\(\frac{1}{x\left(x^2+1\right)}=\frac{a}{x}+\frac{bx+c}{x^2+1}\)
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NV
Nguyễn Việt Lâm
Giáo viên
3 tháng 11 2019
\(x^2-1=\frac{1}{4}\left(a^2+\frac{1}{a^2}+2\right)-1=\frac{1}{4}\left(a-\frac{1}{a}\right)^2\)
\(\Rightarrow\sqrt{x^2-1}=\frac{1}{2}\left(a-\frac{1}{a}\right)\)
Tương tự \(\sqrt{y^2-1}=\frac{1}{2}\left(b-\frac{1}{b}\right)\)
\(A=\frac{\frac{1}{4}\left(a+\frac{1}{a}\right)\left(b+\frac{1}{b}\right)-\frac{1}{4}\left(a-\frac{1}{a}\right)\left(b-\frac{1}{b}\right)}{\frac{1}{4}\left(a+\frac{1}{a}\right)\left(b+\frac{1}{b}\right)+\frac{1}{4}\left(a-\frac{1}{a}\right)\left(b-\frac{1}{b}\right)}=\frac{ab+\frac{a}{b}+\frac{b}{a}+\frac{1}{ab}-ab-\frac{1}{ab}+\frac{a}{b}+\frac{b}{a}}{ab+\frac{a}{b}+\frac{b}{a}+\frac{1}{ab}+ab+\frac{1}{ab}-\frac{a}{b}-\frac{b}{a}}\)
\(=\frac{\frac{a}{b}+\frac{b}{a}}{ab+\frac{1}{ab}}=\frac{a^2+b^2}{a^2b^2+1}\)
b/ \(B=\frac{\left(\sqrt{a+bx}+\sqrt{a-bx}\right)^2}{a+bx-\left(a-bx\right)}=\frac{a+\sqrt{a^2-b^2x^2}}{bx}\)
\(a^2-b^2x^2=a^2-\frac{4a^2m^2}{\left(1+m^2\right)^2}=\frac{a^2\left(m^4+2m^2+1\right)-4a^2m^2}{\left(1+m^2\right)^2}=\frac{a^2\left(1-m^2\right)^2}{\left(1+m^2\right)^2}\)
\(\Rightarrow B=\left(a+\frac{a\left(1-m^2\right)}{1+m^2}\right).\left(\frac{1+m^2}{2am}\right)=\frac{a+am^2+a-am^2}{2am}=\frac{1}{m}\)
KN
3 tháng 10 2019
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24 tháng 12 2016
Có: \(\frac{A}{x-1}+\frac{Bx+C}{x^2+1}=\frac{A\left(x^2+1\right)+\left(Bx+C\right)\left(x-1\right)}{\left(x-1\right)\left(x^2+1\right)}=\frac{Ax^{2\: }+A+Bx^2-Bx+Cx-C}{\left(x-1\right)\left(x^2+1\right)}=\frac{\left(A+B\right)x^2+\left(C-B\right)x+\left(A-C\right)}{\left(x-1\right)\left(x^2+1\right)}\)
Đồng nhất với phân thức \(\frac{x^2+2x-1}{\left(x-1\right)\left(x^2+1\right)}\)
Ta được: \(\begin{cases}A+B=1\\C-B=2\\A-C=-1\end{cases}\)\(\Leftrightarrow\begin{cases}A=1-B\\C-B=2\\1-B-C=-1\end{cases}\)
\(\Leftrightarrow\begin{cases}A=1-B\\C-B=2\\B+C=2\end{cases}\)\(\Leftrightarrow\begin{cases}A=1-B\\B=0\\C=2\end{cases}\)\(\Leftrightarrow\begin{cases}A=1\\B=0\\C=2\end{cases}\)
24 tháng 12 2016
\(VP=\frac{A}{x-1}+\frac{Bx+C}{x^2+1}=\frac{A\left(x^2+1\right)}{\left(x-1\right)\left(x^2+1\right)}+\frac{\left(x-1\right)\left(Bx+C\right)}{\left(x-1\right)\left(x^2+1\right)}\)
\(=\frac{A\left(x^2+1\right)+\left(x-1\right)\left(Bx+C\right)}{\left(x-1\right)\left(x^2+1\right)}\)\(=\frac{Ax^2+A+Bx^2-Bx+Cx-C}{\left(x-1\right)\left(x^2+1\right)}\)
\(=\frac{A\left(x^2+1\right)+Bx\left(x-1\right)+C\left(x-1\right)}{\left(x-1\right)\left(x^2+1\right)}\)\(=\frac{A\left(x^2+1\right)+\left(Bx+C\right)\left(x-1\right)}{\left(x-1\right)\left(x^2+1\right)}\)
\(=\frac{Ax^2+A+Bx+C}{x^2+1}\). Lại có: \(VT=\frac{x^2+2x-1}{\left(x-1\right)\left(x^2+1\right)}=\frac{\left(x-1\right)^2}{\left(x-1\right)\left(x^2+1\right)}=\frac{x-1}{x^2+1}\)
\(\Leftrightarrow\frac{Ax^2+A+Bx+C}{x^2+1}=\frac{x-1}{x^2+1}\Leftrightarrow Ax^2+A+Bx+C=x-1\)
thôi cạn ý tưởng lm tiếp t đi chơi
\(\frac{1}{x\left(x^2+1\right)}=\frac{a}{x}+\frac{bx+c}{x^2+1}\)
\(\Rightarrow\frac{1}{x\left(x^2+1\right)}=\frac{a\left(x^2+1\right)}{x\left(x^2+1\right)}+\frac{x\left(bx+c\right)}{x\left(x^2+1\right)}\)
\(\Rightarrow a\left(x^2+1\right)+x\left(bx+c\right)=1\)
\(\Rightarrow ax^2+a+xbx+xc=1\)