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(x+5)2+(x-2)2-4(x-3)(x+3)
=x2+10x+25+x2-4x+4-4x2+36
=x2+x2-4x2+10x-4x+25+4+36
=-2x2+6x+65
\(\frac{4}{x-3}+\frac{5}{x+3}-\frac{13-9x^2}{x^2-9}\)
ĐKXĐ : \(x\ne\pm3\)
\(=\frac{4}{x-3}+\frac{5}{x+3}-\frac{13-9x^2}{\left(x+3\right)\left(x-3\right)}\)
\(=\frac{4\left(x+3\right)}{\left(x+3\right)\left(x-3\right)}+\frac{5\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}-\frac{13-9x^2}{\left(x+3\right)\left(x-3\right)}\)
\(=\frac{4x+12}{\left(x+3\right)\left(x-3\right)}+\frac{5x-15}{\left(x+3\right)\left(x-3\right)}-\frac{13-9x^2}{\left(x+3\right)\left(x-3\right)}\)
\(=\frac{4x+12+5x-15-13+9x^2}{\left(x+3\right)\left(x-3\right)}\)
\(=\frac{9x^2+9x-16}{\left(x+3\right)\left(x-3\right)}=\frac{9x^2+9x-16}{x^2-9}\)
Ta có: \(\left(2x+3\right)^2+\left(2x+5\right)^2-2\left(2x+3\right)\left(2x+5\right)\)
\(=\left(2x+3-2x-5\right)^2\)
\(=\left(-2\right)^2=4\)
\(3\left(x+1\right)^2+2\left(x-3\right)^3-5\left(x-5\right)\left(x+5\right)\)
\(=3\left(x^2+2x+1\right)+2\left(x^2-6x+9\right)-5\left(x^2-25\right)\)
\(=3x^2+6x+3+2x^2-12x+18-5x^2+125\)
\(=-6x+146\)
\(\left(x-2\right)\left(x^2+2x+4\right)-\left(x^3+5\right)\)
\(=x^3-8-x^3-5\)
=-13
\(\left(x-2\right)\cdot\left(x^2+2x+4\right)-\left(x^3+5\right)\\ =x^2-8-x^3-5\\ =-13\)
Bài 1:
b: \(=\dfrac{x+3-4-x}{x-2}=\dfrac{-1}{x-2}\)
Bài 2:
a: \(=\dfrac{x+1}{2\left(x+3\right)}+\dfrac{2x+3}{x\left(x+3\right)}\)
\(=\dfrac{x^2+x+4x+6}{2x\left(x+3\right)}=\dfrac{x^2+5x+6}{2x\left(x+3\right)}=\dfrac{x+2}{2x}\)
d: \(=\dfrac{3}{2x^2y}+\dfrac{5}{xy^2}+\dfrac{x}{y^3}\)
\(=\dfrac{3y^2+10xy+2x^3}{2x^2y^3}\)
e: \(=\dfrac{x^2+2xy+x^2-2xy-4xy}{\left(x+2y\right)\left(x-2y\right)}=\dfrac{2x^2-4xy}{\left(x+2y\right)\cdot\left(x-2y\right)}=\dfrac{2x}{x+2y}\)
Ta có: ( - 2 ) 5 : ( - 2 ) 3 = ( - 2 ) 5 - 3 = ( - 2 ) 2 = 4 .