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\(a,\left(3x+x\right)\left(x^2-9\right)-\left(x-3\right)\left(x^2+3x+9\right)\)
\(=4x\left(x^2-9\right)-x^3+27\)
\(=4x^3-36x-x^3+27\)
\(=3x^3-36x+27\)
\(\left(x+6\right)^2-2x.\left(x+6\right)+\left(x-6\right).\left(x+6\right)\)
\(=\left(x+6\right).\left(x+6-2x+x-6\right)\)
\(=\left(x+6\right).0\)
\(=0\)
a: \(\dfrac{x^2}{3x+6}+\dfrac{4x+4}{3x+6}=\dfrac{x^2+4x+4}{3x+6}=\dfrac{x+2}{3}\)
b: \(\dfrac{x+3}{x}+\dfrac{x}{3-x}-\dfrac{9}{3x-x^2}\)
\(=\dfrac{x^2-9-x^2+9}{x\left(x-3\right)}\)
=0
Bài 1 :
a) (3+x)(x2-9)-(x-3)(x2+3x+9) = ( x-3)(x+3)2-(x-3)(x2+3x+9)
= (x-3) ( x2+6x+9 - (x2+3x+9)) = (x-3) . 3x = 3x(x-3)
Các câu còn lại mình sẽ gửi bạn sau nếu có thời gian
Nhấn đúng để ủng hộ mình :))
a: \(=3x^3-2x^2+5x\)
b: \(=x^3-2x^2+3x+6x^2-12x+18\)
\(=x^3+4x^2-9x+18\)
c: \(=2x^2-6xy+6xy-15y^2=2x^2-15y^2\)
d: \(=\left(x+3\right)\left(x^2-9\right)-x^3+27\)
\(=x^3-9x+3x^2-27-x^3+27=3x^2-9x\)
a,(3+x)(x2-9)-(x-3)(x2+3x+9)
=(3x2-27+x3-9x)-(x3-27)
=3x2-27+x3-9x-x3+27
=3x2-9x
=3x(x-3)
b,(x+6)2-2x(x+6)+(x-6)(x+6)
=x2+12x+36-2x2-12x+x2-36
=0
a) \(\left(3+x\right)\left(x^2-9\right)-\left(x-3\right)\left(x^2+3x+9\right)\)
\(=\left(3x^2+x^3-27-9x\right)-\left(x^3-27\right)\)
\(=3x^2+x^3-27-9x-x^3+27\)
\(=3x^2-9x\)
b) \(\left(x+6\right)^2-2x\left(x+6\right)+\left(x-6\right)\left(x+6\right)\)
\(=\left(x^2+12x+36\right)-\left(2x^2+12x\right)+\left(x^2-36\right)\)
\(=x^2+12x+36-2x^2-12x+x^2-36\)
\(=0\)
\(a,\left(x+3\right)\left(x-3\right)-\left(x-2\right)\left(x+5\right)=6\)
\(\Leftrightarrow x^2-9-x^2-5x+2x+10=6\)
\(\Leftrightarrow-3x+1=6\Leftrightarrow x=\frac{-5}{3}\)
Vậy x =\(\frac{-5}{3}\)
\(b,\left(3x+2\right)\left(2x+9\right)-\left(x+2\right)\left(6x+1\right)=\left(x+1\right)-\left(x-6\right)\)
\(\Leftrightarrow6x^2+27x+4x+18-6x^2-x-12x-2=x+1-x+6\)
\(\Leftrightarrow18x+16=7\Leftrightarrow x=\frac{-1}{2}\)
Vậy x =\(\frac{-1}{2}\)
a/ \(\left(x+3\right)\left(x-3\right)-\left(x-2\right)\left(x+5\right)=6\)
<=> \(x^2-9-\left(x^2+3x-10\right)=6\)
<=> \(x^2-9-x^2-3x+10=6\)
<=> \(-3x+1=6\)
<=> \(-3x=5\)
<=> \(x=-\frac{5}{3}\)
b/ \(\left(3x+2\right)\left(2x+9\right)-\left(x+2\right)\left(6x+1\right)=\left(x+1\right)-\left(x-6\right)\)
<=> \(6x^2+31x+18-\left(6x^2+13x+2\right)=x+1-x+6\)
<=> \(6x^2+31x+18-6x^2-13x-2=7\)
<=> \(18x+16=7\)
<=> \(18x=-9\)
<=> \(x=-\frac{1}{2}\)
Bài 1:
8: \(=\dfrac{x+3}{x\left(x-3\right)}\)
9: \(=\dfrac{x-2}{x-5}\cdot\dfrac{\left(x-5\right)\left(x+5\right)}{\left(x-2\right)^2}=\dfrac{x+5}{x-2}\)
10: \(=1:\dfrac{a-1}{a}=\dfrac{a}{a-1}\)
12: \(=\dfrac{6\left(x+1\right)}{3x\left(x+1\right)}=\dfrac{2}{x}\)
13: \(\dfrac{3}{x+3}-\dfrac{x-6}{x\left(x+3\right)}\)
\(=\dfrac{3x-x+6}{x\left(x+3\right)}=\dfrac{2x+6}{x\left(x+3\right)}=\dfrac{2}{x}\)
a) (x-3).((3+x)2-(x2+3x+9))
<=> (x-3).(3x+9)
<=> 3.(x-3)2
a) đa thức
\(=\left(x-3\right)\left(x+3\right)^2-\left(x-3\right)\left(x^2+3x+9\right)=\left(x-3\right)\left(x^2+6x+9-x^2-3x-9\right)=3x\left(x-3\right)=3x^2-9x\)
b) đa thức \(=\left(x+6\right)^2-2x\left(x+6\right)+x^2-36=\left(x+6-x\right)^2-36=0\)