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x mũ 2 + 3x / x mũ 2 + 6x + 9 + 3 / x - 3 + 6x / 9 - x mũ 2
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9 tháng 8 2021
1, \(x^3+4x^2+4x=0\Leftrightarrow x\left(x^2+4x+4\right)=0\)
\(\Leftrightarrow x\left(x+2\right)^2=0\Leftrightarrow x=-2;x=0\)
2, \(\left(x+3\right)^2-4=0\Leftrightarrow\left(x+3-2\right)\left(x+3+2\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+5\right)=0\Leftrightarrow x=-5;x=1\)
3, \(x^4-9x^2=0\Leftrightarrow x^2\left(x^2-9\right)=0\)
\(\Leftrightarrow x^2\left(x-3\right)\left(x+3\right)=0\Leftrightarrow x=0;\pm3\)
4, \(x^2-6x+9=81\Leftrightarrow\left(x-3\right)^2=9^2\)
\(\Leftrightarrow\left(x-3-9\right)\left(x-3+9\right)=0\Leftrightarrow\left(x-12\right)\left(x+6\right)=0\Leftrightarrow x=-6;x=12\)
5, em xem lại đề nhé
9 tháng 8 2021
à lag tý @@
5, \(x^3+6x^2+9x-4x=0\Leftrightarrow x^3+6x^2+5x=0\)
\(\Leftrightarrow x\left(x^2+6x+5\right)=0\Leftrightarrow x\left(x^2+x+5x+5\right)=0\)
\(\Leftrightarrow x\left(x+1\right)\left(x+5\right)=0\Leftrightarrow x=-5;x=-1;x=0\)
PL
2 tháng 8 2018
(x + 2)(x - 2) - (x - 2)(x + 5)
= (x - 2)(x + 2 - x - 5)
= (x - 2)-3
= -3x + 6
b) 2x(3x2y + 4x2y - 3)
= 2x(7x2y - 3)
= 14x3y - 6x
3 tháng 8 2018
a) (x+2)(x-2) - (x-2)(x+5 )
= (x-2) (x+2 - x-5)
= -3 (x-2)
c) \(\left(3x+1\right)^2\) - \(\left(1-2x\right)^2\)
= (3x+1 - 1 +2x) (3x+1 +1-2x)
= 5x (x +2)
d) \(x^2\) - 4 - \(\left(x+2\right)^2\)
= (\(x^2\) - 4 ) - ( x+2) (x+2)
= (x-2) (x+2) - (x+2) (x+2)
= (x+2) (x-2 - x-2)
= -4 (x+2)
8 tháng 8 2022
e: \(=x^2-16-2x^2-6x+x^2+6x+9=-7\)
b: \(=\left(6x+1-6x+1\right)^2=2^2=4\)
\(\frac{x^2+3x}{x^2+6x+9}+\frac{3}{x-3}+\frac{6x}{9-x^2}\)
ĐKXĐ : x ≠ ±3
\(=\frac{x\left(x+3\right)}{\left(x+3\right)^2}+\frac{3}{x-3}-\frac{6x}{x^2-9}\)
\(=\frac{x}{x+3}+\frac{3}{x-3}-\frac{6x}{\left(x-3\right)\left(x+3\right)}\)
\(=\frac{x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}+\frac{3\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\frac{6x}{\left(x-3\right)\left(x+3\right)}\)
\(=\frac{x^2-3x}{\left(x-3\right)\left(x+3\right)}+\frac{3x+9}{\left(x-3\right)\left(x+3\right)}-\frac{6x}{\left(x-3\right)\left(x+3\right)}\)
\(=\frac{x^2-3x+3x+9-6x}{\left(x-3\right)\left(x+3\right)}\)
\(=\frac{x^2-6x+9}{\left(x-3\right)\left(x+3\right)}\)
\(=\frac{\left(x-3\right)^2}{\left(x-3\right)\left(x+3\right)}=\frac{x-3}{x+3}\)