a: =-1/2+2/3=-3/6+4/6=1/6

b: =-15/20-16/20=-31/20

c: =-5/9+5/12=-60/108+45/108=-15/108=-5/36

\(a.\dfrac{1}{-2}=\dfrac{-1}{2}\)

\(\dfrac{-3}{6}+\dfrac{4}{6}=\dfrac{1}{6}\)

\(b.\dfrac{-15}{20}-\dfrac{16}{20}=\dfrac{-31}{20}\)

\(c.\dfrac{-5}{9}+\dfrac{5}{12}=\dfrac{-60}{108}+\dfrac{45}{108}=\dfrac{-15}{108}\)

a) 3/13+4/7/26+1/1/2 va 4/3+4/37+55/111

3/13+111/26+3/2

=6

* 4/3+4/37+55/111

=55/37 =1,486

=> a> b

Lik-e ung ho nha

`#3107.101107`

1.

`a,`

`2 - 2/3 + 2,5 + 1/3 + 3 + 1/2`

`= (2 + 2,5 + 3) - (2/3 - 1/3 - 1/2)`

`= 7,5 - (-1/6)`

`= 7,5 + 1/6`

`= 23/3`

`b,`

`9/10 - (6/5 * 3/2 + 7/4)`

`= 9/10 - (18/10 + 7/4)`

`= 9/10 - 18/10 - 7/4`

`= -9/10 - 7/4`

`= -53/20`

`a)A=[2\sqrt{3}+2-2\sqrt{3}+2]/[(2\sqrt{3}-2)(2\sqrt{3}+2)]`

   `A=4/[12-4]=1/2`

Với `x > 0,x ne 1` có:

`B=[x-2\sqrt{x}+1]/[\sqrt{x}(\sqrt{x}-1)]`

`B=[(\sqrt{x}-1)^2]/[\sqrt{x}(\sqrt{x}-1)]=[\sqrt{x}-1]/\sqrt{x}`

`b)B=2/5A`

`=>[\sqrt{x}-1]/\sqrt{x}=2/5 . 1/2`

`<=>5\sqrt{x}-5=\sqrt{x}`

`<=>\sqrt{x}=5/4`

`<=>x=25/16` (t/m)

Vì  Suy ra a > 5 (1)

Ta lại có 0 < a < b nên 

Hay , suy ra a < 10 (2)

Từ (1) và (2) ta có a ∈ {6;7;8;9}

Nếu a = 6 thì  nên b = 30

Nếu a = 7 thì  suy ra b = 17,5 (loại)

Nếu a = 8 thì  suy ra b ≈ 13,3 (loại)

Nếu a = 9 thì  suy ra b = 11,25 (loại)

Vậy chỉ có một cách viết là 

Lời giải:
a.

a. $(x-1)(x+2)-(x-3)(x+1)=5x-3$

$\Leftrightarrow (x^2+x-2)-(x^2-2x-3)=5x-3$

$\Leftrightarrow 3x+1=5x-3$

$\Leftrightarrow 4=2x$

$\Leftrightarrow x=2$

b.

$(2x-1)(x+3)-(x-2)(x+3)=3x+1$

$\Leftrightarrow (2x^2+5x-3)-(x^2-4)=3x+1$

$\Leftrightarrow x^2+5x+1=3x+1$

$\Leftrightarrow x^2+2x=0$

$\Leftrightarrow x(x+2)=0$

$\Leftrightarrow x=0$ hoặc $x=-2$

c.

$x^2(x-1)-x(x-1)(x+1)=0$

$\Leftrightarrow x^2(x-1)-(x^2+x)(x-1)=0$

$\Leftrightarrow (x-1)[x^2-(x^2+x)]=0$

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

$\Leftrightarrow x-1=0$ hoặc $-x=0$

$\Leftrightarrow x=1$ hoặc $x=0$

d.

$4x(x-5)-(2x-3)(2x+3)=9$

$\Leftrightarrow 4x^2-20x-(4x^2-9)=9$

$\Leftrightarrow -20x=0$

$\Leftrightarrow x=0$

a: Ta có: \(\left(x-1\right)\left(x+2\right)-\left(x-3\right)\left(x+1\right)=5x-3\)

\(\Leftrightarrow x^2+2x-x-2-x^2-x+3x+3-5x+3=0\)

\(\Leftrightarrow-2x+4=0\)

\(\Leftrightarrow2x=4\)

hay x=2

b: Ta có: \(\left(2x-1\right)\left(x+3\right)-\left(x-2\right)\left(x+2\right)=3x+1\)

\(\Leftrightarrow2x^2+6x-x-3-x^2+4-3x-1=0\)

\(\Leftrightarrow x^2+2x=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)

c: Ta có: \(x^2\left(x-1\right)-x\left(x-1\right)\left(x+1\right)=0\)

\(\Leftrightarrow x\left(x-1\right)\left(x-x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)

d: Ta có: \(4x\left(x-5\right)-\left(2x-3\right)\left(2x+3\right)=9\)

\(\Leftrightarrow4x^2-20x-4x^2+9=9\)

hay x=0

Bài 1: 

b) ĐKXĐ: \(x\ne3\)

Ta có: \(\dfrac{3-x}{20}=\dfrac{-5}{x-3}\)

\(\Leftrightarrow\dfrac{x-3}{-20}=\dfrac{-5}{x-3}\)

\(\Leftrightarrow\left(x-3\right)^2=100\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=10\\x-3=-10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=13\left(nhận\right)\\x=-7\left(nhận\right)\end{matrix}\right.\)

Vậy: \(x\in\left\{13;-7\right\}\)