Tìm x: \(\dfrac{1}{x^2+9x+20}+\dfrac{1}{x^2+11x+30}+\dfrac{1}{x^2+13x+30}=\dfrac{1}{18}\)
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PV
1
22 tháng 3 2021
$ĐKXĐ:x \neq -4;-5;-6;-7$
$pt⇔\dfrac{1}{x^2+4x+5x+20}+\dfrac{1}{x^2+5x+6x+30}+\dfrac{1}{x^2+6x+7x+42}=\dfrac{1}{18}$
$⇔\dfrac{1}{(x+4)(x+5)}+\dfrac{1}{(x+5)(x+6)}+\dfrac{1}{(x+6)(x+7)}=\dfrac{1}{18}$
$⇔\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}+\dfrac{1}{x+6}-\dfrac{1}{x+7}=\dfrac{1}{18}$
$⇔\dfrac{1}{x+4}-\dfrac{1}{x+7}=\dfrac{1}{18}$
$⇔\dfrac{3}{(x+4)(x+7)}=\dfrac{1}{18}$
$⇔x^2+11x+28=54$
$⇔x^2+11x-26=0$
$⇔x^2-2x+13x-26=0$
$⇔(x-2)(x+13)=0$
$⇔$ \(\left[{}\begin{matrix}x=2\\x=-13\end{matrix}\right.\)(t/m)
Vậy phương trình đã cho có tập nghiệm $S=(2;-13)$
BC
1
30 tháng 4 2021
Ta có:
\(x^2+9x+2x=\left(x+4\right)\left(x+5\right)\)
\(x^2+11x+30=\left(x+6\right)\left(x+5\right)\)
\(x^2+13x+42=\left(x+6\right)\left(x+7\right)\)
ĐK: \(\left\{{}\begin{matrix}x\ne-4\\x\ne-5\\x\ne-6\\x\ne-7\end{matrix}\right.\)
pt \(\Leftrightarrow\dfrac{1}{\left(x+4\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+6\right)}+\dfrac{1}{\left(x+6\right)\left(x+7\right)}=\dfrac{1}{18}\)
\(\Leftrightarrow\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}+\dfrac{1}{x+6}-\dfrac{1}{x+7}=\dfrac{1}{18}\)
\(\Leftrightarrow\dfrac{1}{x+4}-\dfrac{1}{x+7}=\dfrac{1}{18}\)
\(\Leftrightarrow\dfrac{18\left(x+7\right)}{18\left(x+4\right)\left(x+7\right)}-\dfrac{18\left(x+4\right)}{18\left(x+4\right)\left(x+7\right)}=\dfrac{\left(x+4\right)\left(x+7\right)}{18\left(x+4\right)\left(x+7\right)}\)
\(\Rightarrow18\left(x+7\right)-18\left(x+4\right)=\left(x+4\right)\left(x+7\right)\)
\(\Leftrightarrow\left(x+13\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+13=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-13\\x=2\end{matrix}\right.\) (tm)
PA
8 tháng 3 2017
\(\dfrac{1}{x^2+9x+20}+\dfrac{1}{x^2+11x+30}+\dfrac{1}{x^2+13x+42}=\dfrac{1}{18}\) (ĐKXĐ: \(x\notin\left\{-4;-5;-6;-7\right\}\))
<=> \(\dfrac{1}{\left(x+4\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+6\right)}+\dfrac{1}{\left(x+6\right)\left(x+7\right)}-\dfrac{1}{18}=0\)
<=> \(\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}+\dfrac{1}{x+6}-\dfrac{1}{x+7}-\dfrac{1}{18}=0\)
<=> \(\dfrac{1}{x+4}-\dfrac{1}{x+7}-\dfrac{1}{18}=0\)
<=> \(\dfrac{18\left(x+7\right)}{18\left(x+4\right)\left(x+7\right)}-\dfrac{18\left(x+4\right)}{18\left(x+4\right)\left(x+7\right)}-\dfrac{\left(x+4\right)\left(x+7\right)}{18\left(x+4\right)\left(x+7\right)}=0\)
=> \(18\left(x+7\right)-18\left(x+4\right)-\left(x+4\right)\left(x+7\right)=0\)
<=> 18x + 18.7 - 18x - 18.4 - x2 - 7x - 4x - 28 = 0
<=> - x2 - 11x + 26 = 0
<=> (x - 2)(x + 13) = 0
<=> \(\left[{}\begin{matrix}x-2=0\\x+13=0\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=2\\x=-13\end{matrix}\right.\) (nhận)
Vậy S = {-13; 2}
DH
6 tháng 9 2017
Cân thử nào!
\(\dfrac{1}{x^2+9x+20}+\dfrac{1}{x^2+11x+30}+\dfrac{1}{x^2+13x+42}=\dfrac{1}{18}\)
\(\Rightarrow\dfrac{1}{x^2+4x+5x+20}+\dfrac{1}{x^2+5x+6x+30}+\dfrac{1}{x^2+6x+7x+42}=\dfrac{1}{18}\)
\(\Rightarrow\dfrac{1}{\left(x+4\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+6\right)}+\dfrac{1}{\left(x+6\right)\left(x+7\right)}=\dfrac{1}{18}\)
\(\Rightarrow\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}+\dfrac{1}{x+6}-\dfrac{1}{x+7}=\dfrac{1}{18}\)
\(\Rightarrow\dfrac{1}{x+4}-\dfrac{1}{x+7}=\dfrac{1}{18}\)
\(\Rightarrow\dfrac{x+7-x-4}{\left(x+4\right)\left(x+7\right)}=\dfrac{1}{18}\)
\(\Rightarrow\dfrac{3}{x^2+11x+28}=\dfrac{1}{18}\)
\(\Rightarrow x^2+11x+28=54\)
\(\Rightarrow x^2+11x-26=0\)
\(\Rightarrow x^2-2x+13x-26=0\)
\(\Rightarrow x\left(x-2\right)+13\left(x-2\right)=0\)
\(\Rightarrow\left(x-2\right)\left(x+13\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-2=0\\x+13=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=-13\end{matrix}\right.\)
Vậy................
Chúc bạn học tốt!!!
XN
21 tháng 5 2018
Xét mẫu 1: x2+9x+20=x2+4x+5x+20=x(x+4)+5(x+4)=(x+4)(x+5)
Xét mẫu 2: x2+11x+30=x2+5x+6x+30=x(x+5)+6(x+5)=(x+5)(x+6)
Xét mẫu3:x2+13x+42=x2+6x+7x+42=x(x+6)+7(x+6)=(x+6)(x+7)
Vậy .....=\(\dfrac{1}{\text{(x+4)(x+5)}}+\dfrac{1}{\text{(x+5)(x+6)}}+\dfrac{1}{\text{(x+6)(x+7)}}=\dfrac{1}{18}\)
<=>\(\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}+\dfrac{1}{x+6}-\dfrac{1}{x+7}=\dfrac{1}{18}\)
<=>\(\dfrac{1}{x+4}-\dfrac{1}{x+7}=\dfrac{1}{18}\)=.....
NV
Nguyễn Việt Lâm
Giáo viên
15 tháng 2 2019
ĐKXĐ: \(x\ne-4;-5;-6;-7\)
\(\dfrac{1}{\left(x+4\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+6\right)}+\dfrac{1}{\left(x+6\right)\left(x+7\right)}=\dfrac{1}{18}\)
\(\Leftrightarrow\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}+\dfrac{1}{x+6}-\dfrac{1}{x+7}=\dfrac{1}{18}\)
\(\Leftrightarrow\dfrac{1}{x+4}-\dfrac{1}{x+7}=\dfrac{1}{18}\)
\(\Leftrightarrow\dfrac{3}{\left(x+4\right)\left(x+7\right)}=\dfrac{1}{18}\)
\(\Leftrightarrow x^2+11x-26=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+13\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-13\end{matrix}\right.\)
TV
Giải phương trình
\(\dfrac{1}{x^2+9x+20}+\dfrac{1}{x^2+11x+30}+\dfrac{1}{x^2+13x+42}=\dfrac{1}{18}\)
2
14 tháng 2 2019
\(\dfrac{1}{x^2+9x+20}+\dfrac{1}{x^2+11x+30}+\dfrac{1}{x^2+13x+42}=\dfrac{1}{18}\) ĐKXĐ:x\(\ne\)-4,-5,-6,-7
\(\Leftrightarrow\)\(\dfrac{1}{x^2+4x+5x+20}+\dfrac{1}{x^2+5x+6x+30}+\dfrac{1}{x^2+6x+7x+42}=\dfrac{1}{18}\)
\(\Leftrightarrow\)\(\dfrac{1}{x.\left(x+4\right)+5.\left(x+4\right)}+\dfrac{1}{x.\left(x+5\right)+6.\left(x+5\right)}+\dfrac{1}{x.\left(x+6\right)+7.\left(x+6\right)}=\dfrac{1}{18}\)
\(\Leftrightarrow\)\(\dfrac{1}{\left(x+4\right).\left(x+5\right)}+\dfrac{1}{\left(x+5\right).\left(x+6\right)}+\dfrac{1}{\left(x+6\right).\left(x+7\right)}=\dfrac{1}{18}\)
\(\Leftrightarrow\)\(\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}+\dfrac{1}{x+6}-\dfrac{1}{x+7}=\dfrac{1}{18}\)
\(\Leftrightarrow\)\(\dfrac{1}{x+4}-\dfrac{1}{x+7}=\dfrac{1}{18}\)
\(\Leftrightarrow\)\(\dfrac{x+7-x-4}{\left(x+4\right).\left(x+7\right)}=\dfrac{1}{18}\)
\(\Leftrightarrow\)\(\dfrac{3}{\left(x+4\right).\left(x+7\right)}=\dfrac{3}{54}\)
\(\Leftrightarrow\)(x+4).(x+7)=54
\(\Leftrightarrow\)x2+11x+28=54
\(\Leftrightarrow\)x2+11x-26=0
\(\Leftrightarrow\)x2+13x-2x-26=0
\(\Leftrightarrow\)x.(x+13)-2.(x+13)=0
\(\Leftrightarrow\)(x-2).(x+13)=0
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x-2=0\\x+13=0\end{matrix}\right.\)\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=2\left(TM\right)\\x=-13\left(TM\right)\end{matrix}\right.\)
Vậy tập nghiệm của pt trên là S={-13;2}
HD
14 tháng 2 2019
ĐKXĐ: \(x\ne-4;x\ne-5;x\ne-6;x\ne-7\)
\(\Leftrightarrow\dfrac{1}{\left(x+4\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+6\right)}+\dfrac{1}{\left(x+6\right)\left(x+7\right)}=\dfrac{1}{18}\)
\(\Leftrightarrow\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}+\dfrac{1}{x+6}-\dfrac{1}{x+7}=\dfrac{1}{18}\)
\(\Leftrightarrow\dfrac{1}{x+4}-\dfrac{1}{x+7}=\dfrac{1}{18}\)
\(\Leftrightarrow\dfrac{3}{\left(x+4\right)\left(x+7\right)}=\dfrac{1}{18}\)
\(\Leftrightarrow54=x^2+11x+28\)
\(\Leftrightarrow x^2+11x-26=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+13\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+13=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\left(TM\right)\\x=-13\left(TM\right)\end{matrix}\right.\)
BD
giải phương trình
\(\dfrac{1}{x^2+9x+20}+\dfrac{1}{x^2+11x+30}+\dfrac{1}{x^2+13x+42}=\dfrac{1}{18}\)
1
2 tháng 4 2018
\(pt\Leftrightarrow\dfrac{1}{\left(x+4\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+6\right)}+\dfrac{1}{\left(x+6\right)\left(x+7\right)}=\dfrac{1}{18}\)
NV
Nguyễn Việt Lâm
Giáo viên
5 tháng 1 2019
ĐKXĐ: \(x\ne-4;-5;-6;-7\)
\(\dfrac{1}{\left(x+4\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+6\right)}+\dfrac{1}{\left(x+6\right)\left(x+7\right)}=\dfrac{1}{18}\)
\(\Leftrightarrow\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}+\dfrac{1}{x+6}-\dfrac{1}{x+7}=\dfrac{1}{18}\)
\(\Leftrightarrow\dfrac{1}{x+4}-\dfrac{1}{x+7}=\dfrac{1}{18}\)
\(\Leftrightarrow\dfrac{3}{\left(x+4\right)\left(x+7\right)}=\dfrac{1}{18}\)
\(\Leftrightarrow54=x^2+11x+28\)
\(\Leftrightarrow x^2+11x-26=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+13\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-13\end{matrix}\right.\)
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