\(ĐK:x>0;x\ne4\\ B=\dfrac{\sqrt{x}+2}{\sqrt{x}}\cdot\dfrac{x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\dfrac{1}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}+2}\\ B=\dfrac{\sqrt{x}}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}+2}\\ B=\dfrac{x+2\sqrt{x}+\sqrt{x}+2+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\\ B=\dfrac{x+4\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}+4\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

Ta có: \(B=\dfrac{\sqrt{x}+2}{\sqrt{x}}\cdot\dfrac{x}{x-4}+\dfrac{1}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}+2}\)

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

\(=\dfrac{x+3\sqrt{x}+2+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

\(=\dfrac{x+4\sqrt{x}}{x-4}\)

tham khảo:

99/100

\(\left(x-2\right)\left(x-3\right)=\left(x-2\right)\left(x+1\right)\)

\(\Leftrightarrow x^2-5x+6=x^2-x-2\)

\(\Leftrightarrow-4x+8=0\)

\(\Leftrightarrow x=2\)

Vậy ...

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

\(x^2-5x+6=x^2-x-2\)

\(x^2-x^2-5x+x=-6-2\)

\(-4x=-8\)

\(x=2\)

\(\text{Vậy x=2}\)

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

ĐKXĐ : x ≠ 1 ; x ≠ 2 ; x ≠ 3 ; x ≠ 6

pt <=> \(\frac{x^2-5x+6}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}+\frac{2x^2-8x+6}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}+\frac{3x^2-9x+6}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}=\frac{6}{x-6}\)

<=> \(\frac{6x^2-22x+18}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}=\frac{6}{x-6}\)

=> \(\left(x-6\right)\left(6x^2-22x+18\right)=6\left(x-1\right)\left(x-2\right)\left(x-3\right)\)

(bạn tự khai triển rút gọn nhé)

<=> \(6x^3-58x^2+150x-108=6x^3-36x^2+66x-36\)

<=>\(6x^3-58x^2+150x-108-6x^3+36x^2-66x+36=0\)

<=> \(-22x^2+84x-72=0\)

<=> \(11x^2-42x+36=0\)

(pt này lên lớp 9 mới học nên mình dừng tại đây)

`a)C=((2x^2+1)/(x^3-1)-1/(x-1)):(1-(x^2-2)/(x^2+x+1))`

`ĐK:x ne 1`

`C=((2x^2+1-x^2-x-1)/(x^3-1)):((x^2+x+1-x^2+2)/(x^2+x+1))`

`C=((x^2-x)/(x^3-1)):((x+3)/(x^2+x+1))`

`C=x/(x^2+x+1)*(x^2+x+1)/(x+3)`

`C=x/(x+3)`

`b)|1-x|+2=3(x+1)`

`<=>|1-x|+2=3x+3`

`<=>|1-x|=3x+1(x>=-1/3)`

`**1-x=3x+1`

`<=>4x=0<=>x=0(tmđk)`

`**x-1=3x+1`

`<=>2x=-2`

`<=>x=-1(l)`

Thay `x=0` vào C

`=>C=0`

`c)C in ZZ`

`=>x vdots x+3`

`=>x+3-3 vdots x+3`

`=>3 vdots x+3`

`=>x+3 in Ư(3)={+-1,+-3}`

`=>x in {-2,-4,0,-6}`

`d)|C|>C`

Mà `|C|>=0`

`=>C<0`

`<=>x/(x+3)<0`

Để 1 p/s `<=0` thì tử và mẫu trái dấu mà `x<x+3`

`=>` \(\begin{cases}x<0\\x+3>0\\\end{cases}\)

`<=>` \(\begin{cases}x>-3\\x<0\\\end{cases}\)

`<=>-3<x<0`

\(\dfrac{2}{30}\)

`(1 xx 2 xx 3 xx 4)/(3 xx 4 xx 5 xx 6)`

`= (1 xx 2)/(5 xx 6)`

`= 2/12`

`= 1/6`

Bổ sung:

`(1 xx 2)/(5 xx 6)`

`= 2/30`

`= 1/15`