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a: \(=\dfrac{4x-8+2x+4-8}{\left(x-2\right)\left(x+2\right)}=\dfrac{6x-12}{\left(x-2\right)\left(x+2\right)}=\dfrac{6}{x+2}\)
b: \(=\dfrac{-x+7x-4}{3x-2}=\dfrac{6x-4}{3x-2}=2\)
c: \(=\dfrac{x}{2x+1}-\dfrac{1}{\left(2x+1\right)\left(2x-1\right)}-\dfrac{\left(x-2\right)}{2x-1}\)
\(=\dfrac{2x^2-x-1-\left(x-2\right)\left(2x+1\right)}{\left(2x+1\right)\left(2x-1\right)}\)
\(=\dfrac{2x^2-x-1-2x^2-x+4x+2}{\left(2x+1\right)\left(2x-1\right)}\)
\(=\dfrac{2x+1}{\left(2x+1\right)\left(2x-1\right)}=\dfrac{1}{2x-1}\)
d: \(=\dfrac{5}{2x-3}+\dfrac{2}{2x+3}+\dfrac{2x-33}{4x^2-99}\)
\(=\dfrac{10x+15+4x-6+2x-33}{\left(2x-3\right)\left(2x+3\right)}=\dfrac{16x-24}{\left(2x-3\right)\left(2x+3\right)}=\dfrac{8}{2x+3}\)
a/\(\left(x-1\right)\left(x^5+x^4+x^3+x^2+x+1\right).\)
\(=\left(x-1\right)\left[\left(x^5+x^4+x^3\right)+\left(x^2+x+1\right)\right]\)
\(=\left(x-1\right)\left[x^3\left(x^2+x+1\right)+\left(x^2+x+1\right)\right]\)
\(=\left(x-1\right)\left(x^2+x+1\right)\left(x^3+1\right)\)
\(=\left(x-1\right)\left(x^2+x+1\right)\left(x+1\right)\left(x^2-x+1\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(x^2-x+1\right)\left(x^2+x+1\right)\)
\(=\left(x^2-1\right)\left(x^2-x+1\right)\left(x^2+x+1\right)\)
Câu b/ quên làm ạ :> Bù nè
b/ \(2\left(3x-1\right)\left(2x+5\right)-\left(4x-1\right)\left(3x-2\right)\)
\(=2\left(6x^2+15x-2x-5\right)-\left(12x^2-8x-3x+2\right)\)
\(=2\left(6x^2+13x-5\right)-\left(12x^2-11x+2\right)\)
\(=12x^2+26x-10-\left(12x^2-11x+2\right)\)
\(=12x^2+26x-10-12x^2+11x-2\)
\(=37x-12\)
Bài 2:
a: \(\Leftrightarrow\left(x-2\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
Bạn cần viết lại đề bằng công thức toán (biểu tượng $\sum$ góc trái khung soạn thảo). Viết như thế này nhìn khó đọc quá.
a) \(\left(x^2-1\right)\left(x^2+2x\right)=x^4+2x^3-x^2-2x\)
b) \(\left(2x-1\right)\left(3x+2\right)\left(3-x\right)=6x^2-3x+4x-2\left(3-x\right)\)
\(=6x^2-3x+4x-6+2x\)
\(=6x^2+3x-6\)
c) \(\left(x+3\right)\left(x^2+3x-5\right)=x^3+3x^2+3x^2+9x-5x-15\)
\(=x^3+6x^2+4x-15\)
d) \(\left(x+1\right)\left(x^2-x+1\right)=x^3+x^2-x^2-x+x+1\)
\(=x^3+1\)
e) \(\left(2x^3-3x-1\right)\left(5x+2\right)=10x^4-15x^2-5x+4x^3-6x-2\)
\(=10x^4+4x^3-15x^2-11x-2\)
f) \(\left(x^2-2x+3\right)\left(x-4\right)=x^3-2x^2+3x-4x^2+8x-12\)
\(=x^3-6x^2+11x-12\)
a: \(=\dfrac{x^2-2x+1}{x}:\dfrac{x-1-3x^2+3x-3}{\left(x-1\right)\left(x^2-x+1\right)}\)
\(=\dfrac{\left(x-1\right)^2}{x}\cdot\dfrac{\left(x-1\right)\left(x^2-x+1\right)}{-2x^2+4x-4}\)
\(=\dfrac{\left(x-1\right)^3\cdot\left(x^2-x+1\right)}{-2x\left(x^2-2x+2\right)}\)
b: \(=\left[\dfrac{x^2-2x+1}{x^2+x+1}+\dfrac{2x^2-4x+1}{\left(x-1\right)\left(x^2+x+1\right)}+\dfrac{1}{x-1}\right]:\dfrac{2}{x^2+1}\)
\(=\dfrac{x^3-3x^2+3x+1+2x^2-4x+1+x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}\cdot\dfrac{x^2+1}{2}\)
\(=\dfrac{x^3+3}{\left(x-1\right)\left(x^2+x+1\right)}\cdot\dfrac{x^2+1}{2}\)