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PQ
0
PQ
1
HT
24 tháng 7 2016
\(\left(+\right)A=\left(2x+3\right)\left(4x^2-6x+9\right)-2\left(4x^3-1\right).\)
\(A=8x^3-6x^2-18x+27-8x^3+2\)
\(A=6x^2-18x+29\)
\(\left(+\right)B=\left(x-1\right)^3-\left(x+1\right)^3+6\left(x-1\right)\left(x+1\right)\)
\(B=x^3-3x^2+3x+1-x^3-3x^2-3x-1+6\left(x^2-1\right)\)
\(B=-6x^2+6x^2-6\)
\(B=-6\)
NL
0
NM
3
MT
6 tháng 7 2015
C=(x+1)^3+(x-1)^3 -3x(x+1)(x-1)
=(x3+3x2+3x+1)+(x3-3x2+3x-1)-3x(x2-1)
=x3+3x2+3x+1+x3-3x2+3x-1-3x3+3x
=-x3+9x
DL
2
8 tháng 8 2016
\(\left(x-3\right)^3-\left(x+3\right)^3\)
\(=\left(x-3-x-3\right)\left(\left(x-3\right)^2+\left(x-3\right)\left(x+3\right)+\left(x+3\right)^2\right)\)
\(=-6\left(\left(x-3\right)^2+\left(x^2-9\right)+\cdot\left(x+3\right)^2\right)\)
NV
1
NM
Nguyễn Minh Quang
Giáo viên
18 tháng 8 2021
\(C=\left(\frac{x}{2}-y\right)^3-3.2.\left(\frac{x}{2}-y\right)^2+3.2^2\left(\frac{x}{2}-y\right)-2^3=\left(\frac{x}{2}-y-2\right)^3\)
với \(x=206,y=1\Rightarrow C=\left(\frac{x}{2}-y-2\right)^3=\left(\frac{206}{2}-1-2\right)^3=1000\text{ }000\)
23 tháng 10 2020
\(C=\left(1-\frac{x}{x+1}\right)\div\left(\frac{x+3}{x-2}+\frac{2+x}{3-x}+\frac{x+2}{x^2-5x+6}\right)\)
ĐKXĐ : x ≠ -1 ; x ≠ 2 ; x ≠ 3 ; x ≠ 11/5
\(=\left(\frac{x+1}{x+1}-\frac{x}{x+1}\right)\div\left(\frac{x+3}{x-2}-\frac{x+2}{x-3}+\frac{x+2}{\left(x-2\right)\left(x-3\right)}\right)\)
\(=\frac{1}{x+1}\div\left(\frac{\left(x+3\right)\left(x-3\right)}{\left(x-2\right)\left(x-3\right)}-\frac{\left(x+2\right)\left(x-2\right)}{\left(x-2\right)\left(x-3\right)}+\frac{x+2}{\left(x-2\right)\left(x-3\right)}\right)\)
\(=\frac{1}{x+1}\div\left(\frac{x^2-9}{\left(x-2\right)\left(x-3\right)}-\frac{x^2-4}{\left(x-2\right)\left(x-3\right)}+\frac{x+2}{\left(x-2\right)\left(x-3\right)}\right)\)
\(=\frac{1}{x+1}\div\left(\frac{x^2-9-x^2+4+x+2}{\left(x-2\right)\left(x-3\right)}\right)\)
\(=\frac{1}{x+1}\div\frac{x-3}{\left(x-2\right)\left(x-3\right)}\)
\(=\frac{1}{x+1}\times\frac{x-2}{1}\)
\(=\frac{x-2}{x+1}\)
20 tháng 6 2017
Đặt \(\hept{\begin{cases}\left(x+\frac{1}{x}\right)^3=a\\x^3+\frac{1}{x^3}=b\end{cases}}\)
Ta có
\(A=\frac{\left(x+\frac{1}{x}\right)^6-\left(x^6+2+\frac{1}{x^6}\right)}{\left(x+\frac{1}{x}\right)^3+x^3+\frac{1}{x^3}}=\frac{\left(x+\frac{1}{x}\right)^6-\left(x^3+\frac{1}{x^3}\right)^2}{\left(x+\frac{1}{x}\right)^3+x^3+\frac{1}{x^3}}\)
\(=\frac{a^2-b^2}{a+b}=a-b\)
\(=\left(x+\frac{1}{x}\right)^3-\left(x^3+\frac{1}{x^3}\right)\)
\(=x^3+3\left(x+\frac{1}{x}\right)+\frac{1}{x^3}-\left(x^3+\frac{1}{x^3}\right)=\frac{3x^2+3}{x}\)
\(C=\left(x-1\right)^3-\left(x+1\right)^3+6\left(x+1\right)\)
\(=x^3-3x^2+3x-1-x^3-3x^2-3x-1+6x+6\)
\(=-6x^2+6x+4\)
`C = (x -1)^3 - (x+1)^3 + 6(x + 1)`
`= x^3 - 3x^2 + 3x - 1 - (x^3 + 3x^2 + 3x + 1) + 6x + 6`
`= x^3 - 3x^2 + 3x - 1 - x^3 - 3x^2 - 3x - 1 + 6x + 6`
`= - 6x^2 + 6x + 4`