Tìm x<0 sao cho 1/x2+9x+20 + 1/x2+11x+30 + 1/x2+13x+42 = 1/18
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CT
1
NV
Nguyễn Việt Lâm
Giáo viên
1 tháng 3 2023
ĐKXĐ: \(x\ne\left\{-4;-5;-6;-7\right\}\)
\(\dfrac{1}{x^2+9x+20}+\dfrac{1}{x^2+11x+30}+\dfrac{1}{x^2+13x+42}=\dfrac{1}{18}\)
\(\Leftrightarrow\dfrac{1}{\left(x+4\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+6\right)}+\dfrac{1}{\left(x+6\right)\left(x+7\right)}=\dfrac{1}{18}\)
\(\Leftrightarrow\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}+\dfrac{1}{x+6}-\dfrac{1}{x+7}=\dfrac{1}{18}\)
\(\Leftrightarrow\dfrac{1}{x+4}-\dfrac{1}{x+7}=\dfrac{1}{18}\)
\(\Leftrightarrow\dfrac{3}{\left(x+4\right)\left(x+7\right)}=\dfrac{1}{18}\)
\(\Rightarrow\left(x+4\right)\left(x+7\right)=54\)
\(\Leftrightarrow x^2+11x-26=0\)
\(\Leftrightarrow x^2-2x+13x-26=0\)
\(\Leftrightarrow x\left(x-2\right)+13\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+13\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-13\end{matrix}\right.\)
DT
2
3 tháng 3 2017
\(\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+6\right)}+\frac{1}{\left(x+6\right)\left(x+7\right)}\)
\(\Leftrightarrow\frac{1}{\left(x+4\right)}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+6}+\frac{1}{x+6}-\frac{1}{x+7}\)
\(\Leftrightarrow\frac{1}{x+4}-\frac{1}{x+7}=\frac{1}{18}\)
\(\Leftrightarrow\frac{3}{\left(x+4\right)\left(x+7\right)}=\frac{1}{18}\)
\(\Rightarrow x^2+11x-26=0\)
\(\Rightarrow x=2\)
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)
12 tháng 2 2023
\(\Leftrightarrow\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}+...+\dfrac{1}{\left(x+5\right)\left(x+6\right)}=\dfrac{1}{8}\)
=>\(\dfrac{1}{x+2}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+4}+...+\dfrac{1}{x+5}-\dfrac{1}{x+6}=\dfrac{1}{8}\)
=>1/x+2-1/x+6=1/8
=>\(\dfrac{x+6-x-2}{\left(x+2\right)\left(x+6\right)}=\dfrac{1}{8}\)
=>x^2+8x+12=32
=>x^2+8x-20=0
=>(x+10)(x-2)=0
=>x=-10 hoặc x=2
LA
1
23 tháng 4 2018
mình mới trả lời https://hoc24.vn/hoi-dap/question/601446.html
HL
17 tháng 9 2017
\(\dfrac{1}{x^2+9x+20}+\dfrac{1}{x^2+11x+30}+\dfrac{1}{x^2+13x+42}=\dfrac{1}{18}\)
\(\dfrac{1}{\left(x+4\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+6\right)}+\dfrac{1}{\left(x+6\right)\left(x+7\right)}=\dfrac{1}{18}\)
\(\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}+\dfrac{1}{x+6}-\dfrac{1}{x+7}=\dfrac{1}{18}\)
\(\dfrac{1}{x+4}-\dfrac{1}{x+7}=\dfrac{1}{18}\)
\(\dfrac{18\left(x+7-x-4\right)}{18\left(x+4\right)\left(x+7\right)}=\dfrac{\left(x+4\right)\left(x+7\right)}{18\left(x+4\right)\left(x+7\right)}\)
\(18.3=\left(x+4\right)\left(x+7\right)\)
\(x^2+11x+28-54=0\)
\(x^2+11x-26=0\)
\(\left(x-2\right)\left(x+13\right)=0\)
\(\left[{}\begin{matrix}x=2\\x=-13\end{matrix}\right.\)
Theo đề x < 0 nên x = -13
TN
18 tháng 10 2016
Đk:\(x\ne-4;x\ne-5;x\ne-6;x\ne-7\)
\(\Rightarrow\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+6\right)}+\frac{1}{\left(x+6\right)\left(x+7\right)}=\frac{1}{18}\)
\(\Rightarrow\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+6}+\frac{1}{x+6}-\frac{1}{x+7}=\frac{1}{18}\)
\(\Rightarrow\frac{1}{x+4}-\frac{1}{x+7}=\frac{1}{18}\)
\(\Rightarrow\frac{3}{x^2+11x+28}=\frac{1}{18}\)
\(\Rightarrow x^2+11x+28=54\)
\(\Rightarrow x^2+11x-26=0\)
\(\Rightarrow\left(x-2\right)\left(x+13\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=2\\x=-13\end{cases}}\)
Vậy....
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}
\(\dfrac{1}{\left(x^2+13x+42\right)}=\dfrac{1}{\left(\left(x+7\right)\left(x+6\right)\right)}\)
\(\dfrac{1}{\left(x^2+11x+30\right)}=\dfrac{1}{\left(\left(x+5\right)\left(x+6\right)\right)}\)
\(\dfrac{1}{\left(x^2+9x+20\right)}=\dfrac{1}{\left(\left(x+5\right)\left(x+4\right)\right)}\)
Chuyển 1/18 sang ta sẽ có: \(\dfrac{1}{\left(\left(x+7\right)\left(x+6\right)\right)}+\dfrac{1}{\left(\left(x+5\right)\left(x+6\right)\right)}+\dfrac{1}{\left(\left(x+5\right)\left(x+4\right)\right)}-\dfrac{1}{18}=0\)
Mẫu số chung sẽ là \(18\left(x+4\right)\left(x+5\right)\left(x+6\right)\left(x+7\right)\)
Quy đồng và rút gọn ta sẽ được biểu thức: \(\dfrac{-\left(x^2+11x-26\right)}{\left(18\left(x+4\right)\left(x+7\right)\right)}=0\)
Giải phương trình \(-x^2-11x+26\)
Ta sẽ có nghiệm là x = -13 và x = 2.