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

1: \(A=\frac{a\sqrt{a}-1}{a-\sqrt{a}}-\frac{a\sqrt{a}+1}{a+\sqrt{a}}+\left(\sqrt{a}-\frac{1}{\sqrt{a}}\right)\cdot\left(\frac{\sqrt{a}+1}{\sqrt{a}-1}+\frac{\sqrt{a}-1}{\sqrt{a}+1}\right)\)

ĐKXĐ: a>0; a<>1

Ta có: \(\frac{a\sqrt{a}-1}{a-\sqrt{a}}-\frac{a\sqrt{a}+1}{a+\sqrt{a}}\)

\(=\frac{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}-\frac{\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}+1\right)}\)

\(=\frac{a+\sqrt{a}+1-\left(a-\sqrt{a}+1\right)}{\sqrt{a}}=\frac{2\sqrt{a}}{\sqrt{a}}=2\)

Ta có: \(\left(\frac{\sqrt{a}+1}{\sqrt{a}-1}+\frac{\sqrt{a}-1}{\sqrt{a}+1}\right)\)

\(=\frac{\left(\sqrt{a}+1\right)^2+\left(\sqrt{a}-1\right)^2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\)

\(=\frac{a+2\sqrt{a}+1+a-2\sqrt{a}+1}{a-1}\)

\(=\frac{2a+2}{a-1}\)

Ta có: \(A=\frac{a\sqrt{a}-1}{a-\sqrt{a}}-\frac{a\sqrt{a}+1}{a+\sqrt{a}}+\left(\sqrt{a}-\frac{1}{\sqrt{a}}\right)\cdot\left(\frac{\sqrt{a}+1}{\sqrt{a}-1}+\frac{\sqrt{a}-1}{\sqrt{a}+1}\right)\)

\(=2+\frac{a-1}{\sqrt{a}}\cdot\frac{2a+2}{a-1}=2+\frac{2a+2}{\sqrt{a}}=\frac{2a+2\sqrt{a}+2}{\sqrt{a}}\)

2: A=7

=>\(\frac{2a+2\sqrt{a}+2}{\sqrt{a}}=7\)

=>\(2a+2\sqrt{a}+2=7\sqrt{a}\)

=>\(2a-5\sqrt{a}+2=0\)

=>\(\left(2\sqrt{a}-1\right)\left(\sqrt{a}-2\right)=0\)

=>\(\left[\begin{array}{l}2\sqrt{a}-1=0\\ \sqrt{a}-2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}\sqrt{a}=\frac12\\ \sqrt{a}=2\end{array}\right.\Rightarrow\left[\begin{array}{l}a=\frac14\left(nhận\right)\\ a=4\left(nhận\right)\end{array}\right.\)

3: A>6

=>\(\frac{2a+2\sqrt{a}+2}{\sqrt{a}}>6\)

=>\(\frac{a+\sqrt{a}+1}{\sqrt{a}}>3\)

=>\(\frac{a+\sqrt{a}+1-3\sqrt{a}}{\sqrt{a}}>0\)

=>\(\frac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}}>0\) (luôn đúng với mọi a thỏa mãn ĐKXĐ)

a: \(1\frac37-\frac78+2\frac45:2\)

\(=\frac{10}{7}-\frac78+\left(2+\frac45\right):2\)

\(=\frac{80}{56}-\frac{49}{56}+1+\frac25=\frac{31}{56}+\frac75=\frac{155+392}{280}=\frac{547}{280}\)

b: \(812\times X+188\times X=10000\)

=>\(X\times\left(812+188\right)=10000\)

=>\(1000\times X=10000\)

=>X=10000:1000=10

`#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: \(x\in\left\{13;-7\right\}\)