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\(C=\left(\frac{1+a^3}{1+a}-a\right)\left(\frac{2a^2+4}{a^3-8}-\frac{a}{a^2+2a+4}\right)\)
\(=\left(\frac{\left(a+1\right)\left(a^2-a+1\right)}{1-a}-\frac{\left(1-a\right)a}{1-a}\right)\left(\frac{2a^4}{\left(a-2\right)\left(a^2+2a+4\right)}-\frac{a}{a^2+2a+4}\right)\)
\(=\left(\frac{a^3+1-a+a^2}{1-a}\right)\left(\frac{2a^4}{\left(a-2\right)\left(a^2+2a+4\right)}-\frac{a\left(a-2\right)}{\left(a-2\right)\left(a^2+2a+4\right)}\right)\)
\(=\left(\frac{a^3+1-a+a^2}{1-a}\right)\left(\frac{2a^4-a^2+2a}{\left(a-2\right)\left(a^2-2a+4\right)}\right)\)
\(=\left(\frac{a^3+1-a+a^2}{-\left(a-1\right)}\right)\left(\frac{2a\left(a^3-1\right)}{\left(a-2\right)\left(a^2-2a+4\right)}\right)\)
tình nốt nhé, thấy sai sai ở đâu á, kiểm tra lại zùm mk
Bài làm
\(\frac{4x^3y^2-6x^2y^3}{2xy+2xy\left(y-x\right)}=\frac{2x^2y^2\left(2x-3y\right)}{2xy\left(1+y-x\right)}=\frac{xy\left(2x-3y\right)}{1+y-x}\)
Học tốt
\(\frac{4x^3y^2-6x^2y^3}{2xy+2xy\left(y-x\right)}=\frac{2x^2y^2\left(2x-3y\right)}{2xy\left(1+y-x\right)}=\frac{xy\left(2x-3y\right)}{y-x+1}\)
a, \(\frac{x^2}{x+1}+\frac{2x}{x^2-1}+\frac{1}{x+1}+1\)
\(=\frac{x^2\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}-\frac{2x}{\left(x-1\right)\left(x+1\right)}+\frac{x-1}{\left(x+1\right)\left(x-1\right)}+\frac{\left(x+1\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}\)
\(=\frac{x^3-x^2-2x+x-1-x^2-1}{\left(x-1\right)\left(x+1\right)}\)
\(=\frac{x^3-2x^2-x-2}{\left(x-1\right)\left(x+1\right)}\)