giải phương trình: \(\frac{1}{x^2+3x+2}+\frac{1}{x^2+5x+6}+\frac{1}{x^2+7x+12}+...+\frac{1}{x^2+15x+56}=\frac{1}{14}\)
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2 tháng 2 2017
Phân tích mẫu thức thành nhân tử ta có :
1/(x+1)(x+2)+1/(x+2)(x+3)+...+1/(x+7)(x+8)=1/14
1/(x+1)-1/(x+2)+1/(x+2)-1/(x+3)+...+1/(x+7)-1/(x+8)=1/14
1/(x+1)-1/(x+8)=1/14
7/(x+1)(x+8)=1/14
Nhân chéo ta có x^2+9x+8=98
x^2+9x-90=0
(x+15)(x-6)=0
Suy ra x=-15 hoặc x=6
AH
Akai Haruma
Giáo viên
20 tháng 3 2020
Lời giải:
PT \(\Leftrightarrow \frac{1}{(x+1)(x+2)}+\frac{1}{(x+2)(x+3)}+\frac{1}{(x+3)(x+4)}+....+\frac{1}{(x+7)(x+8)}=\frac{1}{14}\)
(ĐK: $x\neq -1;-2;...;-8$)
\(\Leftrightarrow \frac{(x+2)-(x+1)}{(x+1)(x+2)}+\frac{(x+3)-(x+2)}{(x+2)(x+3)}+....+\frac{(x+8)-(x+7)}{(x+7)(x+8)}=\frac{1}{14}\)
\(\Leftrightarrow \frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+....+\frac{1}{x+7}-\frac{1}{x+8}=\frac{1}{14}\)
\(\Leftrightarrow \frac{1}{x+1}-\frac{1}{x+8}=\frac{1}{14}\Leftrightarrow \frac{7}{x^2+9x+8}=\frac{1}{14}\)
\(\Rightarrow x^2+9x+8=98\Leftrightarrow x^2+9x-90=0\Rightarrow x=6\) hoặc $x=-15$ (đều thỏa mãn)
Vậy........
BT
1
30 tháng 5 2017
Ta có : \(\frac{1}{x^2+3x+2}+\frac{1}{x^2+5x+6}+...+\) \(\frac{1}{x^2+15x+56}=\frac{1}{14}\)
<=>\(\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}\)+...+ \(\frac{1}{\left(x+7\right)\left(x+8\right)}=\frac{1}{14}\)
<=> \(\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+...+\frac{1}{x+7}-\frac{1}{x+8}\)= \(\frac{1}{14}\)
<=> \(\frac{1}{x+1}-\frac{1}{x+8}=\frac{1}{14}\)
<=> \(\frac{x+8-x-1}{\left(x+1\right)\left(x+8\right)}=\frac{1}{14}\)
<=>\(\frac{7.14}{14\left(x+1\right)\left(x+8\right)}=\frac{\left(x+1\right)\left(x+8\right)}{14\left(x+1\right)\left(x+8\right)}\)
<=> \(x^2+9x+8=98\)<=> \(x^2+9x-90=0\)
<=> (x-6)(x+15) =0
<=> \(\orbr{\begin{cases}x=6\\x=-15\end{cases}}\)
Vậy phương trình có 2 nghiệm x \(\in\left(6,15\right)\)
==============
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LT
1
4 tháng 8 2017
Để PT đc xác định : \(x^2+3x+2\ne0;x^2+5x+6\ne0;.....;x^2+15x+56\ne0\)
\(\Leftrightarrow\left(x+1\right)\left(x+2\right)\ne0;\left(x+2\right)\left(x+3\right)\ne0;....;\left(x+7\right)\left(x+8\right)\ne0\)
\(\Rightarrow x+1;x+2;x+3;....;x+8\ne0\)
\(\Rightarrow x\ne\left\{-8;-7;...;-3;-2;-1\right\}\)
TXĐ : \(x\ne\left\{-8;-7;...;-3;-2;-1\right\}\)
\(\frac{1}{x^2+3x+2}+\frac{1}{x^2+5x+6}+....+\frac{1}{x^2+15x+56}=\frac{1}{14}\)
\(\Leftrightarrow\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}+....+\frac{1}{\left(x+7\right)\left(x+8\right)}=\frac{1}{14}\)
\(\Leftrightarrow\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+...+\frac{1}{x+7}-\frac{1}{x+8}=\frac{1}{14}\)
\(\Leftrightarrow\frac{1}{x+1}-\frac{1}{x+8}=\frac{1}{14}\)
\(\Leftrightarrow\frac{7}{x^2+9x+8}=\frac{1}{14}\)
\(\Leftrightarrow x^2+9x+8=98\)
\(\Leftrightarrow x^2+9x-90=0\)
\(\Leftrightarrow\left(x-6\right)\left(x+15\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=6\\x=-15\end{cases}}\)(TMĐKXĐ)
Vậy \(x=6\) hoặc \(x=-15\)
22 tháng 3 2022
\(\frac{1}{x^2+3x+2}+\frac{1}{x^2+5x+6}+...+\frac{1}{x^2+15x+56}=\frac{1}{14}\)
\(\Leftrightarrow\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}+...+\frac{1}{\left(x+7\right)\left(x+8\right)}=\frac{1}{14}\)
\(\Leftrightarrow\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+...+\frac{1}{x+7}-\frac{1}{x+8}=\frac{1}{14}\)
\(\Leftrightarrow\frac{1}{x+1}-\frac{1}{x+8}=\frac{1}{14}\)
Làm nốt
22 tháng 3 2022
2/
\(T=8x^2-4x+\frac{1}{4x^2}+15\)
\(=\left(4x^2-4x+1\right)+\left(4x^2+\frac{1}{4x^2}-2\right)+16\)
\(=\left(2x-1\right)^2+\left(\frac{4x^2-1}{2x}\right)^2+16\ge16\)
28 tháng 7 2017
\(\Leftrightarrow\frac{1}{\left(x-1\right)\left(x-2\right)}+\frac{1}{\left(x-2\right)\left(x-3\right)}+\frac{1}{\left(x-3\right)\left(x-4\right)}+\frac{1}{\left(x-4\right)\left(x-5\right)}=\frac{1}{8}\)
\(\Leftrightarrow\frac{1}{x-1}-\frac{1}{x-2}+\frac{1}{x-2}-\frac{1}{x-3}+...+\frac{1}{x-4}-\frac{1}{x-5}=\frac{1}{8}\)
\(\Leftrightarrow\frac{1}{x-1}-\frac{1}{x-5}=\frac{1}{8}\)
\(\Leftrightarrow\frac{x-5-x+1}{\left(x-1\right)\left(x-5\right)}=\frac{1}{8}\)
\(\Leftrightarrow-4.8=x^2-6x+5\)
\(\Leftrightarrow x^2-6x+37=0\)
29 tháng 11 2017
\(\Rightarrow\frac{1}{\left(x+2\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+4\right)}+\frac{1}{\left(x+4\right)\left(x+5\right)}=\frac{1}{6}\)
ĐK:\(x\ne-2;-3;-4;-5\)
MTC:\(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right).6\)
Quy đồng khử mẫu:
ĐKXĐ : Tự tìm nha : )
Ta có : \(\frac{1}{x^2+3x+2}+\frac{1}{x^2+5x+6}+...+\frac{1}{x^2+15x+56}=\frac{1}{14}\)
=> \(\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}+...+\frac{1}{\left(x+7\right)\left(x+8\right)}=\frac{1}{14}\)
=> \(\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+...+\frac{1}{x+7}-\frac{1}{x+8}=\frac{1}{14}\)
=> \(\frac{1}{x+1}-\frac{1}{x+8}=\frac{1}{14}\)
=> \(\frac{x+8}{\left(x+1\right)\left(x+8\right)}-\frac{x+1}{\left(x+8\right)\left(x+1\right)}=\frac{1}{14}\)
=> \(14\left(x+8-x-1\right)=\left(x+1\right)\left(x+8\right)\)
=> \(x^2+x+8x+8=98\)
=> \(x^2+9x-90=0\)
=> \(\left(x+15\right)\left(x-6\right)=0\)
=> \(\left[{}\begin{matrix}x+15=0\\x-6=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=-15\\x=6\end{matrix}\right.\) ( TM )
Vậy phương trình trên có nghiệm là \(S=\left\{6,-15\right\}\)