rút gọn biểu thức: A=x4-x3-x+1 / x4+x3+3x2+2x+2
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ND
Nguyễn Đức Trí
VIP
22 tháng 8 2023
a) \(...=P\left(x\right)=2x^4-x^4+3x^3+4x^2-3x^2+3x-x+3\)
\(P\left(x\right)=x^4+3x^3+x^2+2x+3\)
\(...=Q\left(x\right)=x^4+x^3+3x^2-x^2+4x+4-2\)
\(Q\left(x\right)=x^4+x^3+2x^2+4x+2\)
b) \(P\left(x\right)+Q\left(x\right)=\left(x^4+3x^3+x^2+2x+3\right)+\left(x^4+x^3+2x^2+4x+2\right)\)
\(\Rightarrow P\left(x\right)+Q\left(x\right)=2x^4+4x^3+3x^2+6x+5\)
\(P\left(x\right)-Q\left(x\right)=\left(x^4+3x^3+x^2+2x+3\right)-\left(x^4+x^3+2x^2+4x+2\right)\)
\(\)\(\Rightarrow P\left(x\right)-Q\left(x\right)=x^4+3x^3+x^2+2x+3-x^4-x^3-2x^2-4x-2\)
\(\Rightarrow P\left(x\right)-Q\left(x\right)=2x^3-x^2-2x+1\)
9 tháng 9 2023
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: \(=\dfrac{0.5\left(49x^2-y^2\right)}{0.5x\left(7x-y\right)}=\dfrac{1}{x}\cdot\dfrac{\left(7x-y\right)\left(7x+y\right)}{7x-y}\)
\(=\dfrac{7x+y}{x}\)
PT
1
NH
1
4 tháng 7 2023
1: \(=\dfrac{x-1}{x^2+x+1}+\dfrac{x+1}{x-1}\)
\(=\dfrac{x^2-2x+1+x^3+x^2+x^2+x+x+1}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{x^3+3x^2+2}{\left(x-1\right)\left(x^2+x+1\right)}\)
2: \(=\dfrac{\left(x^2-y^2\right)^2}{\left(x-y\right)\left(x^2+xy+y^2\right)}=\dfrac{\left(x-y\right)\left(x+y\right)^2}{x^2+xy+y^2}\)
PT
1
KN
2
28 tháng 10 2019
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