Bài 3: Rút gọn phân thức

a) \(A=\dfrac{x}{x+3}-\dfrac{2}{x-3}+\dfrac{x^2-1}{9-x^2}\)

\(A=\dfrac{x\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}-\dfrac{2\left(x+3\right)}{\left(x+3\right)\left(x-3\right)}-\dfrac{x^2-1}{\left(x+3\right)\left(x-3\right)}\)

\(A=\dfrac{x^2-3x-2x-6-x^2+1}{\left(x+3\right)\left(x-3\right)}\)

\(A=\dfrac{-5x-5}{\left(x+3\right)\left(x-3\right)}\)

b) \(A:B=\dfrac{-5x-5}{\left(x+3\right)\left(x-3\right)}:\left(2-\dfrac{x+5}{x+3}\right)\)

\(A:B=\dfrac{-5\left(x+1\right)}{\left(x+3\right)\left(x-3\right)}:\dfrac{x+3-x-5}{x+3}\)

\(A:B=\dfrac{-5\left(x+1\right)}{\left(x+3\right)\left(x-3\right)}\cdot\dfrac{x+3}{-2}\)

\(A:B=\dfrac{5\left(x+1\right)}{2\left(x-3\right)}=\dfrac{1}{2}\)

\(\Rightarrow10x+10=2x-6\)

\(\Rightarrow10x-2x=-6-10\)

\(\Rightarrow8x=-16\)

\(\Rightarrow x=-2\)

c) \(x^2-x-2=0\Rightarrow x^2-2x+x-2=0\Rightarrow x\left(x-2\right)+\left(x-2\right)\Rightarrow\left(x+1\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\) Khi x = -1 ta có:

\(A=\dfrac{-5\cdot-1-5}{\left(-1-3\right)\left(-1+3\right)}=\dfrac{5-5}{-4\cdot2}=0\)

Khi x = 2 ta có:

\(A=\dfrac{-5\cdot2-5}{\left(2-3\right)\left(2+3\right)}=\dfrac{-10-5}{-1\cdot5}=\dfrac{-15}{-5}=3\)

\(A=\dfrac{\sqrt{2}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}+\dfrac{\sqrt{2}}{\sqrt{2}+\sqrt{2+\sqrt{3}}}\)

\(=\dfrac{\sqrt{2}\cdot\sqrt{2}}{2-\sqrt{4-2\sqrt{3}}}+\dfrac{\sqrt{2}\cdot\sqrt{2}}{2+\sqrt{4+2\sqrt{3}}}\)

\(=\dfrac{2}{2-\sqrt{\left(\sqrt{3}-1\right)^2}}+\dfrac{2}{2+\sqrt{\left(\sqrt{3}+1\right)^2}}\)

\(=\dfrac{2}{2-\sqrt{3}+1}+\dfrac{2}{2+\sqrt{3}+1}\)

\(=\dfrac{2}{3-\sqrt{3}}+\dfrac{2}{3+\sqrt{3}}\)

\(=\dfrac{2\left(3+\sqrt{3}\right)+2\left(3-\sqrt{3}\right)}{6}=\dfrac{12}{6}=2\)

a) \(A=\dfrac{80x^3-125x}{3\left(x-3\right)-\left(x-3\right)\left(8-4x\right)}\)

\(A=\dfrac{80x^3-125x}{\left(x-3\right)\left(-5+4x\right)}\) có nghĩa \(\Leftrightarrow\left(x-3\right)\left(-5+4x\right)\ne0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-3\ne0\\-5+4x\ne0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ne3\\x\ne\dfrac{5}{4}\end{matrix}\right.\)

\(\Leftrightarrow A=\dfrac{5x\left(16x^2-25\right)}{\left(x-3\right)\left(-5+4x\right)}\)

\(\Leftrightarrow A=\dfrac{5x\left(4x-5\right)\left(4x+5\right)}{\left(x-3\right)\left(4x-5\right)}\)

\(\Leftrightarrow A=\dfrac{5x\left(4x+5\right)}{x-3}\)

b) \(B=\dfrac{32x-8x^2+2x^3}{x^3+64}\) có nghĩa khi và chỉ khi

\(x^3+64\ne0\Leftrightarrow x^3\ne-64\Leftrightarrow x\ne-4\)

\(\Leftrightarrow B=\dfrac{2x\left(x^2-4x+16\right)}{\left(x+4\right)\left(x^2-4x+16\right)}\)

\(\Leftrightarrow B=\dfrac{2x}{x+4}\)

1: \(=\dfrac{-\left[\left(x+5\right)^2-9\right]}{\left(x+2\right)^2}=\dfrac{-\left(x+5-3\right)\left(x+5+3\right)}{\left(x+2\right)^2}\)

\(=\dfrac{-\left(x+2\right)\left(x+8\right)}{\left(x+2\right)^2}=\dfrac{-\left(x+8\right)}{x+2}\)

2: \(=\dfrac{2x\left(x^2-4x+16\right)}{\left(x+4\right)\left(x^2-4x+16\right)}=\dfrac{2x}{x+4}\)

3: \(=\dfrac{5x\left(x^2+1\right)}{\left(x^2-1\right)\left(x^2+1\right)}=\dfrac{5x}{x^2-1}\)

4: \(=\dfrac{3\left(x^2-4x+4\right)}{x\left(x^3-8\right)}=\dfrac{3\left(x-2\right)^2}{x\left(x-2\right)\left(x^2+2x+4\right)}\)

\(=\dfrac{3\left(x-2\right)}{x\left(x^2+2x+4\right)}\)

5: \(=\dfrac{2a\left(a-b\right)}{a\left(c+d\right)-b\left(c+d\right)}=\dfrac{2a\left(a-b\right)}{\left(c+d\right)\left(a-b\right)}=\dfrac{2a}{c+d}\)

6: \(=\dfrac{x\left(x-y\right)}{\left(x-y\right)\left(x+y\right)}\cdot\left(-1\right)=\dfrac{-x}{x+y}\)

7: \(=\dfrac{2\left(1-a\right)}{-\left(1-a^3\right)}=\dfrac{-2\left(1-a\right)}{\left(1-a\right)\left(1+a+a^2\right)}=-\dfrac{2}{1+a+a^2}\)

8: \(=\dfrac{x^4\left(x^3-1\right)}{\left(x^3-1\right)\left(x^3+1\right)}=\dfrac{x^4}{x^3+1}\)

9: \(=\dfrac{\left(x+2-x+2\right)\left(x+2+x-2\right)}{16x}=\dfrac{4\cdot2x}{16x}=\dfrac{1}{2}\)

10: \(=\dfrac{0.5\left(49x^2-y^2\right)}{0.5x\left(7x-y\right)}=\dfrac{1}{x}\cdot\dfrac{\left(7x-y\right)\left(7x+y\right)}{7x-y}\)

\(=\dfrac{7x+y}{x}\)

b: \(\dfrac{5x^2-10xy}{2\left(2y-x\right)^3}=\dfrac{5x\left(x-2y\right)}{-2\left(x-2y\right)^3}=\dfrac{-5x}{2\left(x-2y\right)^2}\)

c: \(\dfrac{x^2+5x+6}{x^2+4x+4}=\dfrac{\left(x+2\right)\left(x+3\right)}{\left(x+2\right)^2}=\dfrac{x+3}{x+2}\)

b) \(\dfrac{-5x}{2\left(x-2y\right)^2}\) 

c) \(\dfrac{x+3}{x+2}\)

\(=\dfrac{x^2+xy-x^2-y^2}{x+y}\cdot\dfrac{x-y+2y}{y\left(x-y\right)}\)

\(=\dfrac{y\left(x-y\right)}{x+y}\cdot\dfrac{x+y}{y\left(x-y\right)}=1\)

\(\left(x^3-x^2\right)^2-4x^2+8x-4=0\\ \Leftrightarrow x^6-2x^5+x^4-4x^2+8x-4=0\\ \Leftrightarrow\left(x-1\right)^2\left(x^2-2\right)\left(x^2+2\right)=0;x^2+2>0\forall x\\ \Rightarrow\left[{}\begin{matrix}x^2-2=0\\x-1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=\sqrt{2}\\x=-\sqrt{2}\end{matrix}\right.\)

a. \(A=\dfrac{1}{x-1}-\dfrac{1}{x+1}+\dfrac{4x+2}{x^2-1}\)

\(A=\dfrac{x+1}{\left(x-1\right)\left(x+1\right)}-\dfrac{x-1}{\left(x-1\right)\left(x+1\right)}+\dfrac{4x+2}{\left(x-1\right)\left(x+1\right)}\)

\(A=\dfrac{\left(x+1\right)-\left(x-1\right)+4x+2}{\left(x-1\right)\left(x+1\right)}\)

\(A=\dfrac{x+1-x+1+4x+2}{\left(x-1\right)\left(x+1\right)}\)

\(A=\dfrac{4x+4}{\left(x-1\right)\left(x+1\right)}=\dfrac{4\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{4}{x-1}\)

b) Ta có: \(A=\dfrac{4}{x-1}=\dfrac{4}{2015}\) (ĐK: \(x\ne\pm1\) )

\(\Leftrightarrow8060=4\left(x-1\right)\)

\(\Leftrightarrow8060=4x-4\)

\(\Leftrightarrow8064=4x\)

\(\Leftrightarrow x=\dfrac{8064}{4}=2016\left(tm\right)\)

c) Ta có: \(\dfrac{4}{x-1}\left(x\ne1\right)\)

Để \(\dfrac{4}{x-1}\) nhận giá trị nguyên thì \(4:\left(x-1\right)\Leftrightarrow x-1\in\text{Ư}\left(4\right)=\left\{1;4;2\right\}\)

Vậy với x ∈ {2; 5; 3; 0; -1; -3} thì biểu thức \(\dfrac{4}{x-1}\) nhận giá trị nguyên

d) Thay \(x=-\dfrac{1}{2}\) vào biểu thức A ta được:

\(\dfrac{4}{-\dfrac{1}{2}-1}=-3\)

Vậy biểu thức A có giá trị -3 tại \(x=-\dfrac{1}{2}\)

Điều kiện xác định là `{(x-3 ne 0),(x(x-3) ne 0):}`

                 `<=>{(x ne 3),(x ne 0):}`

      `=>bb A`

ĐCXĐ: \(\left\{{}\begin{matrix}x\ne0\\x-3\ne0\end{matrix}\right.\)⇔\(\left\{{}\begin{matrix}x\ne0\\x\ne3\end{matrix}\right.\)