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10 tháng 2 2021

mình ghi thiếu: VP=\(\frac{1}{9}\)

10 tháng 2 2021

\(\frac{1}{x^2+4x+3}+\frac{1}{x^2+8x+15}+\frac{1}{x^2+12x+35}=\frac{1}{9}\)

<=> \(\frac{1}{\left(x+1\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+7\right)}=\frac{1}{9}\)

<=> \(\frac{1}{2}\left(\frac{2}{\left(x+1\right)\left(x+3\right)}+\frac{2}{\left(x+3\right)\left(x+5\right)}+\frac{2}{\left(x+5\right)\left(x+7\right)}\right)=\frac{1}{9}\)

<=> \(\frac{1}{x+1}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+7}=\frac{2}{9}\)

<=> \(\frac{1}{x+1}-\frac{1}{x+7}=\frac{2}{9}\)

<=> \(\frac{6}{\left(x+1\right)\left(x+7\right)}=\frac{2}{9}\)

<=> (x + 1)(x + 7) = 27

<=> x2 + 8x + 7 - 27 = 0

<=> x2 + 8x - 20 = 0

<=> x2 - 2x + 10x - 20 = 0

<=> x(x - 2) + 10(x - 2) = 0

<=> (x + 10(x - 2) = 0

<=> \(\orbr{\begin{cases}x=-10\\x=2\end{cases}}\) 

Vậy \(x\in\left\{-10;2\right\}\)là giá trị cần tìm

28 tháng 11 2017

ĐK:\(x\ne-1;-3;-5;-7;-9\)

\(pt\Leftrightarrow\frac{2}{\left(x+1\right)\left(x+3\right)}+\frac{2}{\left(x+3\right)\left(x+5\right)}+\frac{2}{\left(x+5\right)\left(x+7\right)}+\frac{2}{\left(x+7\right)\left(x+9\right)}=\frac{2}{5}\)

\(\Leftrightarrow\frac{1}{x+1}-\frac{1}{x+3}+\frac{1}{x+3}-...-\frac{1}{x+9}=\frac{2}{5}\)

\(\Leftrightarrow\frac{1}{x+1}-\frac{1}{x+9}=\frac{2}{5}\)\(\Leftrightarrow\frac{8}{\left(x+1\right)\left(x+9\right)}=\frac{2}{5}\)

\(\Leftrightarrow2\left(x+1\right)\left(x+9\right)=40\)\(\Leftrightarrow x^2+10x-11=0\)

\(\Rightarrow\orbr{\begin{cases}x-1=0\\x+11=0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=1\\x=-11\end{cases}}\) (thoả)

Vậy....

13 tháng 5 2016

\(\frac{2}{x^2-4x+3}+\frac{2}{x^2-8x+15}+\frac{2}{x^2-12x+35}=-\frac{1}{2}\)(x khác 1;3;5;7)

<=>\(\frac{2}{x^2-3x-x+3}+\frac{2}{x^2-5x-3x+15}+\frac{2}{x^2-5x-7x+35}=-\frac{1}{2}\)

<=>\(\frac{2}{\left(x-1\right)\left(x-3\right)}+\frac{2}{\left(x-3\right)\left(x-5\right)}+\frac{2}{\left(x-5\right)\left(x-7\right)}=-\frac{1}{2}\)

<=>\(\frac{1}{x-3}-\frac{1}{x-1}+\frac{1}{x-5}-\frac{1}{x-3}+\frac{1}{x-7}-\frac{1}{x-5}=-\frac{1}{2}\)

<=>\(\frac{1}{x-7}-\frac{1}{x-1}=-\frac{1}{2}\)

<=>\(2x-2-2x+14=-x^2+8x-7\)

<=>\(x^2-8x+19=0\)

<=>(x-4)2+3=0(vô lí)

Vậy PT vô nghiệm

26 tháng 7 2018

xin lỗi nha, bài đó bằng có một cái 1/5 thôi, tại viết sai

26 tháng 7 2018

ĐK : \(X\ne-1;-3;-7;-9\)

\(\frac{1}{x^2+4x+3}+\frac{1}{x^2+8x+15}+\frac{1}{x^2+12x+35}+\frac{1}{x^2+16x+63}=\frac{1}{5}\)

\(\frac{1}{\left(x+2\right)^2-1}+\frac{1}{\left(x+4\right)^2-1}+\frac{1}{\left(x+6\right)^2-1}+\frac{1}{\left(x-8\right)^2-1}=\frac{1}{5}\)

\(\frac{1}{\left(x+2-1\right)\left(x+2+1\right)}+\frac{1}{\left(x+4-1 \right)\left(x+4+1\right)}+\frac{1}{\left(x+6-1\right)\left(x+6+1\right)}+\frac{1}{\left(x+8-1\right)\left(x+8+1\right)}=\frac{1}{5}\)

\(\frac{1}{\left(x+1\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+7\right)}+\frac{1}{\left(x+7\right)\left(x+9\right)}=\frac{1}{5}\)

\(\frac{1}{2}\cdot\left(\frac{1}{x+1}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+4}+....-\frac{1}{x+9}\right)=\frac{1}{5}\)

\(\frac{1}{2}\cdot\left(\frac{1}{x+1}-\frac{1}{x+9}\right)=\frac{1}{5}\)

\(\frac{1}{x+1}-\frac{1}{x+9}=\frac{1}{5}:\frac{1}{2}=\frac{2}{5}\)

\(\frac{8}{\left(x+1\right)\left(x+9\right)}=\frac{2}{5}\)

\(2\left(x+1\right)\left(x+9\right)=40\)

\(2x^2+20x+18=40\Leftrightarrow x^2+10x+9=20\)

\(\Leftrightarrow x^2+10x-11=0\Leftrightarrow x^2+10x-10-1=0\)

\(\Leftrightarrow\left(x^2-1\right)+\left(10x-10\right)=0\Leftrightarrow\left(x-1\right)\left(x+1\right)+10\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+11\right)=0\)

\(\orbr{\begin{cases}x-1=0\\x++11=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x=-11\end{cases}}}\)( Thõa mãn ) 

Vậy ...............

3 tháng 5 2020

Nguyễn Trần Thành Đạt e cmt ạ

26 tháng 4 2020

đk: ... \(\Rightarrow x\ne-1;-3;-5;-7\)

\(pt\Leftrightarrow\frac{1}{\left(x+1\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+7\right)}=\frac{1}{3}\)

\(\Leftrightarrow\frac{2}{\left(x+1\right)\left(x+3\right)}+\frac{2}{\left(x+3\right)\left(x+5\right)}+\frac{2}{\left(x+5\right)\left(x+7\right)}=\frac{2}{3}\)

\(\Leftrightarrow\frac{1}{x+1}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+7}=\frac{2}{3}\)

\(\Leftrightarrow\frac{1}{x+1}-\frac{1}{x+7}=\frac{2}{3}\)

\(\Leftrightarrow3\left(x+7-x-1\right)=2\left(x+1\right)\left(x+7\right)\)

\(\Leftrightarrow2x^2+16x+14=18\)

\(\Leftrightarrow2x^2+16x-4=0\)

\(\Delta'=64+8=72>0\)

phương trình có 2 nghiệm phân biệt:

\(x_{1,2}=\frac{-b'\pm\sqrt{\Delta}}{a}=\frac{-8\pm\sqrt{72}}{2}=-4\pm3\sqrt{2}\) (tm)

Vậy...

3 tháng 2 2019

a) \(\frac{3\left(2x+1\right)}{4}-\frac{5x+3}{6}+\frac{x+1}{3}=x+\frac{7}{12}\)

\(\frac{3.3\left(2x+1\right)}{12}-\frac{2\left(5x+3\right)}{12}+\frac{4\left(x+1\right)}{12}=\frac{12x+7}{12}\)

\(18x+9-10x-6+4x+4=12x+7\)

\(0x=0\) ( vô số nghiệm )

Vậy x \(\in\)R

b) ĐKXĐ :  x \(\ne\)-1;-3;-5;-7

\(\frac{1}{x^2+4x+3}+\frac{1}{x^2+8x+15}+\frac{1}{x^2+12x+35}=\frac{3}{16}\)

\(\frac{1}{\left(x+1\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+7\right)}=\frac{3}{16}\)

\(\frac{1}{2}\left(\frac{1}{x+1}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+7}\right)=\frac{3}{16}\)

\(\frac{1}{x+1}-\frac{1}{x+7}=\frac{3}{8}\)

\(\left(x+1\right)\left(x+7\right)=16\)

Ta thấy x+1 và x+7 là 2 số cách nhau 6 đơn vị . Mà x + 1 < x + 7

\(\Rightarrow\)\(\hept{\begin{cases}x+1=2\\x+7=8\end{cases}\Rightarrow x=1}\)

hoặc \(\hept{\begin{cases}x+1=-2\\x+7=-8\end{cases}}\Rightarrow\hept{\begin{cases}x=-3\\x=-15\end{cases}}\)( loại )

Vậy x = 1

23 tháng 11 2016

Đk:\(x\ne-1;x\ne-3;x\ne-5;x\ne-7\)

\(\frac{1}{x^2+4x+3}+\frac{1}{x^2+8x+15}+\frac{1}{x^2+12x+35}=\frac{1}{9}\)

\(\Leftrightarrow\frac{1}{\left(x+1\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+7\right)}=\frac{1}{9}\)

\(\Leftrightarrow\frac{1}{2}\left(\frac{2}{\left(x+1\right)\left(x+3\right)}+\frac{2}{\left(x+3\right)\left(x+5\right)}+\frac{2}{\left(x+5\right)\left(x+7\right)}\right)=\frac{1}{9}\)

\(\Leftrightarrow\frac{1}{x+1}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+7}=\frac{2}{9}\)

\(\Leftrightarrow\frac{1}{x+1}-\frac{1}{x+7}=\frac{2}{9}\)\(\Leftrightarrow\frac{6}{\left(x+1\right)\left(x+7\right)}=\frac{2}{9}\)

\(\Leftrightarrow2\left(x^2+8x+7\right)=54\)\(\Leftrightarrow x^2+8x+7=27\)

\(\Leftrightarrow x^2+8x-20=0\)\(\Leftrightarrow\left(x-2\right)\left(x+10\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=-10\end{cases}}\)(thỏa mãn)

8 tháng 3 2016

Mình không ghi lại đề:

\(\frac{1}{\left(x+1\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+7\right)}+\frac{1}{\left(x+7\right)\left(x+9\right)}=\frac{1}{5}\)

\(\frac{2}{\left(x+1\right)\left(x+3\right)}+...+\frac{2}{\left(x+7\right)\left(x+9\right)}=\frac{2}{5}\)

\(\frac{1}{x+1}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+4}+...+\frac{1}{x+7}-\frac{1}{x+9}=\frac{2}{5}\)

\(\frac{1}{x+1}-\frac{1}{x+9}=\frac{2}{5}\)

\(\frac{8}{\left(x+1\right)\left(x+9\right)}=\frac{2}{5}\)

<=>40=2(x+1)(x+9)

<=>\(x^2+10x-11=0\)

<=>\(\left(x-1\right)\left(x+11\right)=0\)

<=>x=1 hoặc x=-11

Ta có:

\(1^2+\left(-11\right)^2=122\)

Ai thấy mình làm đúng thì tích nha.Ai tích mình mình tích lại

8 tháng 3 2016

mk nghĩ là = 122 đó bn

7 tháng 3 2016

\(\frac{1}{x^2+4x+3}=\frac{1}{\left(x+1\right)\left(x+3\right)}=\frac{1}{2}\left(\frac{1}{x+1}-\frac{1}{x+3}\right)\)

\(\frac{1}{x^2+8x+15}=\frac{1}{\left(x+3\right)\left(x+5\right)}=\frac{1}{2}\left(\frac{1}{x+3}-\frac{1}{x+5}\right)\)

...

Cộng theo vế các hạng tử sẽ bị triệt tiêu

7 tháng 3 2016

\(\Leftrightarrow\frac{1}{x^2+16x+63}+\frac{1}{x^2+12x+35}+\frac{1}{x^2+8x+15}+\frac{1}{x^2+4x+3}=\frac{1}{5}\)

\(\Rightarrow\frac{1}{x^2+16x+63}+\frac{1}{x^2+12x+35}+\frac{1}{x^2+8x+15}+\frac{1}{x^2+4x+3}-\frac{1}{5}=0\)

\(\Leftrightarrow-\frac{x^2+10x-11}{5\left(x+1\right)\left(x+9\right)}=0\)

=>x2+10x-11=0

102-(-4(1.11))=144

\(\Rightarrow x_{1,2}=\frac{-b\pm\sqrt{\Delta}}{2a}=\frac{-10\pm\sqrt{144}}{2}\)

x1=[(-10)+12]:2=1

x2=[(-10)-12]:2=-11

tổng nghiệm của pt là 1+(-11)=-10

14 tháng 2 2018

 ĐKXĐ:    \(x\ne-1;\) \(x\ne-3;\)\(x\ne-5;\)\(x\ne-7\)

           \(\frac{1}{x^2+4x+3}+\frac{1}{x^2+8x+15}+\frac{1}{x^2+12x+35}=\frac{3}{16}\)

 \(\Leftrightarrow\)\(\frac{1}{\left(x+1\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+7\right)}=\frac{3}{16}\)

\(\Leftrightarrow\)\(\frac{1}{2}\left(\frac{1}{x+1}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+7}\right)=\frac{3}{16}\)

\(\Leftrightarrow\)\(\frac{1}{x+1}-\frac{1}{x+7}=\frac{3}{8}\)

\(\Leftrightarrow\)\(\frac{6}{\left(x+1\right)\left(x+7\right)}=\frac{3}{8}\)

\(\Rightarrow\)\(3\left(x+1\right)\left(x+7\right)=48\)

\(\Leftrightarrow\)\(x^2+8x+7=16\)

\(\Leftrightarrow\)\(x^2+8x-9=0\)

\(\Leftrightarrow\)\(\left(x+1\right)\left(x-9\right)=0\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=-1\left(KTMĐKXĐ\right)\\x=9\left(TMĐKXĐ\right)\end{cases}}\)

Vậy...

14 tháng 2 2018

\(\frac{1}{x^2+4x+3}+\frac{1}{x^2+8x+15}+\frac{1}{x^2+12x+35}=\frac{3}{16}\)

\(\Leftrightarrow\frac{1}{x^2+x+3x+3}+\frac{1}{x^2+3x+5x+15}+\frac{1}{x^2+5x+7x+35}=\frac{3}{16}\)

\(\Leftrightarrow\frac{1}{\left(x+1\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+7\right)}=\frac{3}{16}\)

\(\Leftrightarrow\frac{\left(x+5\right)\left(x+7\right)}{\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)}+\frac{\left(x+1\right)\left(x+7\right)}{\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)}+\frac{\left(x+1\right)\left(x+3\right)}{\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)}\)

\(=\frac{3\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)}{\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)}\)

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