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Trả lời:
a, \(\left(2x-5\right)^3=\left(2x\right)^3-3.\left(2x\right)^2.5+3.2x.5^2-5^3=8x^3-60x^2+150x-125\)
b, \(\left(2x+3\right)\left(4x^2-6x+9\right)=\left(2x+3\right)\left[\left(2x\right)^2-2x.3+3^2\right]=\left(2x\right)^3+3^3=8x^3+9\)
c, \(\left(\frac{1}{2}x+1\right)^3=\left(\frac{1}{2}x\right)^3+3\left(\frac{1}{2}x\right)^21+3\cdot\frac{1}{2}x.1^2+1^3=\frac{1}{8}x^3+\frac{3}{4}x^2+\frac{3}{2}x+1\)
d, \(\left(x-\frac{2}{3}y\right)\left(x^2+\frac{2}{3}xy+\frac{4}{9}y^2\right)=x^3-\left(\frac{2}{3}y\right)^3=x^3-\frac{8}{27}y^3\)
b) \(\frac{4}{x+2}+\frac{3}{x-2}+\frac{5x+2}{4-x^2}\left(x\ne\pm2\right)\)
\(=\frac{4}{x+2}+\frac{3}{x-2}-\frac{5x-2}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{4\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\frac{5x-2}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{4x-8+3x+6-5x+2}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{2x}{\left(x-2\right)\left(x+2\right)}\)
f) \(x^2+1-\frac{x^4-3x^2+2}{x^2-1}\)
\(=x^2+1-\frac{\left(x^2-2\right)\left(x^2-1\right)}{\left(x+1\right)\left(x-1\right)}\)
\(=x^2+1-\frac{\left(x^2-2\right)\left(x+1\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}\)
\(=x^2+1-\left(x^2-2\right)\)
\(=x^2+1-x^2+2\)
\(=3\)
Bài 1:
a) \(\frac{4}{9}x^2-y^2=\left(\frac{2}{3}x-y\right)\left(\frac{2}{3}x+y\right)\)
b) \(x^2-5=\left(x-\sqrt{5}\right)\left(x+\sqrt{5}\right)\)
c) \(4x^2+6x+9=\left(2x+2\right)^2+5\)ko hiểu ???
d) \(\frac{1}{9}x^2-\frac{4}{3}xy+4=\left(\frac{1}{3}x\right)^2-2.\frac{1}{3}x.2+2^2=\left(\frac{1}{3}x-2\right)^2\)
Bài 2:
a) \(\left(\frac{1}{2}x-\frac{1}{3}y\right)\left(\frac{1}{2}x+\frac{1}{3}y\right)=\frac{1}{4}x^2-\frac{1}{9}y^2\)
b) \(\left(2x-\frac{1}{3}y\right)\left(4x^2+\frac{2}{3}xy+\frac{1}{9}x^2\right)=8x^3-\frac{1}{27}y^3\)
c) \(\left(3x-5y\right)\left(9x^2+15xy+\frac{1}{9}x^2\right)=27x^3-125y^3\)
a) x2(5x3 – x - \(\frac{1}{2}\)) = x2. 5x3 + x2 . (-x) + x2 . ( \(-\frac{1}{2}\) )
= 5x5 – x3 – \(\frac{1}{2}\)x2
b) (3xy – x2 + y) \(\frac{2}{3}\)x2y = \(\frac{2}{3}\)x2y . 3xy + \(\frac{2}{3}\)x2y . (- x2) + \(\frac{2}{3}\)x2y .
y = 2x3y2 – \(\frac{2}{3}\)x4y + \(\frac{2}{3}\)x2y2
c) (4x3– 5xy + 2x)( \(-\frac{1}{2}\)xy) = \(-\frac{1}{2}\)xy . 4x3 + ( \(-\frac{1}{2}\)xy) . (-5xy) + ( \(-\frac{1}{2}\)xy) . 2x
= -2x4y + \(\frac{5}{2}\)x2y2 – x2y.