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\(B=\dfrac{a}{x^2+ax}+\dfrac{a}{x^2+3ax+2a^2}+\dfrac{a}{x^2+5ax+6a^2}+\dfrac{a}{x^2+7ax+12a^2}+\dfrac{a}{x^2+9ax+20a^2}\)
\(=\dfrac{a}{x\left(x+a\right)}+\dfrac{a}{\left(x+a\right)\left(x+2a\right)}+\dfrac{a}{\left(x+2a\right)\left(x+3a\right)}+\dfrac{a}{\left(x+3a\right)\left(x+4a\right)}+\dfrac{a}{\left(x+4a\right)\left(x+5a\right)}\)
\(=\dfrac{5a}{x^2+5ax}\)
\(B=\frac{a}{x\left(x+a\right)}+\frac{a}{\left(x+a\right)\left(x+2a\right)}+\frac{a}{\left(x+2a\right)\left(x+3a\right)}+....+\frac{a}{\left(x+9a\right)\left(x+10a\right)}+\frac{1}{x+10a}\)
\(=\frac{1}{x}-\frac{1}{x+a}+\frac{1}{x+a}-\frac{1}{x+2a}+\frac{1}{x+2a}-\frac{1}{x+3a}+....+\frac{1}{x+9a}-\frac{1}{x+10a}+\frac{1}{x+10a}\)
\(=\frac{1}{x}\)
dài quá, làm từ từ nhé
1, \(\left(a-b\right)^2\left(2a-3b\right)-\left(b-a\right)^2\left(3a-5b\right)+\left(a+b\right)^2\left(a-2b\right)\)
\(=\left(a-b\right)^2\left(2a-3b-3a+5b\right)+\left(a+b\right)^2\left(a-2b\right)\)
\(=\left(a-b\right)^2\left(-a+2b\right)+\left(a+b\right)^2\left(a-2b\right)\)
\(=-\left(a-b\right)^2\left(a-2b\right)+\left(a+b\right)^2\left(a-2b\right)\)
\(=\left(a-2b\right)\left[\left(a+b\right)^2-\left(a-b\right)^2\right]\)
\(=\left(a-2b\right)\left(a+b-a+b\right)\left(a+b+a-b\right)\)
\(=4ab\left(a-2b\right)\)
2, \(x^4-4\left(x^2+5\right)-25=\left(x^2-25\right)-4\left(x^2+5\right)=\left(x^2-5\right)\left(x^2+5\right)-4\left(x^2+5\right)\)
\(=\left(x^2-9\right)\left(x^2+5\right)=\left(x-3\right)\left(x+3\right)\left(x^2+5\right)\)
3,\(\left(2-x\right)^2+\left(x-2\right)\left(x+3\right)-\left(4x^2-1\right)=\left(x-2\right)^2+\left(x-2\right)\left(x+3\right)-\left(4x^2-1\right)\)
\(=\left(x-2\right)\left(x-2+x+3\right)-\left(2x-1\right)\left(2x+1\right)\)
\(=\left(x-2\right)\left(2x+1\right)-\left(2x-1\right)\left(2x+1\right)\)
\(=\left(x-2-2x+1\right)\left(2x+1\right)\)
\(=\left(-x-1\right)\left(2x+1\right)\)
4, câu này đề thiếu
5,\(16\left(xy+6\right)^2-\left(4x^2+y^2-25\right)^2=\left(4xy+24\right)^2-\left(4x^2+y^2-25\right)^2\)
\(=\left(4xy+24-4x^2-y^2+25\right)\left(4xy+24+4x^2+y^2-25\right)\)
\(=\left[49-\left(4x^2-4xy+y^2\right)\right]\left[\left(4x^2+4xy+y^2\right)-1\right]\)
\(=\left[49-\left(2x-y\right)^2\right]\left[\left(2x+y\right)^2-1\right]\)
\(=\left(7-2x+y\right)\left(7+2x-y\right)\left(2x+y-1\right)\left(2x+y+1\right)\)