Phần c khó để tớ giải cho

a) Tìm $x$

$\dfrac{3}{(x+2)(x+5)}+\dfrac{5}{(x+5)(x+10)}+\dfrac{7}{(x+10)(x+17)}=\dfrac{x}{(x+2)(x+17)}$

ĐKXĐ: $x\ne-2;\,-5;\,-10;\,-17$.

Ta có:

$\dfrac{3}{(x+2)(x+5)}=\dfrac{1}{x+2}-\dfrac{1}{x+5}$

$\dfrac{5}{(x+5)(x+10)}=\dfrac{1}{x+5}-\dfrac{1}{x+10}$

$\dfrac{7}{(x+10)(x+17)}=\dfrac{1}{x+10}-\dfrac{1}{x+17}$

Suy ra: $\dfrac{1}{x+2}-\dfrac{1}{x+17}=\dfrac{x}{(x+2)(x+17)}$

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

$\Leftrightarrow x=15$.

Vậy: $x=15$.

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: $x\ne1,3,8,20.$

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

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

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

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

$\Leftrightarrow4=3(x-1)$

$\Leftrightarrow3x=7$

$\Leftrightarrow x=\dfrac73.$

câu a;b: bạn áp dụng công thức \(\frac{a}{n.\left(n+a\right)}=\frac{1}{n+a}-\frac{1}{n}\left(a\inℕ^∗\right)\)

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!

\(\frac{2}{\left(x-1\right)\left(x-3\right)}+\frac{5}{\left(x-3\right)\left(x-8\right)}+\frac{12}{\left(x-8\right)\left(x-20\right)}-\frac{1}{x-20}=-\frac{3}{4}\)

\(=\frac{1}{x-1}-\frac{1}{x-3}+\frac{1}{x-3}-\frac{1}{x-8}+\frac{1}{x-8}-\frac{1}{x-20}-\frac{1}{x-20}=-\frac{3}{4}\)

\(=\frac{1}{x-1}-\frac{2}{x-20}=-\frac{3}{4}\)

\(\frac{-x-18}{\left(x-1\right)\left(x-20\right)}=-\frac{3}{4}\)

\(\frac{-x-18}{x^2-21x+20}=\frac{-3}{4}\)

\(\frac{x+18}{x^2-21x+20}=\frac{3}{4}\)

\(4\left(x+18\right)=3\left(x^2-21x+20\right)\)

\(4x+72=3x^2-63x+60\)

\(3x^2-63x-4x=72-60\)

\(3x^2-67x=12\)

\(x\left(2x-67\right)=12\)

\(\Rightarrow x;2x-67\inƯ\left(12\right)=\left\{-12;-6;-4;-3;-2;-1;1;2;3;4;6;12\right\}\)

Mà 2x - 67 lẻ.

Ta có bảng sau:

Do đó không có \(x\)thỏa mãn.

chỉ có một số \(\frac{-3}{4}\) thôi nha

ĐKXXD : \(x\ne20;8;3;1\)

\(\frac{2}{\left(x-1\right)\left(x-3\right)}+\frac{5}{\left(x-3\right)\left(x-8\right)}+\frac{12}{\left(x-8\right)\left(x-20\right)}-\frac{1}{x-20}=-\frac{3}{4}\)

\(\Leftrightarrow\frac{\left(x-1\right)-\left(x-3\right)}{\left(x-1\right)\left(x-3\right)}+\frac{\left(x-3\right)-\left(x-8\right)}{\left(x-3\right)\left(x-8\right)}+\frac{\left(x-8\right)-\left(x-20\right)}{\left(x-8\right)\left(x-20\right)}-\frac{1}{x-20}=-\frac{3}{4}\)

\(\Leftrightarrow\frac{1}{x-3}-\frac{1}{x-1}+\frac{1}{x-8}-\frac{1}{x-3}+\frac{1}{x-20}-\frac{1}{x-8}+\frac{1}{x-20}=-\frac{3}{4}\)

\(\Leftrightarrow-\frac{1}{x-1}=-\frac{3}{4}\Leftrightarrow x-1=\frac{4}{3}\Rightarrow x=\frac{7}{3}\)

dell biết\