Thực hiện phép tính (x²-6x+9)(-x²-6x-9) Giúp em vs ạ
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KV
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KV
1
PT
1
HN
0
12 tháng 3 2020
\(\frac{6}{x^2-9}+\frac{5x}{x-3}+\frac{x}{x+3}\)
\(=\frac{6x}{\left(x-3\right)\left(x+3\right)}+\frac{5x}{x-3}+\frac{x}{x+3}\)
\(=\frac{6x}{\left(x-3\right)\left(x+3\right)}-\frac{5x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}+\frac{x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\)
\(=\frac{6x+5x\left(x+3\right)+x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\)
\(=\frac{6x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\frac{6x}{x-3}\)
12 tháng 3 2020
\(\frac{6x}{x^2-9}+\frac{5x}{x-3}+\frac{x}{x+3}\left(x\ne\pm3\right)\)
\(=\frac{6x}{\left(x-3\right)\left(x+3\right)}+\frac{5x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}+\frac{x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\)
\(=\frac{6x+5x^2+15x+x^2-3x}{\left(x-3\right)\left(x+3\right)}\)
\(=\frac{6x^2+18x}{\left(x-3\right)\left(x+3\right)}=\frac{6x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\frac{6x}{x-3}\)
M
1
29 tháng 10 2021
a: \(\left(x+5\right)\left(x+1\right)-x^2\)
\(=x^2+6x+5-x^2\)
=6x+5
DH
2
AD
3 tháng 8 2021
\(\frac{\left(x-3\right)^3}{3x^2}:\frac{x^2-6x+9}{6x}\)
\(=\frac{\left(x-3\right)^3}{3x^2}.\frac{6x}{\left(x-3\right)^2}\)
\(=\frac{2\left(x-3\right)}{x}\)
\(=\frac{2x-6}{x}\)
#H
QA
3 tháng 8 2021
Trả lời:
\(\frac{\left(x-3\right)^3}{3x^2}:\frac{x^2-6x+9}{6x}\)
\(=\frac{\left(x-3\right)^3}{3x^2}.\frac{6x}{x^2-6x+9}\)
\(=\frac{\left(x-3\right)^3}{3x^2}.\frac{6x}{\left(x-3\right)^2}\)
\(=\frac{\left(x-3\right)^3.6x}{3x^2.\left(x-3\right)^2}\)
\(=\frac{2\left(x-3\right)}{x}\)
\(=\left[\left(-6x\right)+\left(x^2+9\right)\right]\left[\left(-6x\right)-\left(x^2+9\right)\right]\)
\(=\left(-6x\right)^2-\left(x^2+9\right)^2\)
\(=36x^2-\left(x^4+18x^2+81\right)\)
\(=-x^4+18x^2-81\)
\(=-\left(x^4-18x^2+81\right)\)
\(=-\left(x^2-9\right)^2\)
Ta có: \(\left(x^2-6x+9\right)\left(-x^2-6x-9\right)\)
\(=-\left(x^2-6x+9\right)\left(x^2+6x+9\right)\)
\(=-\left[\left(x-3\right)^2\cdot\left(x+3\right)^2\right]\)
\(=-\left(x^2-9\right)^2\)