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![](https://rs.olm.vn/images/avt/0.png?1311)
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
![](https://rs.olm.vn/images/avt/0.png?1311)
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....
![](https://rs.olm.vn/images/avt/0.png?1311)
1. Câu hỏi của Phạm Tiến Dũng new - Toán lớp 9 - Học toán với OnlineMath
![](https://rs.olm.vn/images/avt/0.png?1311)
1. Câu hỏi của Phạm Tiến Dũng new - Toán lớp 9 - Học toán với OnlineMath
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{1}{x^2+4x+3}=\frac{1}{\left(x+1\right)\left(x+3\right)}=\frac{1}{2}\left(\frac{1}{x+1}-\frac{1}{x+3}\right)\)
\(\frac{1}{x^2+8x+15}=\frac{1}{\left(x+3\right)\left(x+5\right)}=\frac{1}{2}\left(\frac{1}{x+3}-\frac{1}{x+5}\right)\)
...
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
![](https://rs.olm.vn/images/avt/0.png?1311)
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...
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1 bạn phải dùng BDT Bunhiacopxki : ( ax +by )2 <= ( nhỏ hơn bằng ) ( a2 + b2 )( x2 + Y2 )
Ở đây hệ số của x là 1 nên a là 1.
Ta có: ( x + 2y )2 <= ( 12 + (căn2)2 ) ( x2 + ( căn 2 )2y2 )
=> 1 <= 3 ( x2 + 2y2 )
=> x2 + 2y2 >= 1/3
![](https://rs.olm.vn/images/avt/0.png?1311)
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)
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