cho 1/a+1/b+1/c=1/a+b+c.Cm 1/a^3+1/b^3+1/c^3=1/(a^3+b^3+c^3)
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24 tháng 12 2020
a, \(\frac{x+1}{2x+6}+\frac{2x+3}{x^2+3x}=\frac{x+1}{2\left(x+3\right)}+\frac{3x+2}{x\left(x+3\right)}\)
\(=\frac{x^2+x}{2x\left(x+3\right)}+\frac{6x+4}{2x\left(x+3\right)}=\frac{x^2+7x+4}{2x\left(x+3\right)}\)
b, Sua de : \(\frac{3}{2x+6}-\frac{x-6}{2x^2+6x}=\frac{3}{2\left(x+3\right)}-\frac{x-6}{2x\left(x+3\right)}\)
\(=\frac{3x}{2x\left(x+3\right)}-\frac{x-6}{2x\left(x+3\right)}=\frac{2x+6}{2x\left(x+3\right)}=\frac{1}{x}\)
24 tháng 12 2020
a, \(B=\left(\frac{2x+1}{2x-1}+\frac{4}{1-4x^2}-\frac{2x-1}{2x+1}\right):\frac{x^2+2}{2x+1}\)
\(=\left(\frac{2x+1}{2x-1}+\frac{4}{\left(1-2x\right)\left(2x+1\right)}-\frac{2x-1}{2x+1}\right):\frac{x^2+2}{2x+1}\)
\(=\left(\frac{\left(2x+1\right)^2}{\left(2x-1\right)\left(2x+1\right)}-\frac{4}{\left(2x-1\right)\left(2x+1\right)}-\frac{\left(2x-1\right)^2}{\left(2x-1\right)\left(2x+1\right)}\right):\frac{x^2+2}{2x+1}\)
\(=\left(\frac{4x^2+4x+1-4-4x^2+4x-1}{\left(2x-1\right)\left(2x+1\right)}\right):\frac{x^2+2}{2x+1}\)
\(=\frac{8x-4}{\left(2x-1\right)\left(2x+1\right)}.\frac{2x+1}{x^2+2}=\frac{8x-4}{\left(2x-1\right)\left(x^2+2\right)}\)
b, Thay x = -1 ta được : \(\frac{9\left(-1\right)-4}{\left[2\left(-1\right)-1\right]\left[\left(-1\right)^2+2\right]}=-\frac{13}{-9}=\frac{13}{9}\)
24 tháng 12 2020
a, + b, \(A=\frac{x+2}{x-3}+\frac{2x-1}{x-1}-\frac{2x-1}{2x+1}\)DKXD : \(x\ne3;1;-\frac{1}{2}\)
\(=\frac{\left(x+2\right)\left(x-1\right)\left(2x+1\right)}{\left(x-3\right)\left(x-1\right)\left(2x+1\right)}+\frac{\left(4x^2-1\right)\left(x-3\right)}{\left(x-3\right)\left(x-1\right)\left(2x+1\right)}-\frac{\left(2x-1\right)\left(x-1\right)\left(x-3\right)}{\left(2x+1\right)\left(x-1\right)\left(x-3\right)}\)
\(=\frac{2x^3+3x^2-3x-2+4x^3-12x^2-x+4-2x^3+9x^2-10x+3}{MTC}\)
\(=\frac{4x^3-14x+2x^3+5}{MTC}\)
Đề sai ko kiểm tra lại hộ nhé !!!