . 2x + 2 . 9) - (x . 6x + 1x + 2 . 6x + 2 .1) = x + 1 - x + 6

=> (6x² + 31x + 18) - (6x² + 13x + 2) = 7

=> 6x² + 31x + 18 - 6x² - 13x - 2 = 7

=> 18x + 16 = 7  => 18x = -9  => x = -0,5

c) 3 . (2x - 1) . (3x - 1) - (2x - 3) . (9x - 1) = 0

=> 3(2x . 3x - 2x - 3x + 1) - (2x . 9x - 2x - 3. 9x + 3) = 0

=> 3(6x² - 5x + 1) - (18x² - 29x + 3) = 0

=> 18x² - 15x + 3 - 18x² + 29x -3 = 0

=> 14x = 0  => x = 0.

\(a,\Leftrightarrow x^3=\dfrac{20}{3}\Leftrightarrow x=\sqrt[3]{\dfrac{20}{3}}\\ b,\Leftrightarrow x-1=9\Leftrightarrow x=10\\ c,\Leftrightarrow\left[{}\begin{matrix}x-1=5\\x-1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\\ d,\Leftrightarrow2x+1=5\Leftrightarrow x=2\\ e,\Leftrightarrow2x-4=4\Leftrightarrow x=4\)

Câu a) xem lại đề giùm nhé em

b) \(\left(x-1\right)^3=9^3\)

\(x-1=9\)

\(x=10\)

Vậy \(x=10\)

c) \(\left(x-1\right)^2=25\)

\(x-1=5\) hoặc \(x-1=-5\)

* \(x-1=5\)

\(x=6\)

* \(x-1=-5\)

\(x=-4\)

Vậy \(x=-4\); \(x=6\)

d) \(\left(2x+1\right)^3=125\)

\(\left(2x+1\right)^3=5^3\)

\(2x+1=5\)

\(2x=4\)

\(x=2\)

Vậy \(x=2\)

e) Sửa đề: \(\left(2x+4\right)^3=64\)

\(\left(2x+4\right)^3=4^3\)

\(2x+4=4\)

\(2x=0\)

\(x=0\)

Vậy \(x=0\)

\(a,\frac{1}{2}x+\frac{5}{2}=\frac{7}{2}x-\frac{3}{4}\)

\(\Leftrightarrow\frac{1}{2}x+\frac{5}{2}-\frac{7}{2}x=-\frac{3}{4}\)

\(\Leftrightarrow\frac{1}{2}x-\frac{7}{2}x+\frac{5}{2}=-\frac{3}{4}\)

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

\(\Leftrightarrow-3x=-\frac{13}{4}\)

\(\Leftrightarrow x=-\frac{13}{4}:(-3)=-\frac{13}{4}:\frac{-3}{1}=-\frac{13}{4}\cdot\frac{-1}{3}=\frac{13}{12}\)

\(b,\frac{2}{3}x-\frac{2}{5}=\frac{1}{2}x-\frac{1}{3}\)

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

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

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

\(\Leftrightarrow\frac{1}{6}x=\frac{1}{15}\)

\(\Leftrightarrow x=\frac{1}{15}:\frac{1}{6}=\frac{1}{15}\cdot6=\frac{6}{15}=\frac{2}{5}\)

\(c,\frac{1}{3}x+\frac{2}{5}(x+1)=0\)

\(\Leftrightarrow\frac{1}{3}x+\frac{2}{5}x+\frac{2}{5}=0\)

\(\Leftrightarrow\frac{11}{15}x=-\frac{2}{5}\)

\(\Leftrightarrow x=-\frac{6}{11}\)

d,e,f Tương tự

\(\dfrac{5}{x}+1+\dfrac{4}{x}+1=\dfrac{3}{-13}\\ \Rightarrow\dfrac{9}{x}+2=-\dfrac{3}{13}\\ \Rightarrow\dfrac{9}{x}=-\dfrac{59}{13}\\ \Rightarrow x=-\dfrac{207}{59}\)

a. \(\dfrac{5}{x+1}+\dfrac{4}{x+1}=\dfrac{-3}{13}\)

ĐKXĐ: x ≠ -1

⇔ \(\dfrac{65}{13\left(x+1\right)}+\dfrac{52}{13\left(x+1\right)}=\dfrac{-3\left(x+1\right)}{13\left(x+1\right)}\)

⇔ 65 + 52 = -3(x + 1)

⇔ 117 = -3x - 3

⇔ 117 + 3 = -3x

⇔ 120 = -3x 

⇔ x = \(\dfrac{120}{-3}=-40\) (TM)

b. -x + 2 + 2x + 3 + x + \(\dfrac{1}{4}\) + 2x + \(\dfrac{1}{6}\) = \(\dfrac{8}{3}\)

⇔ -x + 2x + x + 2x = \(\dfrac{8}{3}-\dfrac{1}{6}-\dfrac{1}{4}-3-2\)

⇔ 4x = -2,75

⇔ x = \(\dfrac{-2,75}{4}=\dfrac{-11}{16}\)

c. \(\dfrac{3}{2x+1}+\dfrac{10}{4x+2}-\dfrac{6}{6x+2}\) = \(\dfrac{12}{26}\)

⇔  \(\dfrac{3}{2x+1}+\dfrac{10}{2\left(2x+1\right)}-\dfrac{6}{2\left(3x+1\right)}=\dfrac{12}{26}\)

⇔ \(\dfrac{312\left(3x+1\right)}{104\left(2x+1\right)\left(3x+1\right)}\) + \(\dfrac{520\left(3x+1\right)}{104\left(2x+1\right)\left(3x+1\right)}\) - \(\dfrac{312\left(2x+1\right)}{104\left(2x+1\right)\left(3x+1\right)}\)

= \(\dfrac{48\left(2x+1\right)\left(3x+1\right)}{104\left(2x+1\right)\left(3x+1\right)}\)

⇔ 312(3x +1) + 520(3x + 1) - 312(2x + 1) = 48(2x + 1)(3x + 1)

⇔ 936x + 312 + 1560x + 520 - 624x - 312 = (96x + 48)(3x + 1)

⇔ 936x + 312 + 1560x + 520 - 624x - 312 = 288x² + 96x + 144x + 48

⇔ 936x + 1560x - 624x - 96x - 144x - 288x² = 48 - 312 - 520 + 312

⇔ 1632x - 288x² = -472

⇔ -288x² + 1632x + 472 = 0 (Tự giải tiếp, dùng phương pháp tách hạng tử)

⇔ x = 5,942459684 \(\approx\) 6

c: Ta có: \(\dfrac{3}{2x+1}+\dfrac{10}{4x+2}-\dfrac{6}{6x+3}=\dfrac{12}{26}\)

\(\Leftrightarrow\dfrac{3}{2x+1}+\dfrac{5}{2x+1}-\dfrac{2}{2x+1}=\dfrac{6}{13}\)

\(\Leftrightarrow2x+1=13\)

hay x=6

NHANH LÊN MIK ĐANG CẦN GẤP NHA CÁC BAN

\(\frac{1}{6}\)(2x-3) = \(\frac{-1}{2}\)(x-\(\frac{1}{4}\))-\(\frac{2}{3}\)
\(\frac{1}{3}\)x - \(\frac{1}{2}\)=\(\frac{-1x}{2}\)-\(\frac{-1}{8}\)-\(\frac{2}{3}\)
\(\frac{1x}{3}\)-\(\frac{1}{2}\)=\(\frac{-1x}{2}\) - \(\frac{19}{24}\)
\(\frac{1x}{3}\) - \(\frac{1}{2}\) - \(\frac{-1x}{2}\) - \(\frac{19}{24}\) =0
\(\frac{5x}{6}\) - \(\frac{7}{24}\)=0
\(\frac{5x}{6}\) = \(\frac{7}{24}\)
x = \(\frac{7}{20}\)