CT
1
K
Khách
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NN
1
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
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)
LA
1
23 tháng 4 2018
mình mới trả lời https://hoc24.vn/hoi-dap/question/601446.html
LN
3
TN
5 tháng 6 2017
phân tích mẫu thành nhân tử r áp dụng \(\frac{1}{n\left(n+1\right)}=\frac{1}{n}-\frac{1}{n+1}\) sau đó rút gọn quy đồng
5 tháng 6 2017
\(\frac{1}{x^2+9x+20}+\frac{1}{x^2+11x+30}+\frac{1}{x^2+13x+42}=\frac{1}{18}\) \(\left(ĐKXĐ:x\ne0;x\ne-4;x\ne-5;x\ne-6;x\ne-7\right)\)
\(\Leftrightarrow\frac{1}{x^2+4x+5x+20}+\frac{1}{x^2+5x+6x+30}+\frac{1}{x^2+6x+7x+42}=\frac{1}{18}\)
\(\Leftrightarrow\frac{1}{x\left(x+4\right)+5\left(x+4\right)}+\frac{1}{x\left(x+5\right)+6\left(x+5\right)}+\frac{1}{x\left(x+6\right)+7\left(x+6\right)}=\frac{1}{18}\)
\(\Leftrightarrow\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+6\right)}+\frac{1}{\left(x+6\right)\left(x+7\right)}=\frac{1}{18}\)
\(\Leftrightarrow\frac{\left(x+6\right)\left(x+7\right)+\left(x+4\right)\left(x+7\right)+\left(x+4\right)\left(x+5\right)}{\left(x+4\right)\left(x+5\right)\left(x+6\right)\left(x+7\right)}=\frac{1}{18}\)
\(\Leftrightarrow\frac{\left(x^2+13x+42\right)+\left(x^2+11x+28\right)+\left(x^2+9x+20\right)}{\left(x+4\right)\left(x+5\right)\left(x+6\right)\left(x+7\right)}=\frac{1}{18}\)
\(\Leftrightarrow\frac{x^2+13x+42+x^2+11x+28+x^2+9x+20}{\left(x+4\right)\left(x+5\right)\left(x+6\right)\left(x+7\right)}=\frac{1}{18}\)
\(\Leftrightarrow\frac{3x^2+33x+90}{\left(x+4\right)\left(x+5\right)\left(x+6\right)\left(x+7\right)}=\frac{1}{18}\)
\(\Leftrightarrow\frac{3\left(x^2+11x+30\right)}{\left(x+4\right)\left(x+5\right)\left(x+6\right)\left(x+7\right)}=\frac{1}{18}\)
\(\Leftrightarrow\left(x+4\right)\left(x+5\right)\left(x+6\right)\left(x+7\right)=18.3\left(x^2+11x+30\right)\)
\(\Leftrightarrow\left(x+4\right)\left(x+5\right)\left(x+6\right)\left(x+7\right)=54\left(x+5\right)\left(x+6\right)\)
\(\Leftrightarrow\left(x+4\right)\left(x+7\right)=54\)
\(\Leftrightarrow x^2+11x+28-54=0\)
\(\Leftrightarrow x^2+11x-26=0\)
\(\Leftrightarrow x^2+13x-2x-26=0\)
\(\Leftrightarrow x\left(x+13\right)-2\left(x+13\right)=0\)
\(\Leftrightarrow\left(x+13\right)\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+13=0\\x-2=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-13\left(tm\right)\\x=2\left(tm\right)\end{cases}}\)
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
\(pt\Leftrightarrow\dfrac{1}{\left(x+4\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+6\right)}+\dfrac{1}{\left(x+6\right)\left(x+7\right)}=\dfrac{1}{18}\)
ĐKXĐ: \(x\ne\left\{-4;-5;-6;-7\right\}\)
\(\dfrac{1}{x^2+9x+20}+\dfrac{1}{x^2+11x+30}+\dfrac{1}{x^2+13x+42}=\dfrac{1}{18}\)
\(\Leftrightarrow\dfrac{1}{\left(x+4\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+6\right)}+\dfrac{1}{\left(x+6\right)\left(x+7\right)}=\dfrac{1}{18}\)
\(\Leftrightarrow\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}+\dfrac{1}{x+6}-\dfrac{1}{x+7}=\dfrac{1}{18}\)
\(\Leftrightarrow\dfrac{1}{x+4}-\dfrac{1}{x+7}=\dfrac{1}{18}\)
\(\Leftrightarrow\dfrac{3}{\left(x+4\right)\left(x+7\right)}=\dfrac{1}{18}\)
\(\Rightarrow\left(x+4\right)\left(x+7\right)=54\)
\(\Leftrightarrow x^2+11x-26=0\)
\(\Leftrightarrow x^2-2x+13x-26=0\)
\(\Leftrightarrow x\left(x-2\right)+13\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+13\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-13\end{matrix}\right.\)