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Bài 1:
a: \(A=\dfrac{x^2-3+x+3}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{x+3}{x}=\dfrac{x\left(x+1\right)}{x\left(x-3\right)}=\dfrac{x+1}{x-3}\)
b: Để A=3 thì 3x-9=x+1
=>2x=10
hay x=5
Bài 2:
a: \(A=\dfrac{x+x-2-2x-4}{\left(x-2\right)\left(x+2\right)}:\dfrac{x+2-x}{x+2}\)
\(=\dfrac{-6}{x-2}\cdot\dfrac{1}{2}=\dfrac{-3}{x-2}\)
b: Để A nguyên thì \(x-2\in\left\{1;-1;3;-3\right\}\)
hay \(x\in\left\{3;1;5;-1\right\}\)
\(a,=a^3+3a^2b+3ab^2+b^3-a^3+3a^2b-3ab^2+b^3-2b^3=6a^2b\\ b,=\left(6x+1-6x+1\right)^2=2^2=4\)
a.\(A=\dfrac{x^2-4x+4}{x^3-2x^2-\left(4x-8\right)}=\dfrac{\left(x-2\right)^2}{x^2\left(x-2\right)-4\left(x-2\right)}=\dfrac{\left(x-2\right)^2}{\left(x^2-4\right)\left(x-2\right)}=\dfrac{x-2}{\left(x-2\right)\left(x+2\right)}=\dfrac{1}{x+2}\)
\(A=\dfrac{\left(x-2\right)^2}{x^2\left(x-2\right)-4\left(x-2\right)}\left(x\ne\pm2\right)\\ A=\dfrac{\left(x-2\right)^2}{\left(x-2\right)^2\left(x+2\right)}=\dfrac{1}{x+2}\\ B=\dfrac{x+2-x+\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\cdot\dfrac{4\sqrt{x}}{3}\left(x>0\right)\\ B=\dfrac{4\sqrt{x}\left(\sqrt{x}+1\right)}{3\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}=\dfrac{4\sqrt{x}}{3\left(x-\sqrt{x}+1\right)}\)
\(H=\dfrac{a^2\left(a^{-2}b^3\right)^2\cdot b^{-1}}{\left(a^{-1}\cdot b\right)\cdot a^{-5}\cdot b^{-2}}\)
\(=\dfrac{a^2\cdot a^{-4}\cdot b^6\cdot b^{-1}}{a^{-1-5}\cdot b^{1-2}}\)
\(=\dfrac{a^{-2}\cdot b^5}{a^{-4}\cdot b^{-1}}=a^{-2+4}\cdot b^{5+1}=a^2b^6\)
\(H=\dfrac{a^2.a^{-4}.b^6.b^{-1}}{a^{-1}.b.a^{-5}.b^{-2}}=\dfrac{a^{2-4}.b^{6-1}}{a^{-1-5}.b^{1-2}}=\dfrac{a^{-2}.b^5}{a^{-6}.b^{-1}}=a^{-2-\left(-6\right)}.b^{5-\left(-1\right)}=a^4b^6\)
a) \(A=1+3+3^2+...+3^{100}\)
\(3A=3+3^2+3^3+...+3^{101}\)
\(3A-A=\left(3+3^2+3^3+...+3^{101}\right)-\left(1+3+3^2+...+3^{100}\right)\)
\(2A=3^{101}-1\)
\(A=\frac{3^{101}-1}{2}\)
b) \(B=2^{100}-2^{99}+2^{98}-2^{97}+...-2^3+2^2-2+1\)
\(2B=2^{101}-2^{100}+2^{99}-2^{98}+...-2^4+2^3-2^2+2\)
\(B+2B=\left(2^{100}-2^{99}+...-2+1\right)+\left(2^{101}-2^{100}+...-2^2+2\right)\)
\(3B=2^{101}+1\)
\(B=\frac{2^{101}+1}{3}\)
