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....

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

mk nghĩ là = 122 đó bn

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

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

...

Cộng theo vế các hạng tử sẽ bị triệt tiêu

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

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

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

=>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

1. Câu hỏi của Phạm Tiến Dũng new - Toán lớp 9 - Học toán với OnlineMath

1. Câu hỏi của Phạm Tiến Dũng new - Toán lớp 9 - Học toán với OnlineMath

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

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

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

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

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

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

\(=>5\left(x+9\right)-5\left(x+1\right)=2\left(x+1\right)\left(x+9\right)\)

\(=5\left(x+9-x-1\right)-2\left(x+1\right)\left(x+9\right)=0\)

\(=5.8-2\left(x^2+10x+9\right)=0\)

\(=40-2x^2-20x-18=0\)

\(=-2x^2-20x-22=0\)

đến đây dùng máy tính giải hệ phương trình bậc 2 là xong

Đợi tí coi tính ra ko đã

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

Nguyễn Trần Thành Đạt e cmt ạ

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...