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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: \(=\dfrac{x^2+2xy+x^2-2xy-4xy}{\left(x+2y\right)\left(x-2y\right)}=\dfrac{2x^2-4xy}{\left(x+2y\right)\cdot\left(x-2y\right)}=\dfrac{2x}{x+2y}\)
\(\left(x-2\right)\left(x^2+2x+4\right)-\left(x^3+5\right)\)
\(=x^3-8-x^3-5\)
=-13
\(\left(x-2\right)\cdot\left(x^2+2x+4\right)-\left(x^3+5\right)\\ =x^2-8-x^3-5\\ =-13\)
\(\dfrac{5x+2}{x^2-4}+\dfrac{x-5}{x-2}=\dfrac{5x+2+x^2-3x-10}{\left(x-2\right)\left(x+2\right)}\\ =\dfrac{x^2+2x-8}{\left(x-2\right)\left(x+2\right)}=\dfrac{\left(x+4\right)\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{x+4}{x+2}\\ \left(x+4\right)^2-\left(x+3\right)\left(x-2\right)=-13\\ \Leftrightarrow x^2+8x+16-x^2+x+6=-13\\ \Leftrightarrow9x=-13-22=-35\\ \Leftrightarrow x=-\dfrac{35}{9}\)
a/ \(\left(2x+3\right)\left(x-5\right)-\left(x-1\right)^2+x\left(7-x\right)\)
\(=2x^2-2x-15-x^2+2x-1+7x-x^2\)
\(=7x-16\)
\(\frac{4}{x-3}+\frac{5}{x+3}-\frac{13-9x^2}{x^2-9}\)
ĐKXĐ : \(x\ne\pm3\)
\(=\frac{4}{x-3}+\frac{5}{x+3}-\frac{13-9x^2}{\left(x+3\right)\left(x-3\right)}\)
\(=\frac{4\left(x+3\right)}{\left(x+3\right)\left(x-3\right)}+\frac{5\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}-\frac{13-9x^2}{\left(x+3\right)\left(x-3\right)}\)
\(=\frac{4x+12}{\left(x+3\right)\left(x-3\right)}+\frac{5x-15}{\left(x+3\right)\left(x-3\right)}-\frac{13-9x^2}{\left(x+3\right)\left(x-3\right)}\)
\(=\frac{4x+12+5x-15-13+9x^2}{\left(x+3\right)\left(x-3\right)}\)
\(=\frac{9x^2+9x-16}{\left(x+3\right)\left(x-3\right)}=\frac{9x^2+9x-16}{x^2-9}\)
1.(2x+3).(x-5)+2x(3-x)+x-10
=2x^2 -10x+3x-15+6x-2x^2+x-10
=2x-25
2.(-x-2)3+(2x-4).(x2+2x+4)-x2.(x-6)
=-x^3+6x^2-12x-8+2x^3+4x^2+8x-4x^2+8x-16-x^3+6x^2
(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
`@` `\text {Ans}`
`\downarrow`
`(x+3)(x-4)`
`= x(x-4) + 3(x-4)`
`= x^2-4x + 3x - 12`
`= x^2 - x - 12`
=x^2-4x+3x-12
=x^2-x-12