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\(\dfrac{3-3x}{\left(1+x\right)^2}:\dfrac{6x^2-6}{x+1}\)
\(=\dfrac{3\left(1-x\right)}{\left(x+1\right)^2}:\dfrac{6\left(x^2-1\right)}{x+1}\)
\(=\dfrac{-3\left(x-1\right)}{\left(x+1\right)^2}:\dfrac{6\left(x+1\right)\left(x-1\right)}{x+1}\)
\(=\dfrac{-3\left(x-1\right)}{\left(x+1\right)^2}\cdot\dfrac{x+1}{6\left(x+1\right)\left(x-1\right)}\)
\(=\dfrac{-3\left(x-1\right)\left(x+1\right)}{6\left(x+1\right)^3\left(x-1\right)}=\dfrac{-3\left(x+1\right)}{6\left(x+1\right)\left(x+1\right)^2}=\dfrac{-3}{6\left(x+1\right)^2}=\dfrac{-1}{2\left(x+1\right)^2}\)
b) Bạn có thể viết kiểu latex được không ạ ?
a: \(=\dfrac{x^2-5x+x+4}{x\left(x-2\right)}=\dfrac{x^2-4x+4}{x\left(x-2\right)}=\dfrac{x-2}{x}\)
b: \(=\dfrac{x^2-6x+9+4x^2+8x-4x^2-8x}{\left(x-3\right)\left(x+2\right)}\)
\(=\dfrac{x-3}{x+2}\)
a) \(=\dfrac{x\left(x-5\right)+x+4}{x\left(x-2\right)}=\dfrac{x^2-4x+4}{x\left(x-2\right)}=\dfrac{\left(x-2\right)^2}{x\left(x-2\right)}=\dfrac{x-2}{x}\)
b) \(=\dfrac{\left(x-3\right)^2+4x\left(x+2\right)-8x-4x^2}{\left(x+2\right)\left(x-3\right)}=\dfrac{x^2-6x+9+4x^2+8x-8x-4x^2}{\left(x+2\right)\left(x-3\right)}\)
\(=\dfrac{x^2-6x+9}{\left(x+2\right)\left(x-3\right)}=\dfrac{\left(x-3\right)^2}{\left(x+2\right)\left(x-3\right)}=\dfrac{x-3}{x+2}\)
\(a,\left(2x-5\right)\left(5-x\right)=5\left(2x-5\right)-x\left(2x-5\right)=10x-25-2x^2+5x=15x-2x^2-25\\ b,\dfrac{1}{3x-2}-\dfrac{1}{3x+2}=\dfrac{3x+2-3x+2}{\left(3x-2\right)\left(3x+2\right)}=\dfrac{4}{\left(3x-2\right)\left(3x+2\right)}\)
\(c,\dfrac{3}{x-3}-\dfrac{6x}{x^2-9}+\dfrac{x}{x+3}=\dfrac{3\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\dfrac{6x}{\left(x-3\right)\left(x+3\right)}+\dfrac{x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{3x+9-6x+x^2-3x}{\left(x-3\right)\left(x+3\right)}=\dfrac{x^2-6x+9}{\left(x-3\right)\left(x+3\right)}=\dfrac{\left(x-3\right)^2}{\left(x-3\right)\left(x+3\right)}=\dfrac{x-3}{x+3}\)
a:\(2x^2y\left(3x^2-5xy+4y^2\right)-6x^4y-10x^3y^2+8x^2y^3\)
b: \(\left(x^2-9\right)^2-\left(x+3\right)\left(x-3\right)\left(x^2+9\right)\)
\(=x^4-18x^2+81-\left(x^4-81\right)\)
\(=-18x^2+162\)
a, \(\left(x^2-9\right)^2-\left(x-3\right)\left(x+3\right)\left(x^2+9\right)=\left(x^2-9\right)^2-\left(x^2-9\right)\left(x^2+9\right)\)
\(=x^4-18x^2+81-x^4+81=-18x^2+162\)
b, \(\left(x^2+x-3\right)\left(x^2-x+3\right)=\left[x^4-\left(x-3\right)^2\right]\)
\(=x^4-x^2+6x-9\)