Tìm a, b để \(\frac{a}{x}+\frac{b}{1-x}=\frac{1}{x\left(1-x\right)}\)
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LD
29 tháng 7 2017
a) \(A=\left[\frac{\left(x+1\right)^2-\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}\right]:\left[\frac{1}{x+1}+\frac{x}{x-1}+\frac{2}{\left(x-1\right)\left(x+1\right)}\right]\)
\(A=\left[\frac{\left(x+1-x+1\right)\left(x-1+x-1\right)}{\left(x-1\right)\left(x+1\right)}\right]:\left[\frac{x-1}{\left(x-1\right)\left(x+1\right)}+\frac{x\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}+\frac{2}{\left(x-1\right)\left(x+1\right)}\right]\)
\(A=\left[\frac{4x}{\left(x-1\right)\left(x+1\right)}\right]:\left[\frac{x-1+x^2+x+2}{\left(x-1\right)\left(x+1\right)}\right]\)
\(A=\left[\frac{4x}{\left(x-1\right)\left(x+1\right)}\right]:\left[\frac{x^2+2x+1}{\left(x-1\right)\left(x+1\right)}\right]\)
\(A=\left[\frac{4x}{\left(x-1\right)\left(x+1\right)}\right]:\left[\frac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}\right]\)
\(A=\left[\frac{4x}{\left(x-1\right)\left(x+1\right)}\right]:\left(\frac{x+1}{x-1}\right)\)
\(A=\frac{4x}{\left(x-1\right)\left(x+1\right)}\cdot\frac{x-1}{x+1}\)
\(A=\frac{4x\left(x-1\right)}{\left(x-1\right)\left(x+1\right)\left(x+1\right)}\)
\(A=\frac{4x}{2\left(x+1\right)}\)
\(A=\frac{2x}{x+1}\)
b) Thay A = -3 vào biểu thức a ta được:
\(\frac{2x}{x+1}=-3\)
\(\Rightarrow\)\(2x=-3\left(x+1\right)\)
\(\Rightarrow\)\(2x=-3x-3\)
\(\Rightarrow\)\(2x+3x=-3\)
\(\Rightarrow\)\(5x=-3\)
\(\Rightarrow\)\(x=-\frac{3}{5}\)
Vậy khi \(x=-\frac{3}{5}\)thì biểu thức A có giá trị là -3
5 tháng 2 2022
a: \(A=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}-\dfrac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{ab}}\)
\(=\sqrt{a}-\sqrt{b}-\sqrt{a}-\sqrt{b}=-2\sqrt{b}\)
b: \(B=\dfrac{2\sqrt{x}-x-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{x+\sqrt{x}+1}{x-1}\)
\(=\dfrac{-2x+\sqrt{x}-1}{\sqrt{x}-1}\cdot\dfrac{1}{x-1}\)
c: \(C=\dfrac{x-9-x+3\sqrt{x}}{x-9}:\left(\dfrac{3-\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}-2}{\sqrt{x}+3}+\dfrac{x-9}{x+\sqrt{x}-6}\right)\)
\(=\dfrac{3\left(\sqrt{x}-3\right)}{x-9}:\dfrac{9-x+x-4\sqrt{x}+4+x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{3}{\sqrt{x}+3}\cdot\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}{x-4\sqrt{x}+4}\)
\(=\dfrac{3}{\sqrt{x}-2}\)
31 tháng 5 2017
Câu 1:
\(A=\frac{x\left(1-x^2\right)}{1+x^2}:\left[\left(\frac{\left(1-x\right)\left(x^2+x+1\right)}{1-x}+x\right)\left(\frac{\left(1+x\right)\left(x^2-x+1\right)}{1+x}+x\right)\right]\)
\(=\frac{x\left(1-x^2\right)}{x^2+1}:\left[\left(x^2+2x+1\right)\left(x^2-2x+1\right)\right]\)
\(=\frac{x\left(1-x^2\right)}{\left(1+x^2\right)\left(1+x\right)^2\left(x-1\right)^2}=\frac{x}{\left(1+x^2\right)\left(x^2-1\right)}=\frac{x}{x^4-1}\)
Câu 2: thay x vào A có :
\(A=\frac{-\frac{1}{2}}{\frac{1}{4}-1}=\frac{2}{3}\)
Câu c :
2A=1 => \(\frac{x}{x^4-1}=\frac{1}{2}\)ĐK \(\hept{\begin{cases}x\ne1\\x\ne-1\end{cases}}\)
\(\Leftrightarrow x^4-2x-1=0\Leftrightarrow\left(x+1\right)\left(x^3-x^2+x-1\right)=0\)
\(\left(x+1\right)\left(x^2+1\right)\left(x-1\right)=0\Leftrightarrow\orbr{\begin{cases}x=1\\x=-1\end{cases}}\)loại do điều kiện vậy ko có giá trị nào của x thỏa mãn
NM
Nguyễn Minh Quang
Giáo viên
9 tháng 2 2021
Ta có \(A=[\frac{2}{\left(x+1\right)^3}\left(\frac{1}{x}+1\right)+\frac{1}{x^2+2x+1}\left(\frac{1}{x^2}+1\right)]:\frac{x-1}{x^3}\)
\(\Leftrightarrow A=\left[\frac{2}{\left(x+1\right)^3}.\frac{x+1}{x}+\frac{1}{\left(x+1\right)^2}.\frac{x^2+1}{x^2}\right].\frac{x^3}{x-1}\)
\(\Leftrightarrow A=\left[\frac{2x+x^2+1}{x^2\left(x+1\right)^2}\right].\frac{x^3}{x+1}=\frac{x}{x+1}\)
Để \(A=\frac{x}{x+1}< 1\Leftrightarrow\frac{1}{x+1}>0\Leftrightarrow x>-1\)
Để \(A=1-\frac{1}{x+1}\text{ nguyên thì }\frac{1}{x+1}\text{ nguyên hay }x\in\left\{-2,0\right\} \)
ĐKXĐ : \(x\ne0;x\ne1\)
\(\frac{a}{x}+\frac{b}{1-x}=\frac{1}{x\left(1-x\right)}\)
\(\frac{a\left(1-x\right)}{x\left(1-x\right)}+\frac{bx}{x\left(1-x\right)}=\frac{1}{x\left(1-x\right)}\)
\(\frac{a\left(1-x\right)+bx}{x\left(1-x\right)}=\frac{1}{x\left(1-x\right)}\)
\(\Rightarrow a\left(1-x\right)+bx=1\)
\(\Leftrightarrow a-ax+bx=1\)
\(\Leftrightarrow a=ax-bx\)
\(\Leftrightarrow a=x\left(a-b\right)\)
\(\Leftrightarrow x=\frac{a}{a-b}\)