\(\frac{3}{\left(x+2\right)\left(x+5\right)}+\frac{5}{\left(x+5\right)\left(x+10\right)}+\frac{7}{\left(x+10\right)\left(x+17\right)}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)

\(\Rightarrow\frac{1}{x+2}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+10}+\frac{1}{x+10}-\frac{1}{x+17}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)

\(\Rightarrow\frac{1}{x+2}-\frac{1}{x+17}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)

\(\Rightarrow\frac{\left(x+17\right)-\left(x+2\right)}{\left(x+2\right)\left(x+17\right)}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)

\(\Rightarrow\frac{15}{\left(x+2\right)\left(x+17\right)}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)

\(\Rightarrow x=15\)

Vậy x = 15

Thanks Nguyễn Huy Tú - bn thân nhất của tớ nhiều nhé

Điều kiện: $x\ne-2,\,-5,\,-10,\,-17,\,-12.$

$\dfrac3{(x+2)(x+5)}+\dfrac5{(x+5)(x+10)}+\dfrac7{(x+10)(x+17)}=\dfrac{x}{(x+12)(x+17)}$

$\Leftrightarrow\left(\dfrac1{x+2}-\dfrac1{x+5}\right)+\left(\dfrac1{x+5}-\dfrac1{x+10}\right)+\left(\dfrac1{x+10}-\dfrac1{x+17}\right)=\dfrac{x}{(x+12)(x+17)}$

$\Leftrightarrow\dfrac1{x+2}-\dfrac1{x+17}=\dfrac{x}{(x+12)(x+17)}$

$\Leftrightarrow\dfrac{15}{(x+2)(x+17)}=\dfrac{x}{(x+12)(x+17)}$

$\Leftrightarrow15(x+12)=x(x+2)$

$\Leftrightarrow x^2-13x-180=0$

$\Leftrightarrow(x-20)(x+9)=0$

$\Leftrightarrow x=20\ \text{hoặc}\ x=-9.$

Vì $20,\,-9$ đều thỏa mãn điều kiện xác định nên $x=20\ \text{hoặc}\ x=-9.$

a)

Điều kiện: $x\ne-2,-5,-10,-17.$

$\dfrac3{(x+2)(x+5)}+\dfrac5{(x+5)(x+10)}+\dfrac7{(x+10)(x+17)}=\dfrac{x}{(x+2)(x+17)}$

$\Leftrightarrow\dfrac1{x+2}-\dfrac1{x+5}+\dfrac1{x+5}-\dfrac1{x+10}+\dfrac1{x+10}-\dfrac1{x+17}=\dfrac{x}{(x+2)(x+17)}$

$\Leftrightarrow\dfrac1{x+2}-\dfrac1{x+17}=\dfrac{x}{(x+2)(x+17)}$

$\Leftrightarrow\dfrac{15}{(x+2)(x+17)}=\dfrac{x}{(x+2)(x+17)}$

$\Leftrightarrow x=15.$

b)

Điều kiện: $x\ne1,3,8,20.$

$\dfrac2{(x-1)(x-3)}+\dfrac5{(x-3)(x-8)}+\dfrac{12}{(x-8)(x-20)}-\dfrac1{x-20}=-\dfrac34$

$\Leftrightarrow\left(\dfrac1{x-3}-\dfrac1{x-1}\right)+\left(\dfrac1{x-8}-\dfrac1{x-3}\right)+\left(\dfrac1{x-20}-\dfrac1{x-8}\right)-\dfrac1{x-20}=-\dfrac34$

$\Leftrightarrow-\dfrac1{x-1}=-\dfrac34$

$\Leftrightarrow\dfrac1{x-1}=\dfrac34$

$\Leftrightarrow4=3(x-1)$

$\Leftrightarrow3x=7$

$\Leftrightarrow x=\dfrac73.$

bằng 13,34590301 ( mình bấm máy tính bạn nhé :) )

Điều kiện: $x\ne-2,\,-5,\,-10,\,-17.$

$\dfrac2{(x+2)(x+5)}+\dfrac5{(x+5)(x+10)}+\dfrac7{(x+10)(x+17)}=\dfrac{x}{(x+2)(x+17)}$

$\Leftrightarrow\dfrac1{x+2}-\dfrac1{x+5}+\dfrac1{x+5}-\dfrac1{x+10}+\dfrac1{x+10}-\dfrac1{x+17}=\dfrac{x}{(x+2)(x+17)}$

$\Leftrightarrow\dfrac1{x+2}-\dfrac1{x+17}=\dfrac{x}{(x+2)(x+17)}$

$\Leftrightarrow\dfrac{x+17-x-2}{(x+2)(x+17)}=\dfrac{x}{(x+2)(x+17)}$

$\Leftrightrrow\dfrac{15}{(x+2)(x+17)}=\dfrac{x}{(x+2)(x+17)}$

$\Leftrightarrow15=x.$

Vậy $x=15.$

Ta có: \(\frac{3}{\left(x+2\right)\left(x+5\right)}=\frac{1}{x+2}-\frac{1}{x+5}\);  \(\frac{5}{\left(x+5\right)\left(x+10\right)}=\frac{1}{x+5}-\frac{1}{x+10}\)

\(\frac{7}{\left(x+10\right)\left(x+17\right)}=\frac{1}{x+10}-\frac{1}{x+17}\); 

=> Phương trình tương đương:

\(\frac{1}{x+2}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+10}+\frac{1}{x+10}-\frac{1}{x+17}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)

\(\frac{1}{x+2}-\frac{1}{x+17}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)<=> \(\frac{x+17-x-2}{\left(x+2\right)\left(x+17\right)}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)

<=> \(\frac{15}{\left(x+2\right)\left(x+17\right)}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)

=> x=15

Đáp số: x=15

c) \(\frac{x-1}{2009}+\frac{x-2}{2008}=\frac{x-3}{2007}+\frac{x-4}{2006}\)

\(\Leftrightarrow\left(\frac{x-1}{2009}-1\right)+\left(\frac{x-2}{2008}-1\right)=\left(\frac{x-3}{2007}-1\right)+\left(\frac{x-4}{2006}-1\right)\)

\(\Leftrightarrow\frac{x-2010}{2009}+\frac{x-2010}{2008}-\frac{x-2010}{2007}-\frac{x-2010}{2006}=0\)

\(\Leftrightarrow\left(x-2010\right).\left(\frac{1}{2009}+\frac{1}{2008}-\frac{1}{2007}-\frac{1}{2006}\right)=0\)

\(\Leftrightarrow x-2010=0\)

\(\Leftrightarrow x=0+2010\)

\(\Rightarrow x=2010\)

Vậy \(x=2010.\)

Mình chỉ làm câu c) thôi nhé.

Chúc bạn học tốt!

Bài 1:

a) (2x-3). (x+1) < 0

=>2x-3 và x+1 ngược dấu

Mà 2x-3<x+1 với mọi x

\(\Rightarrow\begin{cases}2x-3< 0\\x+1>0\end{cases}\)

\(\Rightarrow\begin{cases}x< \frac{3}{2}\\x>-1\end{cases}\)\(\Rightarrow-1< x< \frac{3}{2}\)

b)\(\left(x-\frac{1}{2}\right)\left(x+3\right)>0\)

\(\Rightarrow x-\frac{1}{2}\) và x+3 cùng dấu

Xét \(\begin{cases}x-\frac{1}{2}>0\\x+3>0\end{cases}\)\(\Rightarrow\begin{cases}x>\frac{1}{2}\\x>-3\end{cases}\)

Xét \(\begin{cases}x-\frac{1}{2}< 0\\x+3< 0\end{cases}\)\(\Rightarrow\begin{cases}x< \frac{1}{2}\\x< -3\end{cases}\)

=>....

Bài 2:

\(S=\frac{1}{2}\left(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{999.1001}\right)\)

\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{999}-\frac{1}{1001}\right)\)

\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{1001}\right)\)

\(=\frac{1}{2}\cdot\frac{998}{3003}\)

\(=\frac{499}{3003}\)

tự làm nhé. bài cô Kiều cho dễ mừ :)