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Bài 1 : Với : \(x>0;x\ne1\)
\(P=\left(1+\frac{1}{\sqrt{x}-1}\right)\frac{1}{x-\sqrt{x}}=\left(\frac{\sqrt{x}}{\sqrt{x}-1}\right).\sqrt{x}\left(\sqrt{x}-1\right)=x\)
Thay vào ta được : \(P=x=25\)
Bài 2 :
a, Với \(x\ge0;x\ne1\)
\(A=\frac{\sqrt{x}}{\sqrt{x}-1}-\frac{2}{\sqrt{x}+1}-\frac{2}{x-1}=\frac{x+\sqrt{x}-2\sqrt{x}+2-2}{x-1}\)
\(=\frac{x-\sqrt{x}}{x-1}=\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{\sqrt{x}}{\sqrt{x}+1}\)
Thay x = 9 vào A ta được : \(\frac{3}{3+1}=\frac{3}{4}\)
a.b. \(A=\dfrac{2}{x-1}+\dfrac{2\left(x+1\right)}{x^2+x+1}+\dfrac{x^2-10x+3}{x^3-1}\) ( x ≠ 1 )
\(A=\dfrac{2\left(x^2+x+1\right)+2\left(x+1\right)\left(x-1\right)+x^2-10x+3}{x^3-1}\)
\(A=\dfrac{2x^2+2x+2+2x^2-2+x^2-10x+3}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(A=\dfrac{5x^2-8x+3}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{5x^2-5x-3x+3}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{5x\left(x-1\right)-3\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{\left(x-1\right)\left(5x-3\right)}{x^2+x+1}=\dfrac{5x-3}{x^2+x+1}\)
c.
\(A=\dfrac{5x-3}{x^2+x+1}\)
\(\Leftrightarrow A\left(x^2+x+1\right)=5x-3\)
\(\Leftrightarrow Ax^2+Ax+A-5x+3=0\)
\(\Leftrightarrow Ax^2+\left(A-5\right)x+A+3=0\)
( \(a=A,b=A-5,c=A+3\) )
* A = 0 \(\Rightarrow x=\dfrac{3}{5}\)
* \(A\ge0\)
\(\Rightarrow\Delta=b^2-4ac\ge0\)
\(\Rightarrow\left(A-5\right)^2-4.A\left(A-3\right)\ge0\)
\(\Rightarrow A^2-10A+25-4A^2-12A\ge0\)
\(\Rightarrow-3A^2-22A+25\ge0\)
\(\Rightarrow-\dfrac{25}{4}\le A\le1\)
\(\Rightarrow Min_A=-\dfrac{25}{3}\Leftrightarrow x=\dfrac{-b}{2a}=\dfrac{\dfrac{25}{3}+5}{2.\left(\dfrac{-25}{3}\right)}=-\dfrac{4}{5}\)
a: \(A=\left(\dfrac{1}{x-1}+\dfrac{x}{\left(x-1\right)\left(x^2+x+1\right)}\cdot\dfrac{x^2+x+1}{x+1}\right)\cdot\dfrac{\left(x+1\right)^2}{2x+1}\)
\(=\left(\dfrac{1}{x-1}+\dfrac{x}{\left(x-1\right)\left(x+1\right)}\right)\cdot\dfrac{\left(x+1\right)^2}{2x+1}\)
\(=\dfrac{x+1+x}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{\left(x+1\right)^2}{2x+1}\)
\(=\dfrac{2x+1}{x-1}\cdot\dfrac{x+1}{2x+1}=\dfrac{x+1}{x-1}\)
b: Thay x=1/2 vào A, ta được:
\(A=\dfrac{\dfrac{1}{2}+1}{\dfrac{1}{2}-1}=\dfrac{3}{2}:\dfrac{-1}{2}=-3\)
c: Để A là số nguyên thì \(x-1+2⋮x-1\)
\(\Leftrightarrow x-1\in\left\{1;-1;2;-2\right\}\)
\(\Leftrightarrow x\in\left\{2;0;3\right\}\)
Bài 2 :
a) Phân thức A xác định \(\Leftrightarrow\hept{\begin{cases}x-2\ne0\\x+2\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ne2\\x\ne-2\end{cases}}}\)
b) \(A=\left(\frac{1}{x-2}-\frac{1}{x+2}\right)\cdot\frac{x^2-4x+4}{4}\)
\(A=\left(\frac{x+2}{\left(x-2\right)\left(x+2\right)}-\frac{x-2}{\left(x-2\right)\left(x+2\right)}\right)\cdot\frac{\left(x-2\right)^2}{4}\)
\(A=\left(\frac{x+2-x+2}{\left(x-2\right)\left(x+2\right)}\right)\cdot\frac{\left(x-2\right)^2}{4}\)
\(A=\frac{4}{\left(x-2\right)\left(x+2\right)}\cdot\frac{\left(x-2\right)^2}{4}\)
\(A=\frac{4\cdot\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)\cdot4}\)
\(A=\frac{x-2}{x+2}\)
c) Thay x = 4 ta có :
\(A=\frac{4-2}{4+2}=\frac{2}{6}=\frac{1}{3}\)
Vậy.........
\(4x^2y^3.\frac{2}{4}x^3y=4x^2y^3.\frac{1}{2}x^3y=2x^5y^4\)
\(\left(5x-2\right)\left(25x^2+10x+4\right)\)
\(=\left(5x-2\right)\left[\left(5x\right)^2+5x.2+2^2\right]\)
\(=\left(5x\right)^3-2^3\)
\(=125x^3-8\)
giúp vs
a) ĐKXĐ: \(x\ne1\)
b) \(A=\frac{2}{x-1}+\frac{2\left(x+1\right)}{x^2+x+1}+\frac{x^2-10x+3}{x^3-1}\)
\(=\frac{2\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{2\left(x+1\right)\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{x^2-10x+3}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\frac{2x^2+2x+2}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{2x^2-2}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{x^2-10x+3}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\frac{5x^2-8x+3}{\left(x-1\right)\left(x^2+x+1\right)}=\frac{\left(x-1\right)\left(5x-3\right)}{\left(x-1\right)\left(x^2+x+1\right)}=\frac{5x-3}{x^2+x+1}\)