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ĐKXĐ : \(x\ne\left\{1;0\right\}\)
a) \(P=\left(\dfrac{\left(x-1\right)^2}{3x+\left(x-1\right)^2}-\dfrac{1-2x^2+4x}{x^3-1}+\dfrac{1}{x-1}\right):\dfrac{2x}{x^3+x}\)
\(P=\left(\dfrac{\left(x-1\right)^2}{x^2+x+1}-\dfrac{1-2x^2+4x}{\left(x-1\right)\left(x^2+x+1\right)}+\dfrac{1}{x-1}\right)\cdot\dfrac{x\left(x^2+1\right)}{2x}\)
\(P=\left(\dfrac{\left(x-1\right)\left(x-1\right)^2}{\left(x-1\right)\left(x^2+x+1\right)}-\dfrac{1-2x^2+4x}{\left(x-1\right)\left(x^2+x+1\right)}+\dfrac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}\right)\cdot\dfrac{x^2+1}{2}\)
\(P=\left(\dfrac{\left(x-1\right)^3-1+2x^2-4x+x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}\right)\cdot\dfrac{x^2+1}{2}\)
\(P=\left(\dfrac{x^3-3x^2+3x-1-1+2x^2-4x+x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}\right)\cdot\dfrac{x^2+1}{2}\)
\(P=\left(\dfrac{x^3-1}{x^3-1}\right)\cdot\dfrac{x^2+1}{2}\)
\(P=1\cdot\dfrac{x^2+1}{2}\)
\(P=\dfrac{x^2+1}{2}\)
b) Vì \(x^2\ge0\forall x\)
\(\Rightarrow P\ge\dfrac{1}{2}\)
Dấu "=" xảy ra \(\Leftrightarrow x=0\)
Mà ĐKXĐ \(x\ne0\)
=> ... đến đây ko biết làm :v 
AI BIẾT LÀM HỘ ĐI
Cái này mk chưa học nên cx chưa rõ cách làm chính xác mong bạn thông cảm :)
a: \(M=\left(\dfrac{x\left(x-2\right)}{2\left(x^2+4\right)}+\dfrac{2x^2}{x^3-2x^2+4x-8}\right)\cdot\dfrac{x^2-x-2}{x^2}\)
\(=\left(\dfrac{x\left(x-2\right)}{2\left(x^2+4\right)}+\dfrac{2x^2}{\left(x^2+4\right)\left(x-2\right)}\right)\cdot\dfrac{\left(x-2\right)\left(x+1\right)}{x^2}\)
\(=\left(\dfrac{x\left(x-2\right)^2+4x^2}{2\left(x^2+4\right)\left(x-2\right)}\right)\cdot\dfrac{\left(x-2\right)\left(x+1\right)}{x^2}\)
\(=\dfrac{x^3-4x^2+4x+4x^2}{2\left(x^2+4\right)}\cdot\dfrac{x+1}{x^2}\)
\(=\dfrac{x\left(x^2+4\right)}{2\left(x^2+4\right)}\cdot\dfrac{x+1}{x^2}=\dfrac{x+1}{2x}\)
b: Thay x=1/2 vào M, ta được:
\(M=\left(\dfrac{1}{2}+1\right):\left(2\cdot\dfrac{1}{2}\right)=\dfrac{3}{2}\)
\(a,\dfrac{x^2-2x}{x^2-4}=\dfrac{x\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{x}{x+2}\)
b) \(\dfrac{x^2+5x+4}{x^2-1}=\dfrac{x^2+x+4x+4}{x^2-1}=\dfrac{\left(x+1\right)\left(x+4\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{x+4}{x-1}\)
c) \(\dfrac{x^4+4}{x\left(x^2+2\right)-2x^2-\left(x-1\right)^2-1}\)
\(=\dfrac{x^4+4x^2-4x^2+4}{x^3+2x-2x^2-x^2+2x-1-1}\)
\(=\dfrac{\left(x^2+2\right)^2-4x^2}{\left(x^3+2x-2x^2\right)-\left(x^2-2x+2\right)}\)
\(=\dfrac{\left(x^2+2-2x\right)\left(x^2+2+2x\right)}{x\left(x^2+2-2x\right)-\left(x^2+2-2x\right)}\)
\(=\dfrac{x^2+2+2x}{x-1}\)
Bài 2:
a) \(\left(\dfrac{2x+1}{2x-1}-\dfrac{2x-1}{2x+1}\right):\dfrac{4x}{10x-5}\)
\(=\dfrac{\left(2x+1\right)^2-\left(2x-1\right)^2}{\left(2x-1\right)\left(2x+1\right)}.\dfrac{5\left(2x-1\right)}{4x}\)
\(=\dfrac{8x}{\left(2x-1\right)\left(2x+1\right)}.\dfrac{5\left(2x-1\right)}{4x}\)
\(=\dfrac{10}{2x+1}\)
b) \(\left(\dfrac{1}{x^2+x}-\dfrac{2-x}{x+1}\right):\left(\dfrac{1}{x}+x-2\right)\)
\(=\dfrac{1-2x+x^2}{x\left(x+1\right)}:\dfrac{1+x^2-2x}{x}\)
\(=\dfrac{1}{x+1}\)
c) Trong ngoặc giữa hai phân số là dấu gì vậy ?
\(a,\frac{x}{xy-y^2}+\frac{2x-y}{xy-x^2}:\left(\frac{1}{x}+\frac{1}{y}\right)\)
\(=\left(\frac{x}{y\left(x-y\right)}+\frac{y-2x}{x\left(x-y\right)}\right):\left(\frac{y}{xy}+\frac{x}{xy}\right)\)
\(=\left(\frac{x-y}{x\left(x-y\right)}\right):\left(\frac{x+y}{xy}\right)\)
\(=\frac{1}{x}.\frac{xy}{x+y}=\frac{y}{x+y}\)
\(A=\left(\dfrac{x^2-2x}{2\left(x^2+4\right)}+\dfrac{2x^2}{x^3-2x^2+4x-8}\right)\cdot\dfrac{x^2-x-2}{x^2}\)
\(=\left(\dfrac{x\left(x-2\right)}{2\left(x^2+4\right)}+\dfrac{2x^2}{\left(x-2\right)\left(x^2+4\right)}\right)\cdot\dfrac{\left(x-2\right)\left(x+1\right)}{x^2}\)
\(=\left(\dfrac{x\left(x^2-4x+4\right)+4x^2}{2\left(x-2\right)\left(x^2+4\right)}\right)\cdot\dfrac{\left(x-2\right)\left(x+1\right)}{x^2}\)
\(=\dfrac{x\left(x^2-4x+4+4x\right)}{2\left(x^2+4\right)}\cdot\dfrac{x+1}{x^2}=\dfrac{x\left(x^2+4\right)}{2\left(x^2+4\right)}\cdot\dfrac{x+1}{x^2}\)
\(=\dfrac{x+1}{2x}\)
( x2−2x / 2x2+8 − 2x2 / 8−4x+2x2−x3 ).(1− 1/x − 2/x2 )
=[ x2−2x / 2(x2+4) − 2x2 / 2(x2+4)−x(x2+4) ]. x2−x−2 / x2
=[x2−2x / 2(x2+4) − 2x2 / (2−x)(x2+3)] . x2−x−2 / x2
=(x2−2x)(2−x)−4x2 / 2(2−x)(x2+4) . x2+x−2x−2 / x2
= −x(x2+4) / 2(2−x)(x2+4). (x+1)(x−2) / x2
=x+1 / 2x
a:
Sửa đè: \(B=\left(2x+1+\dfrac{1}{2x-1}\right):\left(\dfrac{2x^2-6x}{x-3}-\dfrac{4x^2}{2x-1}\right)\)
\(B=\dfrac{4x^2-1+1}{2x-1}:\left(2x-\dfrac{4x^2}{2x-1}\right)\)
\(=\dfrac{4x^2}{2x-1}:\dfrac{4x^2-2x-4x^2}{2x-1}\)
\(=\dfrac{4x^2}{-2x}=-2x\)
b: |x-2|=1
=>x-2=1 hoặc x-2=-1
=>x=1(nhận) hoặc x=3(loại)
Khi x=1 thì A=-2*1=-2