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an có 10000000 quả cam an cho mẹ gấp đôi rồi an co ba số quả lớn hơn mẹ 200 vậy an còn bao nhiêu quả cam
a) \(\frac{5x-1}{3x^2y}+\frac{x-1}{3x^2y}=\frac{5x-1+x-1}{3x^2y}=\frac{6x}{3x^2y}=\frac{2}{xy}\)
b) \(\frac{7}{12xy^2}+\frac{11}{18x^3y}=\frac{7\left(\frac{3}{2}x^2\right)}{18x^3y^2}+\frac{11y}{18x^3y^2}=\frac{10,5x^2+11y}{18x^3y^2}\)
c) \(\frac{x}{x+2}+\frac{7x-16}{\left(x+2\right)\left(4x-7\right)}=\frac{x\left(4x-7\right)}{\left(x+2\right)\left(4x-7\right)}+\frac{7x-16}{\left(x+2\right)\left(4x-7\right)}\)
\(=\frac{4x^2-7x+7x-16}{\left(x+2\right)\left(4x-7\right)}=\frac{4x^2-16}{\left(x+2\right)\left(4x-7\right)}\)
\(\frac{4}{x-1}-\frac{2}{1-x}-\frac{x}{x-1}\)
\(=\frac{4}{x-1}+\frac{-2}{x-1}-\frac{x}{x-1}\)
\(=\frac{2-x}{x-1}\)
ĐKXĐ: \(x\ne1\)
\(\frac{4}{x-1}-\frac{2}{1-x}-\frac{x}{x-1}\)
\(=\frac{4}{x-1}+\frac{2}{x-1}-\frac{x}{x-1}\)
\(=\frac{4+2-x}{x-1}\)
\(=\frac{6-x}{x-1}\)
1,\(\frac{3}{2x+6}-\frac{x-6}{x\left(2x+6\right)}\)
=\(\frac{3x}{x\left(2x+6\right)}+\frac{x-6}{x\left(2x+6\right)}\)
=\(\frac{3x+x-6}{x\left(2x+6\right)}\)=\(\frac{4x-6}{x\left(2x+6\right)}=\frac{2\left(2x-3\right)}{x\left(2x+6\right)}\)
Ta có:\(\frac{1}{1-x}+\frac{1}{1+x}+\frac{2}{1+x^2}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)
\(=\frac{1+x}{\left(1-x\right)\left(1+x\right)}+\frac{1-x}{\left(1-x\right)\left(1+x\right)}+\frac{2}{1+x^2}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)\(=\frac{2}{\left(1-x\right)\left(1+x\right)}+\frac{2}{1+x^2}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)
\(=\frac{2}{1-x^2}+\frac{2}{1+x^2}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)
\(=\frac{2\left(1+x^2\right)}{\left(1-x^2\right)\left(1+x^2\right)}+\frac{2\left(1-x^2\right)}{\left(1-x^2\right)\left(1+x^2\right)}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)
\(=\frac{2+2x^2}{\left(1-x^2\right)\left(1+x^2\right)}+\frac{2-2x^2}{\left(1-x^2\right)\left(1+x^2\right)}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)
\(=\frac{2+2}{\left(1-x^2\right)\left(1+x^2\right)}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)
\(=\frac{4}{1-x^4}+\frac{4}{1+x^4}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)
\(=\frac{4\left(1+x^4\right)}{\left(1-x^4\right)\left(1+x^4\right)}+\frac{4\left(1-x^4\right)}{\left(1-x^4\right)\left(1+x^4\right)}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)
\(=\frac{4+4x^4}{\left(1-x^4\right)\left(1+x^4\right)}+\frac{4-4x^4}{\left(1-x^4\right)\left(1+x^4\right)}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)
\(=\frac{4+4}{\left(1-x^4\right)\left(1+x^4\right)}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)
\(=\frac{8}{1-x^8}+\frac{8}{1+x^8}+\frac{16}{1+x^{16}}\)
\(=\frac{8\left(1+x^8\right)}{\left(1-x^8\right)\left(1+x^8\right)}+\frac{8\left(1-x^8\right)}{\left(1-x^8\right)\left(1+x^8\right)}+\frac{16}{1+x^{16}}\)
\(=\frac{8+8x^8}{\left(1-x^8\right)\left(1+x^8\right)}+\frac{8-8x^8}{\left(1-x^8\right)\left(1+x^8\right)}+\frac{16}{1+x^{16}}\)
\(=\frac{8+8}{\left(1-x^8\right)\left(1+x^8\right)}+\frac{16}{1+x^{16}}\)
\(=\frac{16}{1-x^{16}}+\frac{16}{1+x^{16}}\)
\(=\frac{16\left(1+x^{16}\right)}{\left(1-x^{16}\right)\left(1+x^{16}\right)}+\frac{16\left(1-x^{16}\right)}{\left(1-x^{16}\right)\left(1+x^{16}\right)}\)
\(=\frac{16+16}{\left(1-x^{16}\right)\left(1+x^{16}\right)}\)
\(=\frac{32}{1-x^{32}}\)
ĐKXĐ\(\begin{cases}x^2-4\ne0\\x+2\ne0\\2-x\ne0\end{cases}\)<=>\(\begin{cases}x\ne-2\\x\ne2\end{cases}\)
Với \(x\ne-2,x\ne2\) ta có
\(\frac{x^2}{x^2-4}+\frac{1}{x+2}+\frac{2}{2-x}\)=\(\frac{x^2}{\left(x+2\right)\left(x-2\right)}+\frac{1}{x+2}-\frac{2}{x-2}\)
=\(\frac{x^2}{\left(x+2\right)\left(x-2\right)}+\frac{x-2}{\left(x+2\right)\left(x-2\right)}-\frac{2\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}\)
=\(\frac{x^2+x-2-2x-4}{\left(x+2\right)\left(x-2\right)}\)=\(\frac{x^2-x+6}{\left(x+2\right)\left(x-2\right)}\)
=\(\frac{\left(x+2\right)\left(x-3\right)}{\left(x+2\right)\left(x-2\right)}\)=\(\frac{x-3}{x-2}\)