\(C=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}.....\dfrac{48}{49}.\dfrac{49}{50}=\dfrac{1}{50}\)

\(\dfrac{x+2}{x-2}-\dfrac{2}{x^2-2x}=\dfrac{1}{x}\left(đk:x\ne0,x\ne2\right)\)

\(\Leftrightarrow\dfrac{\left(x+2\right)x-2}{x\left(x-2\right)}=\dfrac{x^2-2x}{x\left(x-2\right)}\)

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

\(\Leftrightarrow4x=2\Leftrightarrow x=\dfrac{1}{2}\)

Cho mình sửa lại nhé:

\(\dfrac{x+2}{x-2}-\dfrac{2}{x^2-2x}=\dfrac{1}{x}\left(đk:x\ne0,x\ne2\right)\)

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

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

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

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

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

Đề ko rõ ràng \(\sqrt{x^2}+x+\dfrac{1}{4}\) hay \(\sqrt{x^2+x+\dfrac{1}{4}}\)??

m??

Lời giải:

$\frac{1}{x}+\frac{1}{y}+\frac{2}{x+y}=\frac{x+y}{xy}+\frac{2}{x+y}$

$=x+y+\frac{2}{x+y}$

$=\frac{x+y}{2}+\frac{x+y}{2}+\frac{2}{x+y}$

$\geq \frac{x+y}{2}+2\sqrt{\frac{x+y}{2}.\frac{2}{x+y}}$ (áp dụng BDT Cô-si)

$\geq \frac{2\sqrt{xy}}{2}+2=\frac{2}{2}+2=3$

Vậy ta có đpcm

Dấu "=" xảy ra khi $x=y=1$

Đề trước đó: 

(x-7)(x+1)-(x-3)^2=(3x-5)(3x+5)-(3x+1)^2+(x-2)^2-x

<=>x^2+x-7x-7-x^2+6x-9=9x^2-25-9x^2-6x-1+x^2-4x+4-x

<=>x^2-11x-6=0

<=>x^2-2x. 11/2 + 121/4-145/4=0

<=>(x-11/2)^2=145/4

<=>|x-11/2|=căn(145)/2

<=>x=[11+-căn(145)]/2

có [x-y]2=1

suy ra [x-y]mũ 2= 1 mũ 2

suy ra x-1=1

x=1+1

x=2

x = 2 nha bạn

đáp án :
(x-1)²= 1
(x-1)² = 1²

x-1=1
x= 1+1
x=2 
vậy x = 2

`(x+3)(x^2-5x+8)=(x+3).x^2`

`<=>(x+3)(x^2-5x+8-x^2)=0`

`<=>(x+3)(8-5x)=0`

`<=>` \(\left[ \begin{array}{l}x+3=0\\8-5x=0\end{array} \right.\) 

`<=>` \(\left[ \begin{array}{l}x=\dfrac85\\x=-3\end{array} \right.\) 

Vậy `S={-3,8/5}`

`(x+3)(x^2-5x+8)=(x+3).x^2`

`<=>(x+3)(x^2-5x+8-x^2)=0`

`<=>(x+3)(-5x+8)=0`

\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\-5x+8=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\dfrac{8}{5}\end{matrix}\right.\)

Vậy `S={-3;8/5}`.

\(\left(x+3\right)\left(x^2-5x+8\right)=\left(x+3\right).x^2\\ \Leftrightarrow\left(x+3\right)\left(x^2-5x+8\right)-x^2\left(x+3\right)=0\\ \Leftrightarrow\left(x+3\right)\left(x^2-5x+8-x^2\right)=0\\ \Leftrightarrow\left(x+3\right)\left(8-5x\right)=0\\\Leftrightarrow x+3=0 \)

hoặc \(8-5x=0\)

\(x+3=0\\ \Leftrightarrow x=-3\)

\(8-5x=0\\ \Leftrightarrow-5x=-8\\ \Leftrightarrow5x=8\\ \Leftrightarrow x=\dfrac{8}{5}\)

Vậy pt có tn là \(S=\left\{-3;\dfrac{8}{5}\right\}\)

\(\dfrac{x}{9}\) < \(\dfrac{4}{7}\) < \(x\) + \(\dfrac{1}{9}\)

\(\dfrac{7x}{63}\) < \(\dfrac{36}{63}\) < \(\dfrac{63x}{63}\) + \(\dfrac{7}{63}\)

7\(x\) < 36 < 63\(x\) + 7

⇒\(\left\{{}\begin{matrix}7x< 36\\63x+7>36\end{matrix}\right.\)⇒\(\left\{{}\begin{matrix}x< \dfrac{36}{7}\\63x>36-7\end{matrix}\right.\)⇒\(\left\{{}\begin{matrix}x< \dfrac{36}{7}\\63x>29\end{matrix}\right.\)⇒\(\left\{{}\begin{matrix}x< \dfrac{36}{7}\\x>\dfrac{29}{63}\end{matrix}\right.\)

\(\dfrac{29}{63}\)<  \(x\) < \(\dfrac{36}{7}\) vì \(x\in\) Z nên \(x\in\) { 1; 2; 3; 4; 5}

⇒ \(\dfrac{x}{9}\) = \(\dfrac{1}{9}\); \(\dfrac{2}{9}\); \(\dfrac{3}{9}\); \(\dfrac{4}{9}\);\(\dfrac{5}{9}\)