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a: A=[(3x^2+3-x^2+2x-1-x^2-x-1)/(x-1)(x^2+x+1)]*(x-2)/2x^2-5x+5
=(x^2+x+1)/(x-1)(x^2+x+1)*(x-2)/2x^2-5x+5
=(x-2)/(2x^2-5x+5)(x-1)
a: \(=\dfrac{x^2+2x+1-x^2+2x-1}{\left(x-1\right)\left(x+1\right)}:\left(\dfrac{1}{x+1}+\dfrac{x}{x-1}+\dfrac{2}{\left(x-1\right)\left(x+1\right)}\right)\)
\(=\dfrac{4x}{\left(x-1\right)\left(x+1\right)}:\dfrac{x-1+x^2+x+2}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{4x}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{\left(x-1\right)\left(x+1\right)}{x^2+2x+1}=\dfrac{4x}{x^2+2x+1}\)
b: \(=\dfrac{x+2}{-\left(x-2\right)}\cdot\dfrac{\left(x-2\right)^2}{4x^2}\cdot\left(\dfrac{2}{2-x}-\dfrac{4}{\left(x+2\right)\left(x^2-2x+4\right)}\cdot\dfrac{x^2-2x+4}{2-x}\right)\)
\(=\dfrac{-\left(x+2\right)\left(x-2\right)}{4x^2}\cdot\left(\dfrac{2}{2-x}-\dfrac{4}{\left(x+2\right)\left(2-x\right)}\right)\)
\(=\dfrac{-\left(x+2\right)\left(x-2\right)}{4x^2}\cdot\dfrac{2x+4-4}{\left(2-x\right)\left(x+2\right)}\)
\(=\dfrac{2x}{4x^2}=\dfrac{1}{2x}\)
\(1.\)
\(a.\)
\(\dfrac{8}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{2}{x^2+3}+\dfrac{1}{x+1}\)
\(=\dfrac{8}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{2\left(x^2-1\right)}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{1\left(x-1\right)\left(x^2+3\right)}{\left(x^2-1\right)\left(x^2+3\right)}\)
\(=\dfrac{8}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{2x^2-2}{\left(x^2+3\right)\left(x^2-1\right)}+\dfrac{x^3-x^2+3x-3}{\left(x^2-1\right)\left(x^2+3\right)}\)
\(=\dfrac{8+2x^2-2+x^3-x^2+3x-3}{\left(x^2+3\right)\left(x^2-1\right)}\)
\(=\dfrac{x^3+x^2+3x+3}{\left(x^2+3\right)\left(x^2-1\right)}\)
\(=\dfrac{x^2\left(x+1\right)+3\left(x+1\right)}{\left(x^2+3\right)\left(x^2-1\right)}\)
\(=\dfrac{\left(x^2+3\right)\left(x+1\right)}{\left(x^2+3\right)\left(x^2-1\right)}\)
\(=x-1\)
\(b.\)
\(\dfrac{x+y}{2\left(x-y\right)}-\dfrac{x-y}{2\left(x+y\right)}+\dfrac{2y^2}{x^2-y^2}\)
\(=\dfrac{x+y}{2\left(x-y\right)}-\dfrac{x-y}{2\left(x+y\right)}+\dfrac{2y^2}{\left(x-y\right)\left(x+y\right)}\)
\(=\dfrac{\left(x+y\right)^2}{2\left(x^2-y^2\right)}-\dfrac{\left(x-y\right)^2}{2\left(x^2-y^2\right)}+\dfrac{4y^2}{2\left(x^2-y^2\right)}\)
\(=\dfrac{x^2+2xy+y^2}{2\left(x^2-y^2\right)}-\dfrac{x^2-2xy+y^2}{2\left(x^2-y^2\right)}+\dfrac{4y^2}{2\left(x^2-y^2\right)}\)
\(=\dfrac{x^2+2xy+y^2-x^2+2xy-y^2+4y^2}{2\left(x^2-y^2\right)}\)
\(=\dfrac{4xy+4y^2}{2\left(x^2-y^2\right)}\)
\(=\dfrac{4y\left(x+y\right)}{2\left(x^2-y^2\right)}\)
\(=\dfrac{2y}{\left(x-y\right)}\)
Tương tự các câu còn lại
Lời giải của bạn Nhật Linh đúng rồi, tuy nhiên cần thêm điều kiện để A có nghĩa: \(x\ne\pm2\)
Bài 2 .
a) \(\dfrac{2x}{x^2+2xy}+\dfrac{y}{xy-2y^2}+\dfrac{4}{x^2-4y^2}\)
\(=\dfrac{2x}{x\left(x+2y\right)}+\dfrac{y}{y\left(x-2y\right)}+\dfrac{4}{\left(x-2y\right)\left(x+2y\right)}\)
\(=\dfrac{2xy\left(x-2y\right)+xy\left(x+2y\right)+4xy}{xy\left(x+2y\right)\left(x-2y\right)}\)
\(=\dfrac{2x^2y-2xy^2+x^2y+2xy^2+4xy}{xy\left(x+2y\right)\left(x-2y\right)}\)
\(=\dfrac{3x^2y+4xy}{xy\left(x+2y\right)\left(x-2y\right)}\)
b) Sai đề hay sao ý
c) \(\dfrac{2x+y}{2x^2-xy}+\dfrac{16x}{y^2-4x^2}+\dfrac{2x-y}{2x^2+xy}\)
\(=\dfrac{2x+y}{x\left(2x-y\right)}+\dfrac{-16x}{\left(2x-y\right)\left(2x+y\right)}+\dfrac{2x-y}{x\left(2x+y\right)}\)
\(=\dfrac{\left(2x+y\right)^2-16x^2+\left(2x-y\right)^2}{x\left(2x-y\right)\left(2x+y\right)}\)
\(=\dfrac{4x^2+4xy+y^2-16x^2+4x^2-4xy+y^2}{x\left(2x-y\right)\left(2x+y\right)}\)
\(=\dfrac{-8x^2}{x\left(2x-y\right)\left(2x+y\right)}\)
d) \(\dfrac{1}{1-x}+\dfrac{1}{1+x}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{2}{1-x^2}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{4}{1-x^4}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
.....
\(=\dfrac{16}{1-x^{16}}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{32}{1-x^{32}}\)
a) \(M=\left(\dfrac{1}{1-x}+\dfrac{2}{x+1}-\dfrac{5-x}{1-x^2}\right):\dfrac{1-2x}{x^2-1}\)
\(\Leftrightarrow M=\left(\dfrac{-1}{x-1}+\dfrac{2}{x+1}+\dfrac{5-x}{x^2-1}\right):\dfrac{1-2x}{x^2-1}\)
\(\Leftrightarrow M=\left(\dfrac{-1}{x-1}+\dfrac{2}{x+1}+\dfrac{5-x}{\left(x-1\right)\left(x+1\right)}\right):\dfrac{1-2x}{x^2-1}\)
\(\Leftrightarrow M=\left(\dfrac{-\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}+\dfrac{2\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}+\dfrac{5-x}{\left(x-1\right)\left(x+1\right)}\right):\dfrac{1-2x}{x^2-1}\)
\(\Leftrightarrow M=\dfrac{-\left(x+1\right)+2\left(x-1\right)+\left(5-x\right)}{\left(x-1\right)\left(x+1\right)}:\dfrac{1-2x}{x^2-1}\)
\(\Leftrightarrow M=\dfrac{-x-1+2x-2+5-x}{\left(x-1\right)\left(x+1\right)}:\dfrac{1-2x}{x^2-1}\)
\(\Leftrightarrow M=\dfrac{2}{\left(x-1\right)\left(x+1\right)}:\dfrac{1-2x}{x^2-1}\)
\(\Leftrightarrow M=\dfrac{2}{\left(x-1\right)\left(x+1\right)}.\dfrac{x^2-1}{1-2x}\)
\(\Leftrightarrow M=\dfrac{2\left(x^2-1\right)}{\left(x-1\right)\left(x+1\right)\left(1-2x\right)}\)
\(\Leftrightarrow M=\dfrac{2\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)\left(1-2x\right)}\)
\(\Leftrightarrow M=\dfrac{2}{1-2x}\)
b) \(M=\dfrac{2}{1-2x}=\dfrac{-2}{3}\)
\(\Rightarrow2.3=\left(1-2x\right).\left(-2\right)\)
\(\Rightarrow6=-2+4x\)
\(\Rightarrow4x=6-\left(-2\right)\)
\(\Rightarrow4x=6+2\)
\(\Rightarrow4x=8\)
\(\Rightarrow x=8:4\)
\(\Rightarrow x=2\)
Vậy \(M=\dfrac{-2}{3}\) thì \(x=2\)
c) Để \(M=\dfrac{2}{1-2x}\in Z\) \(\Leftrightarrow2⋮1-2x\)
\(\Rightarrow1-2x\in U\left(2\right)=\left\{-1;1;-2;2\right\}\)
\(\Rightarrow\left\{{}\begin{matrix}1-2x=-1\Rightarrow x=1\\1-2x=1\Rightarrow x=0\\1-2x=-2\Rightarrow x=1,5\\1-2x=2\Rightarrow x=-0,5\end{matrix}\right.\)
Mà \(x\in Z\)
\(\Rightarrow x\in\left\{1;0\right\}\)
Vậy \(x=1\) hoặc \(x=0\) thì \(M\in Z\)
a) M = \(\left(\dfrac{1}{1-x}+\dfrac{2}{x+1}-\dfrac{5-x}{1-x^2}\right):\dfrac{1-2x}{x^2-1}\)
= \(\left(\dfrac{1}{1-x}+\dfrac{2}{1+x}-\dfrac{5-x}{\left(1-x\right)\left(1+x\right)}\right).\dfrac{x^2-1}{1-2x}\)
= \(\left(\dfrac{1+x}{\left(1-x\right)\left(1+x\right)}+\dfrac{2\left(1-x\right)}{\left(1-x\right)\left(1+x\right)}-\dfrac{5-x}{\left(1-x\right)\left(1+x\right)}\right).\dfrac{\left(x-1\right)\left(x+1\right)}{1-2x}\)
= \(\dfrac{1+x+2-2x-5+x}{\left(1-x\right)\left(1+x\right)}.\dfrac{\left(x-1\right)\left(x+1\right)}{1-2x}\)\(=\dfrac{-2}{\left(1-x\right)\left(1+x\right)}.\dfrac{\left(x-1\right)\left(x+1\right)}{1-2x}\)
= \(\dfrac{2}{\left(x-1\right)\left(x+1\right)}.\dfrac{\left(x-1\right)\left(x+1\right)}{1-2x}\)
=\(\dfrac{2}{1-2x}\)
b) M = \(\dfrac{-2}{3}\Leftrightarrow\dfrac{2}{1-2x}=\dfrac{-2}{3}\)
=> 2 . 3 = -2 (1 - 2x) (tích chéo)
=> 6 = -2 + 4x
=> 6 + 2 - 4x = 0
=> 8 - 4x = 0
=> 4x = 8
=> x = 2 (thỏa mãn đkxđ)
Vậy để M = \(\dfrac{-2}{3}\) thì x = 2
a: \(P=\dfrac{x^2-4-x^2+3}{x\left(x-2\right)}\cdot\dfrac{\left(x-2\right)\left(x+2\right)}{5}\)
\(=\dfrac{-x-2}{5x}\)
b: Để P=x2 thì \(5x^3+x+2=0\)
hay \(x\in\left\{-0.65\right\}\)
c: Để |P|<P thì \(x\in\varnothing\)